If ${\bf{\vec a}},\;{\bf{\vec b}},\;{\bf{\vec c}}$ are three non-coplanar vectors, then $\left( {{\bf{\vec a}} + {\bf{\vec b}} + {\bf{\vec c}}} \right)\cdot\left[ {\left( {{\bf{\vec a}} + {\bf{\vec b}}} \right) \times \left( {{\bf{\vec a}} \times {\bf{\vec c}}} \right)} \right]$ is

$ - \left[ {{\bf{\vec a}}\;{\bf{\vec b}}\;{\bf{\vec c}}} \right]$

$2\left[ {{\bf{\vec a}}\;{\bf{\vec b}}\;{\bf{\vec c}}} \right]$

$\left[ {{\bf{\vec a}}\;{\bf{\vec b}}\;{\bf{\vec c}}} \right]$

The line of intersection of the planes ${\bf{\vec r}}\,\, \cdot \,\,\left( {\widehat {{\bf{\dot i}}} - 3\widehat {\,{\bf{\dot j}}} + {\bf{\hat k}}} \right) = 1$ and ${\bf{\vec r}}\,\, \cdot \,\,\left( {2\,\widehat {{\bf{\dot i}}} + 5\,\widehat {{\bf{\dot j}}} - 3\,{\bf{\hat k}}} \right) = 2$ is parallel to the vector

$4\widehat {\,{\bf{\dot i}}} + 5\widehat {\,{\bf{\dot j}}} + 11\,{\bf{\hat k}}$

$ - \,4\,\widehat {{\bf{\dot i}}} + 5\widehat {\,{\bf{\dot j}}} + 11\,{\bf{\hat k}}$

$ - \,4\,\widehat {{\bf{\dot i}}} - 5\widehat {\,{\bf{\dot j}}} + 11\,{\bf{\hat k}}$

$ - \,4\,\widehat {{\bf{\dot i}}} + 5\,\widehat {{\bf{\dot j}}} - 11\,{\bf{\hat k}}$

For any three vectors ${\bf{\vec a}},{\bf{\vec b}}$ and ${{\bf{\vec c}}}$, $\left( {{\bf{\vec a}} - {\bf{\vec b}}} \right)\cdot\left( {{\bf{\vec b}} + {\bf{\vec c}}\;} \right) \times \left( {\;{\bf{\vec c}} + {\bf{\vec a}}} \right)$ is equal to

$2{\bf{\vec a}}\cdot\left( {{\bf{\vec b}} \times {\bf{\vec c}}\;} \right)$

$\left[ {\;{\bf{\vec a}}\;{\bf{\vec b}}\;{\bf{\vec c}}\;} \right]$

${\left[ {\;{\bf{\vec a}}\;{\bf{\vec b}}\;{\bf{\vec c}}\;} \right]^2}$

SCALAR TRIPLE PRODUCT MCQs for JEE MATHEMATICS — Easy, Page 3

61 Easy

61 of 185

Application +4 | -1

{\text{If}}\,\,\vec a,\,\,\vec b,\vec c\,\,{\text{are}}\,\,{\text{three}}\,\,{\text{vectors}}\,{\text{,}}\,\,{\text{then}}\,\,[\vec a \times \vec b\,\,\,\vec b \times \vec c\,\,\,\vec c \times \vec a] =

View answer and explanation

Correct option: D

\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]^2

Option D (

\begin{{array}}{{lll}} \vec{{a}} & \vec{{b}} & \vec{{c}} \end{{array}}^2

) is correct because the scalar triple product identity allows the given expression to be simplified to the square of the scalar triple product of

\vec{a}

,

\vec{b}

, and

\vec{c}

.

Solution approach

Correct Answer + Core Reason: $\begin{{array}}{{lll}} \vec{{a}} & \vec{{b}} & \vec{{c}} \end{{array}}^2$ . The scalar triple product identity allows simplification to the square of the scalar triple product.

Step-by-Step Solution:

Step 1: Start with the given expression: $[\vec{a} \times \vec{b} \\\, \\, \vec{b} \times \vec{c} \\\, \\, \vec{c} \times \vec{a}]$ .

Step 2: Rewrite the scalar triple product as a dot product: $(\vec{a} \times \vec{b}) \cdot [(\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a})]$ .

Step 3: Apply the vector triple product identity to the term inside the brackets: $(\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a}) = [(\vec{b} \times \vec{c}) \cdot \vec{a}] \vec{c} - [(\vec{b} \times \vec{c}) \cdot \vec{c}] \vec{a}$ .

Step 4: Simplify using the properties of the scalar triple product. Since $(\vec{b} \times \vec{c}) \cdot \vec{c} = 0$ , we have: $(\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a}) = [(\vec{b} \times \vec{c}) \cdot \vec{a}] \vec{c} = [\vec{a} \\, \\, \vec{b} \\, \\, \vec{c}] \vec{c}$ .

Step 5: Substitute this back into the original expression: $(\vec{a} \times \vec{b}) \cdot [(\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a})] = (\vec{a} \times \vec{b}) \cdot ([\vec{a} \\, \\, \vec{b} \\, \\, \vec{c}] \vec{c}) = [\vec{a} \\, \\, \vec{b} \\, \\, \vec{c}] [(\vec{a} \times \vec{b}) \cdot \vec{c}]$ .

Step 6: Recognize that $(\vec{a} \times \vec{b}) \cdot \vec{c} = [\vec{a} \\, \\, \vec{b} \\, \\, \vec{c}]$ . Therefore, the expression becomes: $[\vec{a} \\, \\, \vec{b} \\, \\, \vec{c}] [\vec{a} \\, \\, \vec{b} \\, \\, \vec{c}] = [\vec{a} \\, \\, \vec{b} \\, \\, \vec{c}]^2$ .

Step 7: Final Answer: $\begin{{array}}{{lll}} \vec{{a}} & \vec{{b}} & \vec{{c}} \end{{array}}^2$

Why other options are wrong:
- Option A: $\begin{{array}}{{lll}} \vec{{a}} & \vec{{b}} & \vec{{c}} \end{{array}}$ is incorrect because the expression simplifies to the square of the scalar triple product, not just the scalar triple product itself.
- Option B: $2\begin{{array}}{{lll}} \vec{{a}} & \vec{{b}} & \vec{{c}} \end{{array}}$ is incorrect because the expression simplifies to the square of the scalar triple product, not twice the scalar triple product.
- Option C: $3\begin{{array}}{{lll}} \vec{{a}} & \vec{{b}} & \vec{{c}} \end{{array}}$ is incorrect because the expression simplifies to the square of the scalar triple product, not three times the scalar triple product.

Key Concepts:
- Scalar triple product: $[\vec{a} \\, \\, \vec{b} \\, \\, \vec{c}] = (\vec{a} \times \vec{b}) \cdot \vec{c}$
- Vector triple product identity: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$
- Properties of the scalar triple product.

Memory Aids:
- Remember the scalar triple product and vector triple product identities.

Quick Tips:
- Apply the vector triple product identity carefully.
- Use the properties of the scalar triple product to simplify the expression.

62 Easy

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Application +4 | -1

If

\vec{a}=\hat{i}+\hat{j}+\hat{k}, \vec{b}=\hat{i}+\hat{j}, \vec{c}=\hat{i}

and

(\vec{a} \times \vec{b}) \times \vec{c}=\lambda \vec{a}+\mu \vec{b}

, then

1+m

is equal to

View answer and explanation

Correct option: A

0

Option A (0) is correct because after calculating the vector triple product and comparing coefficients, the sum of lambda and mu is found to be 0.

Solution approach

Correct Answer + Core Reason: 0. The sum of the coefficients λ and μ is 0 after calculating the vector triple product.

Step-by-Step Solution:

Step 1: Given vectors: $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ , $\vec{b} = \hat{i} + \hat{j}$ , $\vec{c} = \hat{i}$

Step 2: Calculate $\vec{a} \cdot \vec{c}$ and $\vec{b} \cdot \vec{c}$ :
$\vec{a} \cdot \vec{c} = (\hat{i} + \hat{j} + \hat{k}) \cdot (\hat{i}) = 1$
$\vec{b} \cdot \vec{c} = (\hat{i} + \hat{j}) \cdot (\hat{i}) = 1$

Step 3: Use the vector triple product formula: $(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a}$

Step 4: Substitute the calculated dot products:
$(\vec{a} \times \vec{b}) \times \vec{c} = (1) \vec{b} - (1) \vec{a} = \vec{b} - \vec{a}$

Step 5: Given $(\vec{a} \times \vec{b}) \times \vec{c} = \lambda \vec{a} + \mu \vec{b}$ , so $\vec{b} - \vec{a} = \lambda \vec{a} + \mu \vec{b}$

Step 6: Comparing coefficients, we get $\lambda = -1$ and $\mu = 1$ .

Step 7: Calculate $\lambda + \mu = -1 + 1 = 0$ .

Step 8: Final Answer: 0

Why other options are wrong:
- Option B: Incorrect. The sum is 0, not 2.
- Option C: Incorrect. The sum is 0, not 1.
- Option D: Incorrect. The sum is 0, not 4.

Key Concepts:
- Vector triple product.
- Dot product of vectors.
- Comparing coefficients.

Memory Aids:
- Remember the vector triple product formula: $(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a}$

Quick Tips:
- Remember the vector triple product formula.
- Be careful with the order of vectors in the triple product.

63 Easy

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Analyzing +4 | -1

Let a, b and c be distinct numbers. If the vectors

a\hat i + a\hat j + c\hat k,{\mkern 1mu} {\mkern 1mu} {\mkern 1mu} \hat i + \hat k

and

c\hat i + c\hat j + b\hat k\,

are coplanar, then

c

is equal to

View answer and explanation

Correct option: D

\sqrt {ab}

Option D (

\sqrt {ab}

) is correct because coplanarity implies the scalar triple product is zero, leading to the relationship

c = \sqrt{ab}

.

Solution approach

Correct Answer + Core Reason: $\sqrt {ab}$ . Coplanarity of vectors implies the determinant formed by their components is zero, leading to $c = \sqrt{ab}$ .

Step-by-Step Solution:

Step 1: Given vectors $a\hat{i} + a\hat{j} + c\hat{k}$ , $\hat{i} + \hat{k}$ , and $c\hat{i} + c\hat{j} + b\hat{k}$ are coplanar.

Step 2: For coplanar vectors, the scalar triple product (determinant) is zero: $\begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} = 0$ .

Step 3: Expand the determinant: $a(0 - c) - a(b - c) + c(c - 0) = 0$ .

Step 4: Simplify the equation: $-ac - ab + ac + c^2 = 0$ .

Step 5: Further simplification: $c^2 - ab = 0$ , which gives $c^2 = ab$ .

Step 6: Solve for $c$ : $c = \sqrt{ab}$ (since a, b, and c are distinct numbers).

Why other options are wrong:
- Option A: $\frac{2}{{\frac{1}{a} + \frac{1}{b}}}$ : This represents the harmonic mean of a and b, which is not the condition for coplanarity in this case.
- Option B: $\frac{{a + b}}{2}$ : This represents the arithmetic mean of a and b, which is not the condition for coplanarity in this case.
- Option C: $\frac{1}{a} + \frac{1}{b}$ : This is incorrect as it does not result from the determinant equation for coplanarity.

Key Concepts:
- Coplanar vectors
- Scalar triple product
- Determinants

Memory Aids:
- Coplanar vectors have a scalar triple product of zero.

Quick Tips:
- Remember the determinant condition for coplanarity of three vectors.
- Simplify the determinant equation carefully to avoid errors.

64 Easy

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Analyzing +4 | -1

If

\vec{a}, \vec{b}, \vec{c}

are non-null non-coplanar vectors, then

[\vec{a}-2 \,\,\vec{b}+\vec{c} \,\,\vec{b}-2 \vec{c}+\vec{a}\,\, \vec{c}-2 \vec{a}+\vec{b}]=

View answer and explanation

Correct option: C

0

Option C (0) is correct because the vectors are coplanar, making the scalar triple product equal to zero.

Solution approach

Correct Answer + Core Reason: 0. The vectors are coplanar, so the scalar triple product is zero.

Step-by-Step Solution:

Step 1: Let $\vec{\alpha} = \vec{a} - 2\vec{b} + \vec{c}$ , $\vec{\beta} = \vec{b} - 2\vec{c} + \vec{a}$ , and $\vec{\gamma} = \vec{c} - 2\vec{a} + \vec{b}$ .

Step 2: We want to find $[\vec{\alpha} \vec{\beta} \vec{\gamma}] = \left|\begin{array}{rrr} 1 & -2 & 1 \\ 1 & 1 & -2 \\ -2 & 1 & 1 \end{array}\right| [\vec{a} \vec{b} \vec{c}]$ .

Step 3: Evaluating the determinant, we get $1(1(1) - (-2)(1)) - (-2)(1(1) - (-2)(-2)) + 1(1(1) - 1(-2)) = 1(1+2) + 2(1-4) + 1(1+2) = 3 - 6 + 3 = 0$ .

Step 4: Therefore, $[\vec{\alpha} \vec{\beta} \vec{\gamma}] = 0 [\vec{a} \vec{b} \vec{c}] = 0$ .

Why other options are wrong:
- Option A: Incorrect. The scalar triple product is zero.
- Option B: Incorrect. The scalar triple product is zero.
- Option D: Incorrect. The scalar triple product is zero.

Key Concepts:
- Scalar triple product
- Coplanar vectors

Memory Aids:
- If three vectors are coplanar, their scalar triple product is zero.

Quick Tips:
- Calculate the determinant to find the scalar triple product.

65 Easy

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Application +4 | -1

The volume of the parallelopiped whose conterminous edges are

\mathbf{i}-\mathbf{j}+\mathbf{k}, 2 \mathbf{i}-4 \mathbf{j}+5 \mathbf{k}

and

3 \mathbf{i}-5 \mathbf{j}+2 \mathbf{k}

is

View answer and explanation

Correct option: D

8

Option D (8) is correct because the volume of the parallelepiped is calculated using the scalar triple product of the given vectors, which equals 8.

Solution approach

Correct Answer + Core Reason: 8. The volume of the parallelepiped is the absolute value of the scalar triple product of the given vectors.

Step-by-Step Solution:

Step 1: The volume $V$ of the parallelepiped formed by vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ is given by the absolute value of their scalar triple product: $V = |[\mathbf{a} \mathbf{b} \mathbf{c}]| = |\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|$ .

Step 2: Express the vectors in component form: $\mathbf{a} = \mathbf{i} - \mathbf{j} + \mathbf{k} = (1, -1, 1)$ , $\mathbf{b} = 2\mathbf{i} - 4\mathbf{j} + 5\mathbf{k} = (2, -4, 5)$ , and $\mathbf{c} = 3\mathbf{i} - 5\mathbf{j} + 2\mathbf{k} = (3, -5, 2)$ .

Step 3: Calculate the scalar triple product as a determinant:
$V = \left| \begin{array}{ccc} 1 & -1 & 1 \\ 2 & -4 & 5 \\ 3 & -5 & 2 \end{array} \right|$

Step 4: Evaluate the determinant:\nV = 1((-4)(2) - (5)(-5)) - (-1)((2)(2) - (5)(3)) + 1((2)(-5) - (-4)(3))\nV = 1(-8 + 25) + 1(4 - 15) + 1(-10 + 12)\nV = 1(17) + 1(-11) + 1(2)\nV = 17 - 11 + 2\nV = 8

Step 5: The volume is the absolute value of the result, which is $|8| = 8$ .

Why other options are wrong:
- Option A (4): This is incorrect. It might be a calculation error in evaluating the determinant.
- Option B (2): This is incorrect. It might be a calculation error in evaluating the determinant.
- Option C (3): This is incorrect. It might be a calculation error in evaluating the determinant.

Key Concepts:
- Scalar triple product
- Volume of a parallelepiped
- Determinant of a 3x3 matrix

Memory Aids:
- The scalar triple product can be remembered as the determinant of a matrix formed by the three vectors.

Quick Tips:
- Always double-check the signs when evaluating determinants.

66 Easy

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Analyzing +4 | -1

If

\mathbf{\vec a},\;\mathbf{\vec b},\;\mathbf{\vec c}

are three non-coplanar vectors, then

\left( {\mathbf{\vec a} + \mathbf{\vec b} + \mathbf{\vec c}} \right)\cdot\left[ {\left( {\mathbf{\vec a} + \mathbf{\vec b}} \right) \times \left( {\mathbf{\vec a} \times \mathbf{\vec c}} \right)} \right]

is

View answer and explanation

Correct option: C

- \left[ {\mathbf{\vec a}\;\mathbf{\vec b}\;\mathbf{\vec c}} \right]

Option C (

\left[ {\mathbf{\vec a}\;\mathbf{\vec b}\;\mathbf{\vec c}} \right]

) is correct because after expanding the scalar triple product, the result simplifies to the negative of the scalar triple product of a, b, and c.

Solution approach

Correct Answer + Core Reason: $- \left[ {\mathbf{\vec a}\;\mathbf{\vec b}\;\mathbf{\vec c}} \right]$ . Expanding the scalar triple product and using properties of determinants, we arrive at this answer.

Step-by-Step Solution:

Step 1: Expand the cross product.
$\left( {\mathbf{\vec a} + \mathbf{\vec b} + \mathbf{\vec c}} \right)\cdot\left[ {\left( {\mathbf{\vec a} + \mathbf{\vec b}} \right) \times \left( {\mathbf{\vec a} \times \mathbf{\vec c}} \right)} \right] = (\vec a + \vec b + \vec c) \cdot [(\vec a \times \vec a) + (\vec b \times \vec a) + (\vec a \times \vec c) + (\vec b \times \vec c)]$

Step 2: Simplify using properties of cross products.\nSince $\vec a \times \vec a = 0$ , we have $(\vec a + \vec b + \vec c) \cdot [(\vec b \times \vec a) + (\vec a \times \vec c) + (\vec b \times \vec c)]$ .

Step 3: Expand the dot product.\nThis becomes $\vec a \cdot (\vec b \times \vec a) + \vec a \cdot (\vec a \times \vec c) + \vec a \cdot (\vec b \times \vec c) + \vec b \cdot (\vec b \times \vec a) + \vec b \cdot (\vec a \times \vec c) + \vec b \cdot (\vec b \times \vec c) + \vec c \cdot (\vec b \times \vec a) + \vec c \cdot (\vec a \times \vec c) + \vec c \cdot (\vec b \times \vec c)$ .

Step 4: Simplify using scalar triple product properties.\nRecall that $\vec a \cdot (\vec b \times \vec a) = 0$ , $\vec a \cdot (\vec a \times \vec c) = 0$ , $\vec b \cdot (\vec b \times \vec a) = 0$ , $\vec b \cdot (\vec b \times \vec c) = 0$ , and $\vec c \cdot (\vec a \times \vec c) = 0$ . Also, $\vec a \cdot (\vec b \times \vec c) = [\vec a \vec b \vec c]$ , $\vec b \cdot (\vec a \times \vec c) = [\vec b \vec a \vec c] = -[\vec a \vec b \vec c]$ , and $\vec c \cdot (\vec b \times \vec a) = [\vec c \vec b \vec a] = -[\vec a \vec b \vec c]$ .\nSo, the expression becomes $0 + 0 + [\vec a \vec b \vec c] + 0 + [- \vec a \vec b \vec c] + 0 + [- \vec a \vec b \vec c] + 0 + 0 = [\vec a \vec b \vec c] - [\vec a \vec b \vec c] - [\vec a \vec b \vec c] = -[\vec a \vec b \vec c]$ .

Why other options are wrong:
- Option A: 0 is incorrect because the expression simplifies to $- \left[ {\mathbf{\vec a}\;\mathbf{\vec b}\;\mathbf{\vec c}} \right]$ .
- Option B: $2\left[ {\mathbf{\vec a}\;\mathbf{\vec b}\;\mathbf{\vec c}} \right]$ is incorrect because the expression simplifies to $- \left[ {\mathbf{\vec a}\;\mathbf{\vec b}\;\mathbf{\vec c}} \right]$ .
- Option D: $\left[ {\mathbf{\vec a}\;\mathbf{\vec b}\;\mathbf{\vec c}} \right]$ is incorrect because the expression simplifies to $- \left[ {\mathbf{\vec a}\;\mathbf{\vec b}\;\mathbf{\vec c}} \right]$ .

Key Concepts:
- Scalar triple product
- Cross product
- Dot product
- Properties of determinants

Memory Aids:
- Remember that $\vec a \cdot (\vec b \times \vec c) = [\vec a \vec b \vec c]$ and $[\vec a \vec b \vec c] = -[\vec b \vec a \vec c] = -[\vec c \vec b \vec a]$

67 Easy

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Application +4 | -1

If

\mathbf{\vec a},\mathbf{\vec b},\mathbf{\vec c}

are the three vectors mutually perpendicular to each other to form a right handed system and

\left| {\mathbf{\vec a}} \right| = 1,\mathrm{\;}\left| {\mathbf{\vec b}} \right| = 3\mathrm{\;and\;}\left| {\mathbf{\vec c}} \right| = 5,\mathrm{\;then\;}\left[ {\mathrm{\;}\mathbf{\vec a} - 2\mathbf{\vec b}\;\mathbf{\vec b} - 3\mathbf{\vec c}\;\mathbf{\vec c} - 4\mathbf{\vec a}\mathrm{\;}} \right]

is equal to

View answer and explanation

Correct option: D

 - 215

Option D (- 215) is correct because the scalar triple product is evaluated using the properties of mutually perpendicular vectors.

Solution approach

Correct Answer + Core Reason: - 215. The scalar triple product is evaluated using the properties of mutually perpendicular vectors.

Step-by-Step Solution:

Step 1: Given $\left| \mathbf{\vec a} \right| = 1,\mathrm{\;}\left| \mathbf{\vec b} \right| = 3\mathrm{\;and\;}\left| \mathbf{\vec c} \right| = 5$

Step 2: Evaluate the scalar triple product: $\left[ {\mathrm{\;}\mathbf{\vec a} - 2\mathbf{\vec b}\;\mathbf{\vec b} - 3\mathbf{\vec c}\;\mathbf{\vec c} - 4\mathbf{\vec a}\mathrm{\;}} \right]$

68 Easy

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Application +4 | -1

The line of intersection of the planes

\mathbf{\vec r}\,\, \cdot \,\,\left( {\widehat {\mathbf{\dot i}} - 3\widehat {\,\mathbf{\dot j}} + \mathbf{\hat k}} \right) = 1

and

\mathbf{\vec r}\,\, \cdot \,\,\left( {2\,\widehat {\mathbf{\dot i}} + 5\,\widehat {\mathbf{\dot j}} - 3\,\mathbf{\hat k}} \right) = 2

is parallel to the vector

View answer and explanation

Correct option: B

4\widehat {\,\mathbf{\dot i}} + 5\widehat {\,\mathbf{\dot j}} + 11\,\mathbf{\hat k}

Option B (

4\widehat {\\,\mathbf{\dot i}} + 5\widehat {\\,\mathbf{\dot j}} + 11\mathbf{\hat k}

) is correct because it is the cross product of the normal vectors of the two planes, which gives the direction vector of the line of intersection.

Solution approach

Correct Answer + Core Reason: $4\widehat {\\,\mathbf{\dot i}} + 5\widehat {\\,\mathbf{\dot j}} + 11\mathbf{\hat k}$ . The line of intersection is parallel to the cross product of the normal vectors of the planes.

Step-by-Step Solution:

Step 1: The normal vectors to the planes are $\mathbf{\vec n_1} = \widehat {\\,\mathbf{\dot i}} - 3\widehat {\\,\mathbf{\dot j}} + \mathbf{\hat k}$ and $\mathbf{\vec n_2} = 2\widehat {\\,\mathbf{\dot i}} + 5\widehat {\\,\mathbf{\dot j}} - 3\mathbf{\hat k}$ .

Step 2: The direction vector of the line of intersection is given by $\mathbf{\vec n_1} \times \mathbf{\vec n_2}$ .

Step 3: Calculate the cross product: $\mathbf{\vec n_1} \times \mathbf{\vec n_2} = \begin{vmatrix} {\widehat {\\,\mathbf{\dot i}}} & {\widehat {\\,\mathbf{\dot j}}} & \mathbf{\hat k} \\ 1 & -3 & 1 \\ 2 & 5 & -3 \end{vmatrix} = (9 - 5)\widehat {\\,\mathbf{\dot i}} - (-3 - 2)\widehat {\\,\mathbf{\dot j}} + (5 + 6)\mathbf{\hat k} = 4\widehat {\\,\mathbf{\dot i}} + 5\widehat {\\,\mathbf{\dot j}} + 11\mathbf{\hat k}$ .

Why other options are wrong:
- Option A: This is incorrect as the cross product is calculated incorrectly.
- Option C: This is incorrect as the cross product is calculated incorrectly.
- Option D: This is incorrect as the cross product is calculated incorrectly.

Key Concepts:
- Planes in 3D space
- Normal vectors
- Cross product
- Line of intersection of planes

Memory Aids:
- Remember that the direction vector of the line of intersection is the cross product of the normal vectors.

Quick Tips:
- Find the normal vectors of the planes and then calculate their cross product.

69 Easy

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Application +4 | -1

If the volume of parallelopiped formed by the vectors

\hat i + \lambda \hat j + \hat k,\hat j + \lambda \hat k

and

\lambda \hat i + \hat k

is minimum, then

\lambda

is equal to

View answer and explanation

Correct option: B

\frac{1}{{\sqrt 3 }}

Option B (

\frac{{1}}{{\sqrt 3 }}

) is correct because it is the value of

\lambda

that minimizes the volume of the parallelepiped formed by the given vectors.

Solution approach

Correct Answer + Core Reason: $\frac{{1}}{{\sqrt 3 }}$ . The value of $\lambda$ that minimizes the volume is found by taking the derivative of the volume with respect to $\lambda$ and setting it to zero.

Step-by-Step Solution:

Step 1: Calculate the volume of the parallelepiped formed by the vectors $\hat i + \lambda \hat j + \hat k$ , $\hat j + \lambda \hat k$ , and $\lambda \hat i + \hat k$ using the scalar triple product: $V = |[\vec{a} \\, \vec{b} \\, \vec{c}]| = \left| {\begin{{array}}{{*{20}{c}}} 1&\lambda &1\\\ 0&1&\lambda \\\ \lambda &0&1 \end{{array}} } \right| = |1(1 - 0) - \lambda(0 - \lambda^2) + 1(0 - \lambda)| = |1 + \lambda^3 - \lambda|$ .

Step 2: To minimize the volume, find the critical points by taking the derivative of $V$ with respect to $\lambda$ and setting it to zero: $\frac{{dV}}{{d\lambda}} = 3\lambda^2 - 1 = 0 \Rightarrow \lambda^2 = \frac{1}{3} \Rightarrow \lambda = \pm \frac{1}{\sqrt{3}}$ .

Step 3: Take the second derivative to determine if the critical points are minima or maxima: $\frac{{d^2V}}{{d\lambda^2}} = 6\lambda$ .

Step 4: Evaluate the second derivative at the critical points:
- For $\lambda = \frac{1}{\sqrt{3}}$ , $\frac{{d^2V}}{{d\lambda^2}} = 6(\frac{1}{\sqrt{3}}) > 0$ , so it's a minimum.
- For $\lambda = -\frac{1}{\sqrt{3}}$ , $\frac{{d^2V}}{{d\lambda^2}} = 6(-\frac{1}{\sqrt{3}}) < 0$ , so it's a maximum.

Step 5: Since we want to minimize the volume, choose $\lambda = \frac{1}{\sqrt{3}}$ .

Why other options are wrong:
- **Option A (- $\frac{{1}}{{\sqrt 3 }}$ ): This value corresponds to a maximum volume, not a minimum.
- Option C ( $\sqrt 3$ ): This value does not satisfy the condition for minimizing the volume (derivative equals zero).
- Option D (- $\sqrt 3$ ): This value does not satisfy the condition for minimizing the volume (derivative equals zero).

Key Concepts:
- Scalar triple product (volume of a parallelepiped)
- Differentiation (finding minima and maxima)
- Second derivative test

Memory Aids:**
- Remember the scalar triple product formula: $V = |[\vec{a} \\, \vec{b} \\, \vec{c}]|$ .
- The second derivative test helps determine if a critical point is a minimum or maximum.

Quick Tips:
- Always check the second derivative to confirm whether you have a minimum or maximum.
- Be careful with signs when taking derivatives.

70 Easy

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Analyzing +4 | -1

If

\vec{a}, \vec{b}, \vec{c}

are non-coplanar vectors and

\lambda

is a real number, then the vectors

\vec{a}+2 \vec{b}+3 \vec{c}, \lambda \vec{b}+4 \vec{c}

and

(2 \lambda-1) \vec{c}

are non-coplanar for

View answer and explanation

Correct option: C

all except two values of

\lambda

Option C (all except two values of

\lambda

) is correct because the vectors are non-coplanar when the scalar triple product is non-zero, which occurs for all values of

\lambda

except 0 and 1/2.

Solution approach

Correct Answer + Core Reason: all except two values of $\lambda$ . The scalar triple product is zero for two values of $\lambda$ .

Step-by-Step Solution:

Step 1: Given $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar vectors and $\vec{\alpha}=\vec{a}+2 \vec{b}+3 \vec{c}, \vec{\beta}=\lambda \vec{b}+4 \vec{c}$ and $\vec{\gamma}=(2 \lambda-1) \vec{c}$ .

Step 2: Calculate the scalar triple product $[\begin{array}{lll}\vec{\alpha} & \vec{\beta} & \vec{\gamma}\end{array}] = \left| {\begin{array}{cccccccccccccccccccc} 1&2&3 \ 0&\lambda &4 \ 0&0&{(2\lambda - 1)} \end{array}} \right|\left[ {\begin{array}{lll} {\vec a}&{\vec b}&{\vec c} \end{array}} \right]$ .

Step 3: Evaluate the determinant: $\left| {\begin{array}{cccccccccccccccccccc} 1&2&3 \ 0&\lambda &4 \ 0&0&{(2\lambda - 1)} \end{array}} \right| = 1(\lambda(2\lambda - 1) - 0) - 2(0 - 0) + 3(0 - 0) = \lambda(2\lambda - 1)$ .

Step 4: So, $[\begin{array}{lll}\vec{\alpha} & \vec{\beta} & \vec{\gamma}\end{array}] = \lambda (2\lambda - 1)[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}]$ .

Step 5: The vectors $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are non-coplanar if $[\begin{array}{lll}\vec{\alpha} & \vec{\beta} & \vec{\gamma}\end{array}] \neq 0$ . Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar, $[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}] \neq 0$ . Therefore, $\lambda (2\lambda - 1) \neq 0$ , which means $\lambda \neq 0$ and $\lambda \neq \frac{1}{2}$ .

Step 6: Hence, $\vec{\alpha}, \vec{\beta}, \vec{\gamma}$ are non-coplanar for all values of $\lambda$ except 0 and $\frac{1}{2}$ .

Why other options are wrong:
- Option A: no value of $\lambda$ : This is incorrect as the vectors are non-coplanar for infinitely many values of $\lambda$ .
- Option B: all except one value of $\lambda$ : This is incorrect as there are two values of $\lambda$ for which the vectors are coplanar.
- Option D: all values of $\lambda$ : This is incorrect as the vectors are coplanar for $\lambda = 0$ and $\lambda = \frac{1}{2}$ .

Key Concepts:
- Scalar triple product and coplanarity
- Determinant of a matrix

Memory Aids:
- Vectors are coplanar if their scalar triple product is zero.

Quick Tips:
- Remember how to calculate the determinant of a 3x3 matrix.
- Use the properties of scalar triple product to simplify the calculation.

71 Easy

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Analyzing +4 | -1

If

\mathbf{u}, \mathbf{v}

and

\mathbf{w}

are non-coplanar vectors and

p, q

are real numbers, then the equality

[3 \mathbf{\,u\,} p\, \mathbf{v\,} p \mathbf{\,w}]-[p\, \mathbf{v\,} \mathbf{w\,} q\, u]

- [2w qv qu]

=0

holds for

View answer and explanation

Correct option: D

exactly one value of

(p, q)

Option D (exactly one value of

(p, q)

) is correct because the equation simplifies to

3p^2 - pq + 2q^2 = 0

, which has only one solution:

p = q = 0

.

Solution approach

Correct Answer + Core Reason: exactly one value of $(p, q)$ . The equation simplifies to a quadratic form with a single solution.

Step-by-Step Solution:

Step 1: Given the equation $[3 \mathbf{\,u\} p\, \mathbf{\,v\} p \mathbf{\,w}]-[p\, \mathbf{\,v\} \mathbf{\,w\} q\, u]$ - [2w qv qu] $=0$ .

Step 2: Expand the scalar triple products using the property $[a \, b \, c] = a \cdot (b \times c)$ .
$3p^2[u \cdot (v \times w)] - pq[v \cdot (w \times u)] - 2q^2[w \cdot (v \times u)] = 0$

Step 3: Use the property $u \cdot (v \times w) = v \cdot (w \times u) = -w \cdot (v \times u)$ .
$3p^2[u \cdot (v \times w)] - pq[u \cdot (v \times w)] + 2q^2[u \cdot (v \times w)] = 0$

Step 4: Factor out $[u \cdot (v \times w)]$ .
$(3p^2 - pq + 2q^2)[u \cdot (v \times w)] = 0$

Step 5: Since $\mathbf{u}, \mathbf{v}, \mathbf{w}$ are non-coplanar, $[u \cdot (v \times w)] \neq 0$ .\nTherefore, $3p^2 - pq + 2q^2 = 0$ .

Step 6: Divide by $q^2$ (assuming $q \neq 0$ ) and let $x = p/q$ .
$3x^2 - x + 2 = 0$

Step 7: Check the discriminant: $D = (-1)^2 - 4(3)(2) = 1 - 24 = -23 < 0$ . This means there are no real solutions for $x$ other than $p=0$ and $q=0$ .\nIf $q=0$ , then $3p^2 = 0$ , so $p=0$ .

Step 8: The only solution is $p = q = 0$ .

Final Answer: The equality holds for exactly one value of $(p, q)$ , which is $(0, 0)$ .

Why other options are wrong:
- Option A: Incorrect because there is only one solution, not two.
- Option B: Incorrect because there is only one solution, not more than two.
- Option C: Incorrect because the equation does not hold for all values of $(p, q)$ .

Key Concepts:
- Scalar triple product
- Non-coplanar vectors
- Discriminant of a quadratic equation

Memory Aids:
- Remember the properties of the scalar triple product: $[a \, b \, c] = a \cdot (b \times c)$
- Remember the condition for non-coplanar vectors: $[u \, v \, w] \neq 0$

Quick Tips:
- Simplify the equation using properties of scalar triple products.
- Check the discriminant to determine the number of real solutions.

72 Easy

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Application +4 | -1

Let the vectors

\vec{a}, \vec{b}, \vec{c}

represent three coterminous edges of a parallelopiped of volume

\text{V}

. Then the volume of the parallelopiped, whose coterminous edges are represented by

\vec{a}, \vec{b}+\vec{c}

and

\vec{a}+2 \vec{b}+3 \vec{c}

is equal to

View answer and explanation

Correct option: C

\text{V}

Option C (V) is correct because the volume of the parallelepiped is given by the scalar triple product, which simplifies to V after expansion.

Solution approach

Correct Answer + Core Reason: V. The volume of the parallelepiped is given by the scalar triple product, which simplifies to V.

Step-by-Step Solution:

Step 1: The volume of the parallelepiped formed by vectors $\vec{a}, \vec{b}+\vec{c}, \vec{a}+2\vec{b}+3\vec{c}$ is given by the scalar triple product: $V' = |[\vec{a} \,\, \vec{b}+\vec{c} \,\, \vec{a}+2\vec{b}+3\vec{c}]|$ .

Step 2: Expand the scalar triple product:
$V' = |\vec{a} \cdot ((\vec{b}+\vec{c}) \times (\vec{a}+2\vec{b}+3\vec{c}))|$
$V' = |\vec{a} \cdot (\vec{b} \times \vec{a} + 2\vec{b} \times \vec{b} + 3\vec{b} \times \vec{c} + \vec{c} \times \vec{a} + 2\vec{c} \times \vec{b} + 3\vec{c} \times \vec{c})|$

Step 3: Simplify using the properties of the cross product ( $\vec{b} \times \vec{b} = 0$ and $\vec{c} \times \vec{c} = 0$ ) and rearrange terms:
$V' = |\vec{a} \cdot (\vec{b} \times \vec{a} + 3\vec{b} \times \vec{c} + \vec{c} \times \vec{a} + 2\vec{c} \times \vec{b})|$
$V' = |\vec{a} \cdot ( - \vec{a} \times \vec{b} + 3\vec{b} \times \vec{c} + \vec{c} \times \vec{a} - 2\vec{b} \times \vec{c})|$
$V' = |\vec{a} \cdot ( - \vec{a} \times \vec{b} + \vec{b} \times \vec{c} + \vec{c} \times \vec{a})|$

Step 4: Rewrite in terms of scalar triple products:
$V' = | -[\vec{a} \,\, \vec{a} \,\, \vec{b}] + [\vec{a} \,\, \vec{b} \,\, \vec{c}] + [\vec{a} \,\, \vec{c} \,\, \vec{a}] |$

Step 5: Since the scalar triple product is zero if any two vectors are the same, we have $[\vec{a} \,\, \vec{a} \,\, \vec{b}] = 0$ and $[\vec{a} \,\, \vec{c} \,\, \vec{a}] = 0$ .
$V' = | [\vec{a} \,\, \vec{b} \,\, \vec{c}] |$

Step 6: Given that the volume of the original parallelepiped is $V = |[\vec{a} \,\, \vec{b} \,\, \vec{c}]|$ , we have $V' = |V| = V$ .

Why other options are wrong:
- Option A (2V): This is incorrect because the scalar triple product simplifies to V, not 2V. The coefficients in the linear combination do not directly translate to a factor of 2 in the volume.
- Option B (6V): This is incorrect because the scalar triple product simplifies to V, not 6V. The coefficients in the linear combination do not directly translate to a factor of 6 in the volume.
- Option D (3V): This is incorrect because the scalar triple product simplifies to V, not 3V. The coefficients in the linear combination do not directly translate to a factor of 3 in the volume.

Key Concepts:
- Scalar triple product
- Volume of a parallelepiped
- Properties of cross product
- Properties of dot product

Memory Aids:
- Scalar triple product represents the volume of a parallelepiped.
- The determinant form of the scalar triple product can be used for calculation.

Quick Tips:
- Remember that if any two vectors in a scalar triple product are the same, the result is zero.
- Use determinant properties to simplify the calculation of the scalar triple product.

73 Easy

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Application +4 | -1

If

\vec a,\vec b,\vec c

are non-coplanar vectors then

\frac{{\vec a\cdot\left( {\vec b \times \vec c} \right)}}{{\left( {\vec c \times \vec a} \right)\cdot\vec b}} + \frac{{\vec b\cdot\left( {\vec a \times \vec c} \right)}}{{\vec c\cdot\left( {\vec a \times \vec b} \right)}}

is equal to

View answer and explanation

Correct option: A

0

Option A (0) is correct because the expression simplifies to 1 - 1 = 0 using properties of scalar triple product.

Solution approach

Correct Answer + Core Reason: 0. The expression simplifies to zero using properties of the scalar triple product.

Step-by-Step Solution:

Step 1: Rewrite the expression using the scalar triple product notation:
$\frac{{\vec a\cdot\left( {\vec b \times \vec c} \right)}}{{\left( {\vec c \times \vec a} \right)\cdot\vec b}} + \frac{{\vec b\cdot\left( {\vec a \times \vec c} \right)}}{{\vec c\cdot\left( {\vec a \times \vec b} \right)}} = \frac{{\left[ {\vec a\;\vec b\;\vec c} \right]}}{{\left[ {\vec c\;\vec a\;\vec b} \right]}} + \frac{{\left[ {\vec b\;\vec a\;\vec c} \right]}}{{\left[ {\vec c\;\vec a\;\vec b} \right]}}$

Step 2: Use the property that $\left[ {\vec c\;\vec a\;\vec b} \right] = \left[ {\vec a\;\vec b\;\vec c} \right]$ and $\left[ {\vec b\;\vec a\;\vec c} \right] = -\left[ {\vec a\;\vec b\;\vec c} \right]$ :
$= \frac{{\left[ {\vec a\;\vec b\;\vec c} \right]}}{{\left[ {\vec a\;\vec b\;\vec c} \right]}} + \frac{{ - \left[ {\vec a\;\vec b\;\vec c} \right]}}{{\left[ {\vec a\;\vec b\;\vec c} \right]}}$

Step 3: Simplify the expression:
$= 1 - 1 = 0$

Why other options are wrong:
- Option B: This is incorrect because it assumes the second term is positive, leading to 1 + 1 = 2.
- Option C: This is incorrect because it assumes the entire expression simplifies to 1.
- Option D: This is incorrect because the expression has a defined value of 0.

Key Concepts:
- Scalar triple product properties
- Cyclic permutation of vectors in scalar triple product

Memory Aids:
- Remember that swapping two vectors in a scalar triple product changes the sign.
- $\left[ {\vec a\;\vec b\;\vec c} \right] = \vec a \cdot (\vec b \times \vec c)$

Quick Tips:
- Always remember the properties of the scalar triple product to simplify expressions.

74 Easy

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Application +4 | -1

If

\mathbf{\vec a}\cdot\widehat \mathbf{i} = 4

then

\left( {\mathbf{\vec a} \times \widehat \mathbf{j}} \right)\cdot\left( {2\widehat \mathbf{j} - 3\mathbf{\hat k}} \right)

is equal to

View answer and explanation

Correct option: D

- 12

Option D (

- 12

) is correct because it is the result of the scalar triple product calculation using the given dot product.

Solution approach

Correct Answer + Core Reason: $- 12$ . The result is obtained by using properties of scalar triple product and the given dot product.

Step-by-Step Solution:

Step 1: Given $\mathbf{\vec a}\cdot\widehat \mathbf{i} = 4$ .

Step 2: Rewrite the expression using the scalar triple product property: $\left( {\mathbf{\vec a} \times \widehat \mathbf{j}} \right)\cdot\left( {2\widehat \mathbf{j} - 3\mathbf{\hat k}} \right) = \mathbf{\vec a}\cdot\left\{ {\widehat \mathbf{j} \times \left( {2\widehat \mathbf{j} - 3\mathbf{\hat k}} \right)} \right\}$ .

Step 3: Simplify the expression: ${\mathbf{\vec a}}\cdot\left\{ {\widehat \mathbf{j} \times \left( {2\widehat \mathbf{j} - 3\mathbf{\hat k}} \right)} \right\} = \mathbf{\vec a}\cdot\left\{ {2\left( {\widehat \mathbf{j} \times \widehat \mathbf{j}} \right) - 3\left( {\widehat \mathbf{j} \times \mathbf{\hat k}} \right)} \right\} = \mathbf{\vec a}\cdot\left\{ { - 3\left( {\widehat \mathbf{j} \times \mathbf{\hat k}} \right)} \right\}$ .

Step 4: Use the fact that $\widehat \mathbf{j} \times \mathbf{\hat k} = \widehat \mathbf{i}$ : ${\mathbf{\vec a}}\cdot\left\{ { - 3\widehat \mathbf{i}} \right\} = - 3\left( {\mathbf{\vec a}\cdot\widehat \mathbf{i}} \right)$ .

Step 5: Substitute the given value $\mathbf{\vec a}\cdot\widehat \mathbf{i} = 4$ : $- 3\left( {\mathbf{\vec a}\cdot\widehat \mathbf{i}} \right) = - 3(4) = - 12$ .

Why other options are wrong:
- Option A (12): This is incorrect because it has the wrong sign. The correct answer is -12.
- Option B (2): This is incorrect because it does not follow from the correct application of the scalar triple product and the given dot product.
- Option C (0): This is incorrect because it does not follow from the correct application of the scalar triple product and the given dot product.

Key Concepts:
- Scalar triple product
- Dot product
- Cross product

Memory Aids:
- Remember the cyclic property of the scalar triple product: $\vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{b} \cdot (\vec{c} \times \vec{a}) = \vec{c} \cdot (\vec{a} \times \vec{b})$ .
- Remember the cross product of unit vectors: $\hat{i} \times \hat{j} = \hat{k}$ , $\hat{j} \times \hat{k} = \hat{i}$ , $\hat{k} \times \hat{i} = \hat{j}$ .

Quick Tips:
- When dealing with scalar triple products, use the properties to simplify the expression before calculating.

75 Easy

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Application +4 | -1

Shortest distance between lines:

\frac{x-1}{2}=\frac{2 y-2}{3}=\frac{z-3}{1}

and

\frac{x-2}{3}=\frac{y-1}{2}=\frac{z+2}{4}

is

View answer and explanation

Correct option: A

\frac{13}{\sqrt{165}}

Option A (

1+\frac{e^2}{4}

) is correct because it satisfies the given differential equation and initial condition when solving for y(e).

Solution approach

Correct Answer + Core Reason: $1+\frac{{e^2}}{{4}}$ . The solution to the differential equation is found using an integrating factor, and the constant of integration is determined using the initial condition.

Step-by-Step Solution:

Step 1: The differential equation is $x \ln x \frac{{dy}}{{dx}} + y = x^2 \ln x$ .

Step 2: Divide by $x \ln x$ : $\frac{{dy}}{{dx}} + \frac{{y}}{{x \ln x}} = x$ .

Step 3: The integrating factor (IF) is $e^{\int \frac{{1}}{{x \ln x}} dx} = e^{\ln |\ln x|} = |\ln x|$ .

Step 4: The solution is $y(\text{{IF}}) = \int x \cdot |\ln x| dx$ .

Step 5: Assuming $x > 1$ , we have $y \ln x = \int x \ln x dx$ .

Step 6: Integrate by parts: $y \ln x = \ln x \cdot \frac{{x^2}}{{2}} - \int \frac{{1}}{{x}} \cdot \frac{{x^2}}{{2}} dx = \frac{{x^2 \ln x}}{{2}} - \frac{{x^2}}{{4}} + c$ .

Step 7: Given $y(2) = 2$ , substitute $x = 2$ and $y = 2$ : $2 \ln 2 = \frac{{4 \ln 2}}{{2}} - \frac{{4}}{{4}} + c \Rightarrow 2 \ln 2 = 2 \ln 2 - 1 + c \Rightarrow c = 1$ .

Step 8: The solution is $y \ln x = \frac{{x^2 \ln x}}{{2}} - \frac{{x^2}}{{4}} + 1$ .

Step 9: We want to find $y(e)$ . Substitute $x = e$ : $y \ln e = \frac{{e^2 \ln e}}{{2}} - \frac{{e^2}}{{4}} + 1 \Rightarrow y = \frac{{e^2}}{{2}} - \frac{{e^2}}{{4}} + 1 = \frac{{e^2}}{{4}} + 1$ .

Why other options are wrong:
- **Option B ( $1-\frac{e^2}{4}$ ): This is incorrect because it does not satisfy the differential equation and initial condition.
- Option C ( $\frac{e^2}{2}$ ): This is incorrect because it does not satisfy the differential equation and initial condition.
- Option D ( $1+\frac{e^2}{2}$ ): This is incorrect because it does not satisfy the differential equation and initial condition.

Key Concepts:
- Differential equations
- Integrating factor
- Integration by parts

Memory Aids:
- Remember the formula for the integrating factor.

Quick Tips:**
- Solve the differential equation step by step and use the initial condition to find the constant of integration.

76 Easy

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Application +4 | -1

The shortest distance between the lines

\frac{{x - 3}}{3} = \frac{{y - 8}}{{ - 1}} = \frac{{z - 3}}{1}

and

\frac{{x + 3}}{{ - 3}} = \frac{{y + 7}}{2} = \frac{{z - 6}}{4}

is

View answer and explanation

Correct option: A

3\sqrt {30}

Option A (

3\sqrt {30}

) is correct because it is the result of applying the shortest distance formula between two skew lines.

Solution approach

Correct Answer + Core Reason: $3\sqrt {30}$ . The shortest distance is calculated using the determinant formula for skew lines.

Step-by-Step Solution:

Step 1: Identify the points and direction vectors of the two lines.\nLine 1: Point $A(3, 8, 3)$ , Direction vector $\vec{{b_1}} = (3, -1, 1)$ .\nLine 2: Point $B(-3, -7, 6)$ , Direction vector $\vec{{b_2}} = (-3, 2, 4)$ .

Step 2: Find the vector connecting the two points: $\vec{{AB}} = B - A = (-3 - 3, -7 - 8, 6 - 3) = (-6, -15, 3)$ .

Step 3: Calculate the scalar triple product: $\vec{{AB}} \cdot (\vec{{b_1}} \times \vec{{b_2}})$ . This is the determinant of the matrix formed by the components of the vectors.
$\left| {\begin{{array}}{{{*{{20}}{c}}}} -6&-15&3\\3&-1&1\-3&2&4 \end{{array}}} \right| = -6(-4 - 2) - (-15)(12 + 3) + 3(6 - 3) = 36 + 225 + 9 = 270$ .

Step 4: Calculate the magnitude of the cross product of the direction vectors: $|\vec{{b_1}} \times \vec{{b_2}}|$ .
$\vec{{b_1}} \times \vec{{b_2}} = \left| {\begin{{array}}{{{*{{20}}{c}}}} i&j&k\\3&-1&1\-3&2&4 \end{{array}}} \right| = i(-4 - 2) - j(12 + 3) + k(6 - 3) = -6i - 15j + 3k = (-6, -15, 3)$ .
$|\vec{{b_1}} \times \vec{{b_2}}| = \sqrt{{(-6)^2 + (-15)^2 + 3^2}} = \sqrt{{36 + 225 + 9}} = \sqrt{{270}}$ .

Step 5: Calculate the shortest distance using the formula: $d = \frac{{|\vec{{AB}} \cdot (\vec{{b_1}} \times \vec{{b_2}})|}}{{|\vec{{b_1}} \times \vec{{b_2}}|}} = \frac{{270}}{{\sqrt{{270}}}} = \sqrt{{270}} = 3\sqrt{{30}}$ .

Why other options are wrong:
- Option B: This value is incorrect due to an error in the determinant or cross product calculation.
- Option C: This value is incorrect due to an error in the determinant or cross product calculation.
- Option D: This value is incorrect due to an error in the determinant or cross product calculation.

Key Concepts:
- Shortest distance between skew lines
- Scalar triple product
- Vector cross product
- Determinants

Memory Aids:
- Remember the formula for the shortest distance between skew lines.

Quick Tips:
- Double-check the signs when calculating the determinant and cross product.
- Simplify the expression after each step to avoid errors.

77 Easy

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Application +4 | -1

For any three vectors,

\vec{a}, \vec{b}\,\, \vec{c}

, the value of

[\vec{a}+\vec{b} \quad \vec{b}+\vec{c} \,\,\vec{c}+\vec{a}]

, is

View answer and explanation

Correct option: B

2\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]

Option B (

2\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]

) is correct because the scalar triple product can be expanded and simplified using determinant properties.

Solution approach

Correct Answer + Core Reason: $2\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]$ . The scalar triple product expands to twice the original volume.

Step-by-Step Solution:

Step 1: Define the new vectors: $\vec{\alpha}=\vec{a}+\vec{b}, \vec{\beta}=\vec{b}+\vec{c}$ , and $\vec{\gamma}=\vec{c}+\vec{a}$ .

Step 2: Express the new vectors in terms of $\vec{a}, \vec{b}, \vec{c}$ : $\vec{\alpha}=\vec{a}+\vec{b}+0 \vec{c}, \vec{\beta}=0 \vec{a}+\vec{b}+\vec{c}$ and $\vec{\gamma}=\vec{a}+0 \vec{b}+\vec{c}$ .

Step 3: Calculate the scalar triple product: $\left[\begin{array}{lll}\vec{\alpha} & \vec{\beta} & \vec{\gamma}\end{array}\right] = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr} 1&1&0 \\ 0&1&1 \\ 1&0&1 \end{array}} \right|\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]$

Step 4: Evaluate the determinant: $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr} 1&1&0 \\ 0&1&1 \\ 1&0&1 \end{array}} \right| = 1(1-0) - 1(0-1) + 0 = 1 + 1 = 2$

Step 5: Final Answer: $\left[\begin{array}{lll}\vec{\alpha} & \vec{\beta} & \vec{\gamma}\end{array}\right]=2\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]$

Why other options are wrong:
- Option A: This is incorrect because the scalar triple product does not vanish.
- Option C: This is incorrect because the scalar triple product is twice the original volume.
- Option D: This is incorrect because the scalar triple product is not negative.

Key Concepts:
- Scalar triple product
- Determinants
- Vector addition

Memory Aids:
- Remember the determinant expansion for scalar triple products.

Quick Tips:
- Express all vectors in terms of a common basis before calculating the scalar triple product.

78 Easy

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Analyzing +4 | -1

Let

G_1, G_2, G_3

be the centroids of the triangular faces

O B C, O C A, O A B

of a tetrahedron

O A B C

. If

V_1

denote the volume of the tetrahedron

\mathrm{OABC}

and

V_2

that of the parallelopiped with

O G_1, O G_2, O G_3

as three concurrent edges, then

View answer and explanation

Correct option: A

4 V_1=9 V_2

Option A (

4 V_1=9 V_2

) is correct because it relates the volume of the tetrahedron to the volume of the parallelepiped formed by the centroids of its faces.

Solution approach

Correct Answer + Core Reason: $4 V_1=9 V_2$ . The volumes are related by considering the position vectors of the centroids and applying scalar triple product properties.

Step-by-Step Solution:

Step 1: Define the position vectors of A, B, and C as $\vec{a}, \vec{b},$ and $\vec{c}$ respectively, with O as the origin.

Step 2: Determine the position vectors of the centroids $G_1, G_2,$ and $G_3$ : $\overrightarrow{OG_1} = \frac{{\vec{b} + \vec{c}}}{{3}}, \overrightarrow{OG_2} = \frac{{\vec{c} + \vec{a}}}{{3}}, \overrightarrow{OG_3} = \frac{{\vec{a} + \vec{b}}}{{3}}$ .

Step 3: Calculate the volume of the tetrahedron $V_1 = \frac{1}{6} [\vec{a} \vec{b} \vec{c}]$ .

Step 4: Calculate the volume of the parallelepiped $V_2 = [\overrightarrow{OG_1} \overrightarrow{OG_2} \overrightarrow{OG_3}] = \left[ {\frac{{\vec{b} + \vec{c}}}{3} \frac{{\vec{c} + \vec{a}}}{3} \frac{{\vec{a} + \vec{b}}}{3}} \right]$ .

Step 5: Simplify $V_2 = \frac{1}{{27}} [(\vec{b} + \vec{c}) \,(\vec{c} + \vec{a}) \,(\vec{a} + \vec{b})] = \frac{2}{{27}} [\vec{a} \vec{b} \vec{c}]$ .

Step 6: Relate $V_1$ and $V_2$ : $V_2 = \frac{2}{{27}} \times 6V_1 = \frac{4}{{9}}V_1$ , which implies $4V_1 = 9V_2$ .

Why other options are wrong:
- Option B: Incorrect algebraic manipulation of the volume relationship.
- Option C: Incorrect algebraic manipulation of the volume relationship.
- Option D: Incorrect algebraic manipulation of the volume relationship.

Key Concepts:
- Centroid of a triangle
- Volume of a tetrahedron
- Volume of a parallelepiped
- Scalar triple product

Memory Aids:
- Volume of tetrahedron = $\frac{1}{6} [\vec{a} \vec{b} \vec{c}]$
- Volume of parallelepiped = $[\vec{a} \vec{b} \vec{c}]$

Quick Tips:
- Remember the formulas for the volume of a tetrahedron and a parallelepiped in terms of vectors.
- Use the properties of the scalar triple product to simplify calculations.

79 Easy

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Analyzing +4 | -1

For three vectors

\vec{u}, \vec{v}, \vec{w}

which of the following expressions is not equal to any of the remaining three?

View answer and explanation

Correct option: C

\vec{v} \cdot(\vec{u} \times \vec{w})

Option C (

\vec{v} \cdot(\vec{u} \times \vec{w})

) is correct because it is the only expression that is not equal to the other three due to the properties of the scalar triple product.

Solution approach

Correct Answer + Core Reason: $\vec{v} \cdot(\vec{u} \times \vec{w})$ . This expression is different because it results in the negative of the other three due to the cyclic property of the scalar triple product.

Step-by-Step Solution:

Step 1: Recall the properties of the scalar triple product:
$\vec{u} \cdot (\vec{v} \times \vec{w}) = (\vec{v} \times \vec{w}) \cdot \vec{u}$
$(\vec{u} \times \vec{v}) \cdot \vec{w} = \vec{u} \cdot (\vec{v} \times \vec{w})$

Step 2: Analyze the given options:\nOption A: $\vec{u} \cdot (\vec{v} \times \vec{w})$ \nOption B: $(\vec{v} \times \vec{w}) \cdot \vec{u} = \vec{u} \cdot (\vec{v} \times \vec{w})$ \nOption C: $\vec{v} \cdot (\vec{u} \times \vec{w}) = - \vec{u} \cdot (\vec{v} \times \vec{w})$ \nOption D: $(\vec{u} \times \vec{v}) \cdot \vec{w} = \vec{u} \cdot (\vec{v} \times \vec{w})$

Step 3: Compare the options. Options A, B, and D are equal due to the properties of the scalar triple product. Option C is the negative of the other three.

Why other options are wrong:
- Option A: This is incorrect because it is equal to options B and D.
- Option B: This is incorrect because it is equal to options A and D.
- Option D: This is incorrect because it is equal to options A and B.

Key Concepts:
- Scalar triple product
- Cyclic property of scalar triple product

Memory Aids:
- Remember that the scalar triple product is invariant under cyclic permutations: $\vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{b} \cdot (\vec{c} \times \vec{a}) = \vec{c} \cdot (\vec{a} \times \vec{b})$
- Interchanging any two vectors in the scalar triple product changes its sign.

Quick Tips:
- Use the properties of the scalar triple product to simplify expressions.
- Remember that the positions of the dot and cross products can be interchanged in the scalar triple product.

80 Easy

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Application +4 | -1

The value of scalar triple product

\mathbf{i}-2 \mathbf{j}+3 \mathbf{k}, 2 \mathbf{i}+\mathbf{j}-\mathbf{k}

and

\mathbf{j}+\mathbf{k}

is

View answer and explanation

Correct option: A

12

Option A (12) is correct because the scalar triple product is calculated correctly using the determinant of the matrix formed by the components of the vectors.

Solution approach

Correct Answer + Core Reason: 12. The scalar triple product is calculated using the determinant.

Step-by-Step Solution:

Step 1: Set up the determinant with the components of the vectors.\nLet $\mathbf{a} = \mathbf{i} - 2\mathbf{j} + 3\mathbf{k}$ , $\mathbf{b} = 2\mathbf{i} + \mathbf{j} - \mathbf{k}$ , and $\mathbf{c} = \mathbf{j} + \mathbf{k}$ .\nScalar triple product = $\begin{vmatrix} 1 & -2 & 3 \\ 2 & 1 & -1 \\ 0 & 1 & 1 \end{vmatrix}$

Step 2: Calculate the determinant.
= 1 * (1 * 1 - (-1) * 1) - (-2) * (2 * 1 - (-1) * 0) + 3 * (2 * 1 - 1 * 0)
= 1 * (1 + 1) + 2 * (2 - 0) + 3 * (2 - 0)
= 1 * 2 + 2 * 2 + 3 * 2
= 2 + 4 + 6
= 12

Why other options are wrong:
- Option B (10): This is incorrect because there was an error in calculating the determinant.
- Option C (14): This is incorrect because there was an error in calculating the determinant.
- Option D (16): This is incorrect because there was an error in calculating the determinant.

Key Concepts:
- Scalar triple product
- Determinant of a 3x3 matrix

Memory Aids:
- Remember the formula for the scalar triple product: $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$

Quick Tips:
- Double-check your determinant calculations to avoid errors.
- The scalar triple product can be used to find the volume of a parallelepiped.

81 Easy

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Understanding +4 | -1

\frac{{\mathbf{\vec a}\cdot\left( {\;\mathbf{\vec b} \times \;\mathbf{\vec c}} \right)}}{{\mathbf{\vec b}\cdot\left( {\;\mathbf{\vec c} \times \mathbf{\vec a}} \right)}} + \frac{{\mathbf{\vec b}\cdot\left( {\;\mathbf{\vec a} \times \mathbf{\vec b}} \right)}}{{\mathbf{\vec a}\cdot\left( {\mathbf{\vec b} \times \;\mathbf{\vec c}} \right)}}\;\mathrm{is\;equal\;to}

View answer and explanation

Correct option: A

1

Option A (1) is correct because the expression simplifies to 1 using properties of scalar triple products.

Solution approach

Correct Answer + Core Reason: 1. The expression simplifies to 1 using properties of scalar triple products.

Step-by-Step Solution:

Step 1: Recognize the scalar triple product notation.

Step 2: Recall that $\mathbf{\vec a} \cdot (\;\mathbf{\vec b} \times \;\mathbf{\vec c}) = \mathbf{\vec b} \cdot (\;\mathbf{\vec c} \times \mathbf{\vec a}) = \mathbf{\vec c} \cdot (\;\mathbf{\vec a} \times \mathbf{\vec b}) = [ \mathbf{\vec a}\,\mathbf{\vec b}\,\mathbf{\vec c} ]$ .

Step 3: Also, $\mathbf{\vec b} \cdot (\;\mathbf{\vec a} \times \mathbf{\vec b}) = 0$ since $\mathbf{\vec a} \times \mathbf{\vec b}$ is perpendicular to both $\mathbf{\vec a}$ and $\mathbf{\vec b}$ .

Step 4: Substitute these properties into the given expression:
$\frac{{\mathbf{\vec a}\cdot\left( {\;\mathbf{\vec b} \times \;\mathbf{\vec c}} \right)}}{{\mathbf{\vec b}\cdot\left( {\;\mathbf{\vec c} \times \mathbf{\vec a}} \right)}} + \frac{{\mathbf{\vec b}\cdot\left( {\;\mathbf{\vec a} \times \mathbf{\vec b}} \right)}}{{\mathbf{\vec a}\cdot\left( {\mathbf{\vec b} \times \;\mathbf{\vec c}} \right)}} = \frac{{\left[ {\mathbf{\vec a}\,\mathbf{\vec b}\,\mathbf{\vec c}} \right]}}{{\left[ {\mathbf{\vec b}\,\mathbf{\vec c}\,\mathbf{\vec a}} \right]}} + \frac{{\left[ {\mathbf{\vec b}\,\mathbf{\vec a}\,\mathbf{\vec b}} \right]}}{{\left[ {\mathbf{\vec a}\,\mathbf{\vec b}\,\mathbf{\vec c}} \right]}} = \frac{{\left[ {\mathbf{\vec a}\,\mathbf{\vec b}\,\mathbf{\vec c}} \right]}}{{\left[ {\mathbf{\vec a}\,\mathbf{\vec b}\,\mathbf{\vec c}} \right]}} + \frac{0}{{\left[ {\mathbf{\vec a}\,\mathbf{\vec b}\,\mathbf{\vec c}} \right]}} = 1 + 0 = 1$

Why other options are wrong:
- Option B: Incorrect. The expression simplifies to 1, not 2.
- Option C: Incorrect. The expression simplifies to 1, not 0.
- Option D: Incorrect. The expression simplifies to 1, not infinity.

Key Concepts:
- Scalar triple product
- Properties of vector products

Memory Aids:
- Cyclic permutation of vectors in scalar triple product does not change the value.
- If two vectors are the same in a scalar triple product, the result is zero.

Quick Tips:
- Remember the cyclic property of the scalar triple product: [a b c] = [b c a] = [c a b].

82 Easy

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Application +4 | -1

If

\mathbf{a}, \mathbf{b}

and

\mathbf{c}

are non-coplanar vectors and

\mathbf{d}=\lambda \mathbf{a}+\mu \mathbf{b}+v \mathbf{c}

, then

\lambda

is equal to

View answer and explanation

Correct option: B

\frac{\text { [bcd] }}{\text { [bca] }}

Option B (

\frac{\text { [bcd] }}{\text { [bca] }}

) is correct because by taking the dot product of

\mathbf{d}

with

\mathbf{b} \times \mathbf{c}

, we can isolate

\lambda

and express it in terms of scalar triple products.

Solution approach

Correct Answer + Core Reason: $\frac{\text { [bcd] }}{\text { [bca] }}$ . By taking the dot product of $\mathbf{d}$ with $\mathbf{b} \times \mathbf{c}$ , we can isolate $\lambda$ .

Step-by-Step Solution:

Step 1: Start with the given equation: $\mathbf{d} = \lambda \mathbf{a} + \mu \mathbf{b} + \nu \mathbf{c}$ .

Step 2: Take the dot product of both sides with $\mathbf{b} \times \mathbf{c}$ : $\mathbf{d} \cdot (\mathbf{b} \times \mathbf{c}) = (\lambda \mathbf{a} + \mu \mathbf{b} + \nu \mathbf{c}) \cdot (\mathbf{b} \times \mathbf{c})$ .

Step 3: Distribute the dot product: $\mathbf{d} \cdot (\mathbf{b} \times \mathbf{c}) = \lambda \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) + \mu \mathbf{b} \cdot (\mathbf{b} \times \mathbf{c}) + \nu \mathbf{c} \cdot (\mathbf{b} \times \mathbf{c})$ .

Step 4: Recognize that $\mathbf{b} \cdot (\mathbf{b} \times \mathbf{c}) = 0$ and $\mathbf{c} \cdot (\mathbf{b} \times \mathbf{c}) = 0$ because the scalar triple product of linearly dependent vectors is zero.\nSo, $\mathbf{d} \cdot (\mathbf{b} \times \mathbf{c}) = \lambda \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ .

Step 5: Express the dot products as scalar triple products: $[\,\mathbf{d} \, \mathbf{b} \, \mathbf{c} = \lambda [\,\mathbf{a} \, \mathbf{b} \, \mathbf{c}$ .

Step 6: Solve for $\lambda$ : $\lambda = \frac{[\,\mathbf{d} \, \mathbf{b} \, \mathbf{c}}{[\,\mathbf{a} \, \mathbf{b} \, \mathbf{c}}$ .\nSince $[\,\mathbf{d} \, \mathbf{b} \, \mathbf{c} = [\,\mathbf{b} \, \mathbf{c} \, \mathbf{d}$ and $[\,\mathbf{a} \, \mathbf{b} \, \mathbf{c} = [\,\mathbf{b} \, \mathbf{c} \, \mathbf{a}$ , we can rewrite $\lambda$ as $\lambda = \frac{[\,\mathbf{b} \, \mathbf{c} \, \mathbf{d}}{[\,\mathbf{b} \, \mathbf{c} \, \mathbf{a}}$ .

Why other options are wrong:
- Option A: $\frac{[\,\mathrm{dbc}}{[\,\mathrm{bac}}$ : Incorrect order of vectors in the scalar triple product in the denominator.
- Option C: $\frac{[\,b d c}{[\,a b c}$ : Correct numerator, but the order of vectors in the denominator is incorrect.
- Option D: $\frac{[\,\mathbf{c b d}}{[\,\mathbf{a b c}}$ : Incorrect numerator. The order of vectors in the numerator is incorrect.

Key Concepts:
- Scalar triple product
- Dot product
- Linear dependence of vectors

Memory Aids:
- Scalar triple product: $[\,\mathbf{a} \, \mathbf{b} \, \mathbf{c} = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$
- Cyclic permutation of vectors in scalar triple product does not change its value.

Quick Tips:
- Remember that the scalar triple product is zero if any two vectors are linearly dependent.
- Use the properties of the scalar triple product to simplify expressions.

83 Easy

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Application +4 | -1

If

\vec{a}, \vec{b}, \vec{c}

are three non-coplanar vectors represented by non-current edges of a parallelopiped of volume 4 units, then the value of

(\vec a + \vec b) \cdot (\vec b \times \vec c) + (\vec b + \vec c) \cdot (\vec c \times \vec a) + (\vec c + \vec a) \cdot (\vec a \times \vec b){\text{, is }}

View answer and explanation

Correct option: C

\pm 12

Option C (

\pm 12

) is correct because the expression simplifies to three times the scalar triple product, which is related to the volume of the parallelepiped.

Solution approach

Correct Answer + Core Reason: $\pm 12$ . The expression simplifies to 3 times the volume of the parallelopiped.

Step-by-Step Solution:

Step 1: State the volume of the parallelepiped: $\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]=\pm 4$ .

Step 2: Expand the given expression: $(\vec{a}+\vec{b}) \cdot(\vec{b} \times \vec{c})+(\vec{b}+\vec{c}) \cdot(\vec{c} \times \vec{a})+(\vec{c}+\vec{a}) \cdot(\vec{a} \times \vec{b})$ .

Step 3: Simplify using the properties of the scalar triple product: $\vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{b} \cdot (\vec{b} \times \vec{c}) + \vec{b} \cdot (\vec{c} \times \vec{a}) + \vec{c} \cdot (\vec{c} \times \vec{a}) + \vec{c} \cdot (\vec{a} \times \vec{b}) + \vec{a} \cdot (\vec{a} \times \vec{b})$ .

Step 4: Note that $\vec{b} \cdot (\vec{b} \times \vec{c}) = 0$ , $\vec{c} \cdot (\vec{c} \times \vec{a}) = 0$ , and $\vec{a} \cdot (\vec{a} \times \vec{b}) = 0$ .

Step 5: The expression becomes: $\vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{b} \cdot (\vec{c} \times \vec{a}) + \vec{c} \cdot (\vec{a} \times \vec{b}) = 3\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]$ .

Step 6: Substitute the volume: $3\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right] = 3(\pm 4) = \pm 12$ .

Why other options are wrong:
- Option A: This is incorrect because the expression simplifies to $\pm 12$ .
- Option B: This is incorrect because the expression simplifies to $\pm 12$ .
- Option D: This is incorrect because the expression simplifies to $\pm 12$ .

Key Concepts:
- Scalar triple product
- Volume of a parallelepiped
- Properties of dot and cross products

Memory Aids:
- Remember that $\vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{b} \cdot (\vec{c} \times \vec{a}) = \vec{c} \cdot (\vec{a} \times \vec{b})$ .

Quick Tips:
- Use the properties of scalar triple products to simplify expressions.

84 Easy

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Application +4 | -1

The value of

a

so that the volume of parallelopiped formed by

\widehat \mathbf{i} + a\widehat \mathbf{j} + \mathbf{\hat k},\;\widehat \mathbf{j} + a\mathbf{\hat k}

and

a\widehat \mathbf{i} + \mathbf{\hat k}

becomes minimum is

View answer and explanation

Correct option: C

1/\sqrt 3

Option C (

1/\sqrt 3

) is correct because it minimizes the volume of the parallelepiped, which is found by taking the derivative and setting it to zero.

Solution approach

Correct Answer + Core Reason: $1/\sqrt 3$ . The volume is minimized by finding the critical points of the volume function.

Step-by-Step Solution:

Step 1: Find the volume of the parallelepiped using the scalar triple product: $V = |(\hat{i} + a\hat{j} + \hat{k}) \cdot ((\hat{j} + a\hat{k}) \times (a\hat{i} + \hat{k}))|$ .

Step 2: Calculate the cross product: $(\hat{j} + a\hat{k}) \times (a\hat{i} + \hat{k}) = \hat{j} \times a\hat{i} + \hat{j} \times \hat{k} + a\hat{k} \times a\hat{i} + a\hat{k} \times \hat{k} = -a\hat{k} + \hat{i} + a^2\hat{j} + 0 = \hat{i} + a^2\hat{j} - a\hat{k}$ .

Step 3: Calculate the scalar triple product: $(\hat{i} + a\hat{j} + \hat{k}) \cdot (\hat{i} + a^2\hat{j} - a\hat{k}) = 1 \times 1 + a \times a^2 + 1 \times (-a) = 1 + a^3 - a$ .

Step 4: Find the volume function: $f(a) = |1 + a^3 - a|$ . To minimize the volume, we need to minimize $|1 + a^3 - a|$ .

Step 5: Find the derivative of $g(a) = 1 + a^3 - a$ : $g'(a) = 3a^2 - 1$ .

Step 6: Set the derivative equal to zero and solve for $a$ : $3a^2 - 1 = 0 \Rightarrow a^2 = \frac{1}{3} \Rightarrow a = \pm \frac{1}{\sqrt{3}}$ .

Step 7: Find the second derivative: $g''(a) = 6a$ .

Step 8: Evaluate the second derivative at $a = \frac{1}{\sqrt{3}}$ : $g''(\frac{1}{\sqrt{3}}) = 6 \times \frac{1}{\sqrt{3}} = \frac{6}{\sqrt{3}} > 0$ , which indicates a minimum.

Step 9: Evaluate the second derivative at $a = -\frac{1}{\sqrt{3}}$ : $g''(-\frac{1}{\sqrt{3}}) = 6 \times -\frac{1}{\sqrt{3}} = -\frac{6}{\sqrt{3}} < 0$ , which indicates a maximum.

Step 10: Therefore, the value of $a$ that minimizes the volume is $a = \frac{1}{\sqrt{3}}$ .

Step 11: Final Answer: $1/\sqrt 3$

Why other options are wrong:
- Option A (-3): This is incorrect because it does not satisfy the condition for minimum volume based on the derivative test.
- Option B (3): This is incorrect because it does not satisfy the condition for minimum volume based on the derivative test.
- **Option D ( $\sqrt 3$ ): This is incorrect because it does not satisfy the condition for minimum volume based on the derivative test.

Key Concepts:
- Scalar triple product and volume of a parallelepiped
- Finding minimum using derivatives

Memory Aids:
- Remember the scalar triple product formula.

Quick Tips:**
- Use derivatives to find the minimum or maximum of a function.
- Check the second derivative to confirm whether it's a minimum or maximum.

85 Easy

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Application +4 | -1

If

\mathbf{a} \cdot \mathbf{i}=4

, then

(a \times \mathbf{j}) \cdot(2 \mathbf{j}-3 \mathbf{k})=

View answer and explanation

Correct option: D

-12

Option D (-12) is correct because the expression (a x j) . (2j - 3k) simplifies to -3(a . i), and given a . i = 4, the result is -12.

Solution approach

Correct Answer + Core Reason: -12. Using vector identities and the given information, the expression simplifies to -12.

Step-by-Step Solution:

Step 1: We are given that $\mathbf{a} \cdot \mathbf{i} = 4$ and we need to find $(\mathbf{a} \times \mathbf{j}) \cdot (2\mathbf{j} - 3\mathbf{k})$ .

Step 2: Use the vector identity $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ :
$(\mathbf{a} \times \mathbf{j}) \cdot (2\mathbf{j} - 3\mathbf{k}) = \mathbf{a} \cdot (\mathbf{j} \times (2\mathbf{j} - 3\mathbf{k}))$

Step 3: Calculate the cross product $\mathbf{j} \times (2\mathbf{j} - 3\mathbf{k})$ :
$\mathbf{j} \times (2\mathbf{j} - 3\mathbf{k}) = 2(\mathbf{j} \times \mathbf{j}) - 3(\mathbf{j} \times \mathbf{k})$ \nSince $\mathbf{j} \times \mathbf{j} = 0$ and $\mathbf{j} \times \mathbf{k} = \mathbf{i}$ , we have:
$= 0 - 3\mathbf{i} = -3\mathbf{i}$

Step 4: Calculate the dot product $\mathbf{a} \cdot (-3\mathbf{i})$ :
$\mathbf{a} \cdot (-3\mathbf{i}) = -3(\mathbf{a} \cdot \mathbf{i})$

Step 5: Substitute the given value $\mathbf{a} \cdot \mathbf{i} = 4$ :
$-3(\mathbf{a} \cdot \mathbf{i}) = -3(4) = -12$

Why other options are wrong:
- Option A (12): This value is incorrect. It might arise from an error in sign when calculating the dot product.
- Option B (2): This value is incorrect. It might arise from an error in calculating the cross product or the dot product.
- Option C (0): This value is incorrect. It might arise from assuming the expression equals zero without proper calculation.

Key Concepts:
- Vector identities
- Cross product
- Dot product

Memory Aids:
- Remember the vector identity $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ .
- Remember the cross products of unit vectors.

Quick Tips:
- Use vector identities to simplify expressions.
- Double-check the signs when calculating the cross product and the dot product.

86 Easy

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Application +4 | -1

The length of the shortest distance between the two lines

\mathbf{\vec r} = \left( { - 3\widehat \mathbf{i} + 6\widehat \mathbf{j}} \right) + s\left( { - 4\widehat \mathbf{i} + 3\widehat \mathbf{j} + 2\mathbf{\hat k}} \right)

and

\mathbf{\vec r} = \left( { - 2\widehat \mathbf{i} + 7\mathbf{\hat k}} \right) + t\left( { - 4\widehat \mathbf{i} + \widehat \mathbf{j} + \mathbf{\hat k}} \right)\mathrm{\;}

is

View answer and explanation

Correct option: D

9 units

Option D (9 units) is correct because the shortest distance between the two lines is calculated using the formula involving the vector difference and the cross product of the direction vectors.

Solution approach

Correct Answer + Core Reason: 9 units. The shortest distance between two skew lines is calculated using the formula involving the vector difference and the cross product of the direction vectors.

Step-by-Step Solution:

Step 1: Identify the vectors $\mathbf{\vec a}_1$ , $\mathbf{\vec a}_2$ , $\mathbf{\vec b}_1$ , and $\mathbf{\vec b}_2$ from the given equations.
$\mathbf{\vec a}_1 = -3\widehat{\mathbf{i}} + 6\widehat{\mathbf{j}}$ , $\mathbf{\vec a}_2 = -2\widehat{\mathbf{i}} + 7\mathbf{\hat k}$
$\mathbf{\vec b}_1 = -4\widehat{\mathbf{i}} + 3\widehat{\mathbf{j}} + 2\mathbf{\hat k}$ , $\mathbf{\vec b}_2 = -4\widehat{\mathbf{i}} + \widehat{\mathbf{j}} + \mathbf{\hat k}$

Step 2: Calculate $\mathbf{\vec a}_2 - \mathbf{\vec a}_1$ .
$\mathbf{\vec a}_2 - \mathbf{\vec a}_1 = (-2 - (-3))\widehat{\mathbf{i}} + (0 - 6)\widehat{\mathbf{j}} + (7 - 0)\mathbf{\hat k} = \widehat{\mathbf{i}} - 6\widehat{\mathbf{j}} + 7\mathbf{\hat k}$

Step 3: Calculate $\mathbf{\vec b}_1 \times \mathbf{\vec b}_2$ .
$\mathbf{\vec b}_1 \times \mathbf{\vec b}_2 = \left| {\begin{{array}}{{*{20}{c}}} {\widehat{\mathbf{i}}}&{\widehat{\mathbf{j}}}&{\mathbf{\hat k}}\\ { - 4}&3&2\\ { - 4}&1&1 \end{{array}}} \right| = (3 - 2)\widehat{\mathbf{i}} - (-4 - (-8))\widehat{\mathbf{j}} + (-4 - (-12))\mathbf{\hat k} = \widehat{\mathbf{i}} - 4\widehat{\mathbf{j}} + 8\mathbf{\hat k}$

Step 4: Calculate the magnitude of $\mathbf{\vec b}_1 \times \mathbf{\vec b}_2$ .
$\left| {\mathbf{\vec b}_1 \times \mathbf{\vec b}_2} \right| = \sqrt{{1^2 + (-4)^2 + 8^2}} = \sqrt{{1 + 16 + 64}} = \sqrt{{81}} = 9$

Step 5: Calculate the scalar triple product $(\mathbf{\vec a}_2 - \mathbf{\vec a}_1) \cdot (\mathbf{\vec b}_1 \times \mathbf{\vec b}_2)$ .
$(\mathbf{\vec a}_2 - \mathbf{\vec a}_1) \cdot (\mathbf{\vec b}_1 \times \mathbf{\vec b}_2) = (\widehat{\mathbf{i}} - 6\widehat{\mathbf{j}} + 7\mathbf{\hat k}) \cdot (\widehat{\mathbf{i}} - 4\widehat{\mathbf{j}} + 8\mathbf{\hat k}) = (1)(1) + (-6)(-4) + (7)(8) = 1 + 24 + 56 = 81$

Step 6: Calculate the shortest distance $d$ .
$d = \left| {\frac{{(\mathbf{\vec a}_2 - \mathbf{\vec a}_1) \cdot (\mathbf{\vec b}_1 \times \mathbf{\vec b}_2)}}{{\left| {\mathbf{\vec b}_1 \times \mathbf{\vec b}_2} \right|}}} \right| = \left| {\frac{{81}}{9}} \right| = 9$

Why other options are wrong:
- Option A (7 units): This is incorrect because it's not the correct shortest distance calculated using the formula.
- Option B (13 units): This is incorrect because it's not the correct shortest distance calculated using the formula.
- Option C (8 units): This is incorrect because it's not the correct shortest distance calculated using the formula.

Key Concepts:
- Shortest distance between two skew lines
- Vector subtraction
- Cross product of vectors
- Scalar triple product

Memory Aids:
- Remember the formula for the shortest distance between two skew lines.

Quick Tips:
- Be careful with the signs when calculating the vector differences and cross products.

87 Easy

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Analyzing +4 | -1

For any three vectors

\mathbf{\vec a},\mathbf{\vec b}

and

{\mathbf{\vec c}}

,

\left( {\mathbf{\vec a} - \mathbf{\vec b}} \right)\cdot\left( {\mathbf{\vec b} + \mathbf{\vec c}\;} \right) \times \left( {\;\mathbf{\vec c} + \mathbf{\vec a}} \right)

is equal to

View answer and explanation

Correct option: D

0

Option D (0) is correct because the scalar triple product simplifies to zero due to vector properties and cancellations.

Solution approach

Correct Answer + Core Reason: 0. The scalar triple product simplifies to zero due to vector properties.

Step-by-Step Solution:

Step 1: Expand the cross product: $\left( {\mathbf{\vec a} - \mathbf{\vec b}} \right)\cdot\left( {\mathbf{\vec b} + \mathbf{\vec c}\\; } \right) \times \left( {\;\mathbf{\vec c} + \mathbf{\vec a}} \right) = \left( {\mathbf{\vec a} - \mathbf{\vec b}} \right)\cdot\left[ {\mathbf{\vec b} \times \mathbf{\vec c} + \mathbf{\vec b} \times \mathbf{\vec a} + \mathbf{\vec c} \times \mathbf{\vec c} + \mathbf{\vec c} \times \mathbf{\vec a}\\; } \right]$

Step 2: Simplify using properties of cross product: Since $\mathbf{\vec c} \times \mathbf{\vec c} = 0$ , the expression becomes $\left( {\mathbf{\vec a} - \mathbf{\vec b}} \right)\cdot\left[ {\mathbf{\vec b} \times \mathbf{\vec c} + \mathbf{\vec b} \times \mathbf{\vec a} + \mathbf{\vec c} \times \mathbf{\vec a}\\; } \right]$

Step 3: Expand the dot product: ${\mathbf{\vec a}\cdot\left( {\mathbf{\vec b} \times \mathbf{\vec c}} \right) + \mathbf{\vec a}\cdot\left( {\mathbf{\vec b} \times \mathbf{\vec a}} \right) + \mathbf{\vec a}\cdot\left( {\mathbf{\vec c} \times \mathbf{\vec a}} \right) - \mathbf{\vec b}\cdot\left( {\mathbf{\vec b} \times \mathbf{\vec c}} \right) - \mathbf{\vec b}\cdot\left( {\mathbf{\vec b} \times \mathbf{\vec a}} \right) - \mathbf{\vec b}\cdot\left( {\mathbf{\vec c} \times \mathbf{\vec a}} \right)}$

Step 4: Simplify using scalar triple product properties: $\mathbf{\vec a}\cdot\left( {\mathbf{\vec b} \times \mathbf{\vec a}} \right) = 0$ and $\mathbf{\vec a}\cdot\left( {\mathbf{\vec c} \times \mathbf{\vec a}} \right) = 0$ and $\mathbf{\vec b}\cdot\left( {\mathbf{\vec b} \times \mathbf{\vec c}} \right) = 0$ and $\mathbf{\vec b}\cdot\left( {\mathbf{\vec b} \times \mathbf{\vec a}} \right) = 0$ . The expression reduces to $\mathbf{\vec a}\cdot\left( {\mathbf{\vec b} \times \mathbf{\vec c}} \right) - \mathbf{\vec b}\cdot\left( {\mathbf{\vec c} \times \mathbf{\vec a}} \right)$ .

Step 5: Further simplification: $\mathbf{\vec a}\cdot\left( {\mathbf{\vec b} \times \mathbf{\vec c}} \right) - \mathbf{\vec b}\cdot\left( {\mathbf{\vec c} \times \mathbf{\vec a}} \right) = \left[ {\;\mathbf{\vec a}\;\mathbf{\vec b}\;\mathbf{\vec c}\mathrm{\;}} \right] - \left[ {\;\mathbf{\vec b}\;\mathbf{\vec c}\;\mathbf{\vec a}\mathrm{\;}} \right] = \left[ {\;\mathbf{\vec a}\;\mathbf{\vec b}\;\mathbf{\vec c}\mathrm{\;}} \right] - \left[ {\;\mathbf{\vec a}\;\mathbf{\vec b}\;\mathbf{\vec c}\mathrm{\;}} \right] = 0$

Final Answer: 0

Why other options are wrong:
- Option A: This option is incorrect because it only considers one term of the expansion and doesn't account for cancellations.
- Option B: This option is incorrect because it represents the scalar triple product, but the overall expression evaluates to zero.
- Option C: This option is incorrect because it squares the scalar triple product, which is not the result of the given expression.
- Option D: This option is the correct answer.

Key Concepts:
- Scalar triple product
- Vector cross product
- Vector dot product
- Properties of vector operations

Memory Aids:
- Remember the cyclic property of the scalar triple product: $\left[ {\;\mathbf{\vec a}\;\mathbf{\vec b}\;\mathbf{\vec c}\mathrm{\;}} \right] = \left[ {\;\mathbf{\vec b}\;\mathbf{\vec c}\;\mathbf{\vec a}\mathrm{\;}} \right] = \left[ {\;\mathbf{\vec c}\;\mathbf{\vec a}\;\mathbf{\vec b}\mathrm{\;}} \right]$

Quick Tips:
- Expand the cross product first.
- Use properties of dot and cross products to simplify.
- Remember that $\vec a \times \vec a = 0$ and $\vec a \cdot (\vec b \times \vec a) = 0$

88 Easy

88 of 185

Application +4 | -1

If the volume of the parallelopiped with

\mathbf{\vec a},\;\mathbf{\vec b}

and

{\mathbf{\vec c}}

as coterminous edges is 40 cu units, then the volume of the parallelopiped having

\mathbf{\vec b} + \mathbf{\vec c},\;\mathbf{\vec c} + \mathbf{\vec a}

and

\mathbf{\vec a} + \mathbf{\vec b}

as coterminous edges inn cubic units is

View answer and explanation

Correct option: A

80

Option A (80) is correct because the volume of the new parallelepiped is twice the volume of the original parallelepiped.

Solution approach

Correct Answer + Core Reason: 80. The volume of the new parallelepiped is twice the volume of the original.

Step-by-Step Solution:

Step 1: Recall that the volume of a parallelepiped formed by vectors $\vec{a}$ , $\vec{b}$ , and $\vec{c}$ is given by the scalar triple product: $V = |[\vec{a} \,\vec{b} \,\vec{c}]|$ .

Step 2: The volume of the new parallelepiped is given by $V' = |[\vec{b} + \vec{c} \,\vec{c} + \vec{a} \,\vec{a} + \vec{b}]|$ .

Step 3: Expand the scalar triple product: $V' = |(\vec{b} + \vec{c}) \cdot ((\vec{c} + \vec{a}) \times (\vec{a} + \vec{b}))|$ .

Step 4: Calculate the cross product: $(\vec{c} + \vec{a}) \times (\vec{a} + \vec{b}) = \vec{c} \times \vec{a} + \vec{c} \times \vec{b} + \vec{a} \times \vec{a} + \vec{a} \times \vec{b} = \vec{c} \times \vec{a} + \vec{c} \times \vec{b} + 0 + \vec{a} \times \vec{b}$ .

Step 5: Substitute this back into the scalar triple product: $V' = |(\vec{b} + \vec{c}) \cdot (\vec{c} \times \vec{a} + \vec{c} \times \vec{b} + \vec{a} \times \vec{b})| = |\vec{b} \cdot (\vec{c} \times \vec{a}) + \vec{b} \cdot (\vec{c} \times \vec{b}) + \vec{b} \cdot (\vec{a} \times \vec{b}) + \vec{c} \cdot (\vec{c} \times \vec{a}) + \vec{c} \cdot (\vec{c} \times \vec{b}) + \vec{c} \cdot (\vec{a} \times \vec{b})|$ .

Step 6: Simplify using the properties of the scalar triple product: $\vec{b} \cdot (\vec{c} \times \vec{b}) = 0$ and $\vec{c} \cdot (\vec{c} \times \vec{a}) = 0$ and $\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$ and $\vec{c} \cdot (\vec{c} \times \vec{b}) = 0$ .\nSo, $V' = |\vec{b} \cdot (\vec{c} \times \vec{a}) + \vec{c} \cdot (\vec{a} \times \vec{b})| = |[\vec{b} \,\vec{c} \,\vec{a}] + [\vec{c} \,\vec{a} \,\vec{b}]|$ .

Step 7: Use the property that $[\vec{b} \,\vec{c} \,\vec{a}] = [\vec{c} \,\vec{a} \,\vec{b}] = [\vec{a} \,\vec{b} \,\vec{c}]$ : $V' = |[\vec{a} \,\vec{b} \,\vec{c}] + [\vec{a} \,\vec{b} \,\vec{c}]| = |2[\vec{a} \,\vec{b} \,\vec{c}]|$ .

Step 8: Substitute the given volume: $V' = |2 \times 40| = 80$ .

Step 9: Final Answer: 80

Why other options are wrong:
- Option B (120): This is incorrect due to an error in calculating the scalar triple product or applying its properties.
- Option C (160): This is incorrect due to an error in calculating the scalar triple product or applying its properties.
- Option D (40): This is incorrect because it assumes the volume remains the same, which is not true when the vectors are transformed in this way.

Key Concepts:
- Scalar triple product and volume of a parallelepiped
- Properties of scalar triple product

Memory Aids:
- Remember the properties of the scalar triple product and how it relates to the volume of a parallelepiped.

Quick Tips:
- Use the properties of the scalar triple product to simplify the calculation.
- Be careful with the order of vectors in the scalar triple product.

89 Easy

89 of 185

Application +4 | -1

The shortest distance between the lines whose equations are

\vec{r}=t(\hat{i}+\hat{j}+\hat{k})

and

\vec{r}=\hat{k}+s(\hat{i}-2 \hat{j}+3

\hat{\mathrm{k}})

is :

View answer and explanation

Correct option: B

\frac{3}{\sqrt{38}}

Option B (

\frac{3}{\sqrt{38}}

) is correct. The shortest distance between two lines is given by the formula involving the cross product of the direction vectors and the vector connecting a point on each line.

Solution approach

Correct Answer + Core Reason: $\frac{3}{\sqrt{38}}$ . The shortest distance is calculated using the formula for the distance between two skew lines.

Step-by-Step Solution:

Step 1: Identify the direction vectors and a point on each line:\nLine 1: $\vec{r} = t(\hat{i} + \hat{j} + \hat{k})$ . Direction vector $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ . Point on the line: $\vec{p_1} = \vec{0}$ .\nLine 2: $\vec{r} = \hat{k} + s(\hat{i} - 2\hat{j} + 3\hat{k})$ . Direction vector $\vec{b} = \hat{i} - 2\hat{j} + 3\hat{k}$ . Point on the line: $\vec{p_2} = \hat{k}$ .

Step 2: Find the vector connecting the two points:
$\vec{p_2} - \vec{p_1} = \hat{k} - \vec{0} = \hat{k}$ .

Step 3: Calculate the cross product of the direction vectors:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{vmatrix} = (3 - (-2))\hat{i} - (3 - 1)\hat{j} + (-2 - 1)\hat{k} = 5\hat{i} - 2\hat{j} - 3\hat{k}$ .

Step 4: Calculate the shortest distance using the formula:
$d = \frac{|(\vec{a} \times \vec{b}) \cdot (\vec{p_2} - \vec{p_1})|}{|\vec{a} \times \vec{b}|} = \frac{|(5\hat{i} - 2\hat{j} - 3\hat{k}) \cdot \hat{k}|}{\sqrt{5^2 + (-2)^2 + (-3)^2}} = \frac{|-3|}{\sqrt{25 + 4 + 9}} = \frac{3}{\sqrt{38}}$ .

Why other options are wrong:
- Option A: 3: This is incorrect because it doesn't account for the magnitudes of the vectors involved in the calculation.
- Option C: $\frac{\sqrt{3}}{14}$ : This is incorrect because it is not the correct application of the shortest distance formula.
- Option D: $\frac{2}{\sqrt{13}}$ : This is incorrect because it is not the correct application of the shortest distance formula.

Key Concepts:
- Shortest distance between two lines
- Cross product of vectors
- Dot product of vectors

Memory Aids:
- Remember the formula for the shortest distance between two skew lines.

Quick Tips:
- Identify the direction vectors and points on each line correctly.
- Use the correct formula for the shortest distance.

90 Easy

90 of 185

Analyzing +4 | -1

If

\vec{a}, \vec{b}, \vec{c}

are three non-coplanar, non-zero vectors, then

(\vec{a} \cdot \vec{a})(\vec{b} \times \vec{c})+(\vec{a} \cdot \vec{b})(\vec{c} \times \vec{a})+(\vec{a} \cdot \vec{c})(\vec{a} \times \vec{b})

is equal to

View answer and explanation

Correct option: B

\left[\begin{array}{lll}\vec{b} & \vec{c} & \vec{a}\end{array}\right] \vec{a}

Option B (

\left[\begin{array}{lll}\vec{b} & \vec{c} & \vec{a}\end{array}\right] \vec{a}

) is correct because the given expression simplifies to the scalar triple product multiplied by the vector

\vec{a}

.

Solution approach

Correct Answer + Core Reason: $\left[\begin{array}{lll}\vec{b} & \vec{c} & \vec{a}\end{array}\right] \vec{a}$ . The expression simplifies to the scalar triple product multiplied by $\vec{a}$ .

Step-by-Step Solution:

Step 1: We are given the expression $(\vec{a} \cdot \vec{a})(\vec{b} \times \vec{c})+(\vec{a} \cdot \vec{b})(\vec{c} \times \vec{a})+(\vec{a} \cdot \vec{c})(\vec{a} \times \vec{b})$ .

Step 2: We can rewrite this as $(\vec{a} \cdot \vec{a})(\vec{b} \times \vec{c})+(\vec{a} \cdot \vec{b})(\vec{c} \times \vec{a})+(\vec{a} \cdot \vec{c})(\vec{a} \times \vec{b})$ .

Step 3: Let's express $\vec{a}$ as a linear combination of $\vec{b} \times \vec{c}$ , $\vec{c} \times \vec{a}$ , and $\vec{a} \times \vec{b}$ : $\vec{a} = x(\vec{b} \times \vec{c}) + y(\vec{c} \times \vec{a}) + z(\vec{a} \times \vec{b})$ .

Step 4: Taking the dot product with $\vec{a}$ , $\vec{b}$ , and $\vec{c}$ respectively, we get:
$\vec{a} \cdot \vec{a} = x[\vec{a} \vec{b} \vec{c}] \Rightarrow x = \frac{{\vec{a} \cdot \vec{a}}}{{[\vec{a} \vec{b} \vec{c}]}}$
$\vec{a} \cdot \vec{b} = y[\vec{a} \vec{b} \vec{c}] \Rightarrow y = \frac{{\vec{a} \cdot \vec{b}}}{{[\vec{a} \vec{b} \vec{c}]}}$
$\vec{a} \cdot \vec{a} = z[\vec{a} \vec{b} \vec{c}] \Rightarrow z = \frac{{\vec{a} \cdot \vec{c}}}{{[\vec{a} \vec{b} \vec{c}]}}$

Step 5: Substituting these values back into the original expression, we get:
$[\vec{a} \vec{b} \vec{c}] (x(\vec{b} \times \vec{c}) + y(\vec{c} \times \vec{a}) + z(\vec{a} \times \vec{b})) = [\vec{a} \vec{b} \vec{c}] \vec{a}$ .\nSince $[\vec{a} \vec{b} \vec{c}] = [\vec{b} \vec{c} \vec{a}]$ , the expression equals $[\vec{b} \vec{c} \vec{a}] \vec{a}$ .

Why other options are wrong:
- Option A: Incorrect. The expression does not simplify to this form.
- Option C: Incorrect. The expression does not simplify to this form.
- Option D: Incorrect. The expression simplifies to a specific form involving $\vec{a}$ .

Key Concepts:
- Scalar triple product
- Linear combination of vectors

Memory Aids:
- Remember the properties of scalar triple products.

Quick Tips:
- Express vectors as linear combinations of non-coplanar vectors.

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