$\left( {{\bf{\vec a}} - {\bf{\vec b}}} \right)\cdot\left\{ {\left( {{\bf{\vec b}} - {\bf{\vec c}}} \right) \times \left( {{\bf{\vec c}} - {\bf{\vec a}}\;} \right)} \right\}$ is equal to

$2\;{\bf{\vec a}}\cdot{\bf{\vec b}} \times {\bf{\vec c}}$

${\bf{\vec a}}\cdot{\bf{\vec b}} \times \;{\bf{\vec c}}$

${\bf{\vec a}}\cdot\;{\bf{\vec b}}\;$

SCALAR TRIPLE PRODUCT MCQs for JEE MATHEMATICS — Easy, Page 6

151 Easy

151 of 185

Application +4 | -1

The shortest distance between the lines

\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3}

and

\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5}

is

View answer and explanation

Correct option: B

6 \sqrt{3}

Option B (

6 \sqrt{3}

) is correct because it is the shortest distance between the two lines calculated using the formula involving the cross product of direction vectors and the difference of position vectors.

Solution approach

Correct Answer + Core Reason: $6 \sqrt{3}$ . The shortest distance between two skew lines is calculated using the formula $d=\left|\frac{\left(\overrightarrow{a_{1}}-\overrightarrow{a_{2}}\right) \cdot\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)}{\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|}\right|$ .

Step-by-Step Solution:

Step 1: Identify the direction vectors and position vectors of the two lines:\nLine 1: $\frac{x-5}{1}=\frac{y-2}{2}=\frac{z-4}{-3}$ . Here, $\overrightarrow{b_{1}}= \hat{i}+2 \hat{j}-3 \hat{k}$ and $\overrightarrow{a_{1}}=5 \hat{i}+2 \hat{j}+4 \hat{k}$ .\nLine 2: $\frac{x+3}{1}=\frac{y+5}{4}=\frac{z-1}{-5}$ . Here, $\overrightarrow{b_{2}}= \hat{i}+4 \hat{j}-5 \hat{k}$ and $\overrightarrow{a_{2}}=-3 \hat{i}-5 \hat{j}+1 \hat{k}$ .

Step 2: Calculate $\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}$ : $\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\1 & 2 & -3 \\1 & 4 & -5\end{array}\right|= \hat{i}(2(-5)-(-3)(4)) - \hat{j}(1(-5)-(-3)(1)) + \hat{k}(1(4)-2(1)) = \hat{i}(-10+12) - \hat{j}(-5+3) + \hat{k}(4-2) = 2 \hat{i}+2 \hat{j}+2 \hat{k}$ .

Step 3: Calculate $\overrightarrow{a_{1}}-\overrightarrow{a_{2}}$ : $\overrightarrow{a_{1}}-\overrightarrow{a_{2}}=(5-(-3)) \hat{i}+(2-(-5)) \hat{j}+(4-1) \hat{k}=8 \hat{i}+7 \hat{j}+3 \hat{k}$ .

Step 4: Calculate $\left(\overrightarrow{a_{1}}-\overrightarrow{a_{2}}\right) \cdot\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right)$ : $(8 \hat{i}+7 \hat{j}+3 \hat{k}) \cdot(2 \hat{i}+2 \hat{j}+2 \hat{k})=8(2)+7(2)+3(2)=16+14+6=36$ .

Step 5: Calculate $\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|$ : $\left|2 \hat{i}+2 \hat{j}+2 \hat{k}\right|=\sqrt{2^{2}+2^{2}+2^{2}}=\sqrt{12}=2 \sqrt{3}$ .

Step 6: Calculate the shortest distance $d$ : $d=\left|\frac{36}{2 \sqrt{3}}\right|=\frac{18}{\sqrt{3}}=\frac{18 \sqrt{3}}{3}=6 \sqrt{3}$ .

Why other options are wrong:
- Option A: $5 \sqrt{3}$ : This is incorrect as it doesn't follow from the correct calculations of the cross product and dot product.
- Option C: $4 \sqrt{3}$ : This is incorrect as it doesn't follow from the correct calculations of the cross product and dot product.
- Option D: $7 \sqrt{3}$ : This is incorrect as it doesn't follow from the correct calculations of the cross product and dot product.

Key Concepts:
- Shortest distance between two skew lines.
- Vector algebra (cross product and dot product).
- Position vectors and direction vectors.

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

Quick Tips:
- Always identify the direction vectors and position vectors correctly.
- Double-check the cross product and dot product calculations.

152 Easy

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Application +4 | -1

If

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

are non-coplanar non-null vetors stch that

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

, then

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

View answer and explanation

Correct option: B

16

Option B (16) is correct because

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

, and thus the square of that is

\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]^4 = 2^4 = 16

.

Solution approach

Correct Answer + Core Reason: 16. The expression simplifies to the fourth power of the scalar triple product.

Step-by-Step Solution:

Step 1: Recall the identity for the scalar triple product of cross products:
$[\vec{a} \times \vec{b} \\,\\, \vec{b} \times \vec{c} \\,\\, \vec{c} \times \vec{a}] = ([\vec{a} \\,\\, \vec{b} \\,\\, \vec{c}])^2$

Step 2: Substitute the given value:
$\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right] = 2$

Step 3: Calculate the square of the scalar triple product of cross products:
$\left\{{ \left[\begin{array}{llllllllllllllllllll} {\vec a \times \vec b}&{\vec b \times \vec c}&{\vec c \times \vec a} \end{array} \right]} \right\}^2 = {\left\{{ \left[\begin{array}{llllllllllllllllllll} {\vec a}&{\vec b}&{\vec c} \end{array} \right]} \right\}^4 = {2^4} = 16$

Final Answer: 16

Why other options are wrong:
- Option A (4): This is incorrect because it only calculates the square of the original scalar triple product, not the fourth power.
- Option C (8): This is incorrect because it does not correctly apply the identity and the given value.
- Option D (none of these): This is incorrect because 16 is the correct answer.

Key Concepts:
- Scalar triple product
- Vector cross product
- Identity for scalar triple product of cross products

Memory Aids:
- Remember the identity: $[\vec{a} \times \vec{b} \\,\\, \vec{b} \times \vec{c} \\,\\, \vec{c} \times \vec{a}] = ([\vec{a} \\,\\, \vec{b} \\,\\, \vec{c}])^2$

Quick Tips:
- Recognize the standard identities for scalar triple products to simplify calculations.

153 Easy

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Analyzing +4 | -1

If the vectors

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

are non-coplanar und

l, m, n

are distinct scalars such that

[l \vec{a}+m \vec{b}+n \vec{c} \,\,l \vec{b}+m \vec{c}+n \vec{a} \,\,l \vec{c}+m \vec{a}+n \vec{b}]=0

then

View answer and explanation

Correct option: B

l+m+n=0

Option B (

l+m+n=0

) is correct because the determinant equation simplifies to

(l+m+n)(l^2+m^2+n^2-l m-m n-n l)=0

, and since

l, m, n

are distinct,

l+m+n=0

.

Solution approach

Correct Answer + Core Reason: $l+m+n=0$ . The determinant equation simplifies to $(l+m+n)(l^2+m^2+n^2-l m-m n-n l)=0$ , leading to $l+m+n=0$ .

Step-by-Step Solution:

Step 1: Start with the given equation:
$l \vec{a}+m \vec{b}+n \vec{c} \quad l \vec{b}+m \vec{c}+n \vec{a} \quad l \vec{c}+m \vec{a}+n \vec{b}=0$

Step 2: Rewrite the equation:
$l\vec a + m\vec b + n\vec c \quad n\vec a + l\vec b + m\vec c \quad m\vec a + n\vec b + l\vec c = 0$

Step 3: Express the scalar triple product as a determinant:
$\left| {\begin{array}{cccccccccccccccccccc}\n l&m&n \\ n&l&m \\ m&n&l\n\end{array}} \right|\vec{a} \quad \vec{b} \quad \vec{c}=0$

Step 4: Since $\vec{a}, \vec{b}, \vec{c}$ are non-coplanar, $\vec{a} \quad \vec{b} \quad \vec{c} \neq 0$ . Therefore, the determinant must be zero:
$\left| {\begin{array}{cccccccccccccccccccc}\n l&m&n \\ n&l&m \\ m&n&l\n\end{array}} \right|=0$

Step 5: Evaluate the determinant:
$l(l^2 - mn) - m(nl - m^2) + n(n^2 - lm) = 0$
$l^3 - lmn - mnl + m^3 + n^3 - lmn = 0$
$l^3 + m^3 + n^3 - 3lmn = 0$

Step 6: Factor the equation:
$(l + m + n)(l^2 + m^2 + n^2 - lm - mn - nl) = 0$

Step 7: Since $l, m, n$ are distinct scalars, $l^2 + m^2 + n^2 - lm - mn - nl \neq 0$ . Therefore:
$l + m + n = 0$

Final Answer: $l+m+n=0$

Why other options are wrong:
- **Option A ( $l m+m n+n l=0$ ):** This is incorrect because the correct condition is $l+m+n=0$ .
- **Option C ( $l^2+m^2+n^2=0$ ):** This is incorrect because the correct condition is $l+m+n=0$ .
- **Option D ( $l^3+m^3+n^3=0$ ):** This is incorrect because the correct condition is $l+m+n=0$ .

Key Concepts:
- Scalar triple product
- Non-coplanar vectors
- Determinants
- Factoring algebraic expressions

Memory Aids:
- Remember the determinant expansion formula.
- Remember the factorization of $l^3 + m^3 + n^3 - 3lmn$ .

Quick Tips:
- When dealing with scalar triple products and non-coplanar vectors, the determinant is a key tool.
- Look for opportunities to factor algebraic expressions to simplify equations.

154 Easy

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Application +4 | -1

Let

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

be three non-coplanar vectors and let

\mathbf{\vec p},\,\,\mathbf{\vec q}\mathrm{\;}\,\,\,\mathrm{and}\,\mathrm{\;}\mathbf{\vec r}

be vector defined by the relations.

\mathbf{\vec p} = \frac{{\mathbf{\vec b} \times \mathbf{\vec c}}}{{\left[ {\;\mathbf{\vec a}\,\;\mathbf{\vec b}\,\;\mathbf{\vec c}\;} \right]}},\,\,\mathbf{\vec q} = \frac{{\mathbf{\vec c} \times \mathbf{\vec a}}}{{\left[ {\mathbf{\vec a}\;\,\mathbf{\vec b}\;\,\mathbf{\vec c}\;} \right]}}\,\,\mathrm{and\;}\,\,\mathbf{\vec r} = \frac{{\mathbf{\vec a} \times \mathbf{\vec b}}}{{\left[ {\mathbf{\vec a}\;\,\mathbf{\vec b}\;\,\mathbf{\vec c}\;} \right]}}

. Then, the value of the expression

\left( {\mathbf{\vec a} + \mathbf{\vec b}} \right)\,\, \cdot \,\,\mathbf{\vec p} + \left( {\mathbf{\vec b} + \mathbf{\vec c}} \right)\,\, \cdot \,\,\mathbf{\vec q} + \left( {\mathbf{\vec c} + \mathbf{\vec a}} \right)\,\, \cdot \,\,\mathbf{\vec r}

is equal to

View answer and explanation

Correct option: D

3

Option D (3) is correct because by using the properties of reciprocal vectors, we can simplify the expression to equal 3.

Solution approach

Correct Answer + Core Reason: 3. The value of the expression is 3, which is found using the properties of reciprocal vectors.

Step-by-Step Solution:

Step 1: Given $\mathbf{\vec p} = \frac{{\mathbf{\vec b} \times \mathbf{\vec c}}}{{\left[ {\;\mathbf{\vec a}\,\mathbf{\vec b}\,\mathbf{\vec c}\} \right]}}$ , $\mathbf{\vec q} = \frac{{\mathbf{\vec c} \times \mathbf{\vec a}}}{{\left[ {\;\mathbf{\vec a}\,\mathbf{\vec b}\,\mathbf{\vec c}\} \right]}}$ and $\mathbf{\vec r} = \frac{{\mathbf{\vec a} \times \mathbf{\vec b}}}{{\left[ {\;\mathbf{\vec a}\,\mathbf{\vec b}\,\mathbf{\vec c}\} \right]}}$ . We need to find the value of $\left( {\mathbf{\vec a} + \mathbf{\vec b}} \right) \cdot \mathbf{\vec p} + \left( {\mathbf{\vec b} + \mathbf{\vec c}} \right) \cdot \mathbf{\vec q} + \left( {\mathbf{\vec c} + \mathbf{\vec a}} \right) \cdot \mathbf{\vec r}$ .

Step 2: Recall that $\mathbf{\vec a} \cdot \mathbf{\vec p} = \frac{{\mathbf{\vec a} \cdot \left( {\mathbf{\vec b} \times \mathbf{\vec c}} \right)}}{{\left[ {\;\mathbf{\vec a}\,\mathbf{\vec b}\,\mathbf{\vec c}\} \right]}} = 1$ , $\mathbf{\vec b} \cdot \mathbf{\vec q} = \frac{{\mathbf{\vec b} \cdot \left( {\mathbf{\vec c} \times \mathbf{\vec a}} \right)}}{{\left[ {\;\mathbf{\vec a}\,\mathbf{\vec b}\,\mathbf{\vec c}\} \right]}} = 1$ , and $\mathbf{\vec c} \cdot \mathbf{\vec r} = \frac{{\mathbf{\vec c} \cdot \left( {\mathbf{\vec a} \times \mathbf{\vec b}} \right)}}{{\left[ {\;\mathbf{\vec a}\,\mathbf{\vec b}\,\mathbf{\vec c}\} \right]}} = 1$ .

Step 3: Also, $\mathbf{\vec b} \cdot \mathbf{\vec p} = \frac{{\mathbf{\vec b} \cdot \left( {\mathbf{\vec b} \times \mathbf{\vec c}} \right)}}{{\left[ {\;\mathbf{\vec a}\,\mathbf{\vec b}\,\mathbf{\vec c}\} \right]}} = 0$ , $\mathbf{\vec c} \cdot \mathbf{\vec q} = \frac{{\mathbf{\vec c} \cdot \left( {\mathbf{\vec c} \times \mathbf{\vec a}} \right)}}{{\left[ {\;\mathbf{\vec a}\,\mathbf{\vec b}\,\mathbf{\vec c}\} \right]}} = 0$ , and $\mathbf{\vec a} \cdot \mathbf{\vec r} = \frac{{\mathbf{\vec a} \cdot \left( {\mathbf{\vec a} \times \mathbf{\vec b}} \right)}}{{\left[ {\;\mathbf{\vec a}\,\mathbf{\vec b}\,\mathbf{\vec c}\} \right]}} = 0$ .

Step 4: Substitute these values into the expression: $\left( {\mathbf{\vec a} + \mathbf{\vec b}} \right) \cdot \mathbf{\vec p} + \left( {\mathbf{\vec b} + \mathbf{\vec c}} \right) \cdot \mathbf{\vec q} + \left( {\mathbf{\vec c} + \mathbf{\vec a}} \right) \cdot \mathbf{\vec r} = \mathbf{\vec a} \cdot \mathbf{\vec p} + \mathbf{\vec b} \cdot \mathbf{\vec p} + \mathbf{\vec b} \cdot \mathbf{\vec q} + \mathbf{\vec c} \cdot \mathbf{\vec q} + \mathbf{\vec c} \cdot \mathbf{\vec r} + \mathbf{\vec a} \cdot \mathbf{\vec r} = 1 + 0 + 1 + 0 + 1 + 0 = 3$ .

Step 5: Final Answer: 3

Why other options are wrong:
- Option A: 0: This is incorrect because it doesn't account for the dot products of the reciprocal vectors with the original vectors.
- Option B: 1: This is incorrect because it only accounts for one of the three terms in the expression.
- Option C: 2: This is incorrect because it only accounts for two of the three terms in the expression.

Key Concepts:
- Scalar triple product
- Reciprocal vectors
- Dot product of vectors

Memory Aids:
- Remember that $\mathbf{\vec a} \cdot \left( {\mathbf{\vec b} \times \mathbf{\vec c}} \right) = \left[ {\;\mathbf{\vec a}\,\mathbf{\vec b}\,\mathbf{\vec c}\} \right]$
- Remember that $\mathbf{\vec a} \cdot \mathbf{\vec p} = 1$ if $\mathbf{\vec p}$ is the reciprocal vector of $\mathbf{\vec a}$ .

Quick Tips:
- Recognize the reciprocal vector form to simplify the dot products.
- Use the properties of the scalar triple product to evaluate the dot products.

155 Easy

155 of 185

Application +4 | -1

Let

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

be three unit vectors such that

\vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c}=0.

If the angle between

\vec{b}

and

\vec{c}

is

\frac{\pi}{3},

then the volume of the parallelopiped whose three coterminous edges are

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

is

View answer and explanation

Correct option: A

\frac{\sqrt{3}}{2}

cubic units

Option A (

\frac{\sqrt{3}}{2}

cubic units) is correct because the volume of the parallelepiped is calculated using the scalar triple product, considering the given dot product and unit vector conditions.

Solution approach

Correct Answer + Core Reason: $\frac{\sqrt{3}}{2}$ cubic units. The volume is calculated using the scalar triple product.

Step-by-Step Solution:

Step 1: State the given conditions: $\vec{a} \cdot \vec{b}=\vec{a} \cdot \vec{c}=0$ and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{3}$ . Also, $|\vec{a}| = |\vec{b}| = |\vec{c}| = 1$ .

Step 2: Deduce that $\vec{a} \perp \vec{b}$ and $\vec{a} \perp \vec{c}$ , which implies $\vec{a}|| \vec{b} \times \vec{c}$ .

Step 3: Calculate the magnitude of the cross product: $|\vec{b} \times \vec{c}|=|\vec{b}||\vec{c}| \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2} \quad[\because|\vec{b}|=|\vec{c}|=1]$ .

Step 4: Calculate the volume: $|[\vec{a} \,\vec{b} \,\vec{c}]|=|\vec{a} \cdot(\vec{b} \times \vec{c})|=|\vec{a}||\vec{b} \times \vec{c}|=\frac{\sqrt{3}}{2}$ .

Why other options are wrong:
- Option B: This is incorrect because the volume is $\frac{\sqrt{3}}{2}$ .
- Option C: This is incorrect because the volume is $\frac{\sqrt{3}}{2}$ .
- Option D: This is incorrect because the volume is $\frac{\sqrt{3}}{2}$ .

Key Concepts:
- Scalar triple product
- Volume of a parallelepiped
- Dot product
- Cross product

Memory Aids:
- Remember the relationship between the scalar triple product, dot product, and cross product.

Quick Tips:
- Use the given conditions to simplify the calculations.

156 Easy

156 of 185

Analyzing +4 | -1

If

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

are three non-coplanar vectors, then

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

equals

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because the triple scalar product simplifies to

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

.

Solution approach

Correct Answer + Core Reason: $\vec{u} \cdot(\vec{v} \times \vec{w})$ . The triple scalar product simplifies to $\vec{u} \cdot(\vec{v} \times \vec{w})$ .

Step-by-Step Solution:

Step 1: Start with the given expression:
$(\vec{u}+\vec{v}-\vec{w}) \cdot(\vec{u}-\vec{v}) \times(\vec{v}-\vec{w})$

Step 2: Express the expression as a scalar triple product:
$\vec{u}+\vec{v}-\vec{w} \quad \vec{u}-\vec{v} \quad \vec{v}-\vec{w}$

Step 3: Expand the scalar triple product:
$\vec{u}+\vec{v}-\vec{w} \quad \vec{u}-\vec{v}+0 \vec{w} \quad 0 \vec{u}+\vec{v}-\vec{w}$

Step 4: Express the scalar triple product as a determinant:
$\left| {\begin{array}{cccccccccccccccccccc}\n 1&1&{ - 1} \\ 1&{ - 1}&0 \\ 0&1&{ - 1}\n\end{array}} \right|\vec{u} \quad \vec{v} \quad \vec{w}$

Step 5: Evaluate the determinant:
$1(1 - 0) - 1(-1 - 0) + (-1)(1 - 0) = 1 + 1 - 1 = 1$

Step 6: Simplify the expression:
$1\vec{u} \quad \vec{v} \quad \vec{w} = \vec{u} \cdot(\vec{v} \times \vec{w})$

Final Answer: $\vec{u} \cdot(\vec{v} \times \vec{w})$

Why other options are wrong:
- **Option B ( $\vec{u} \cdot(\vec{w} \times \vec{v})$ ):** This is incorrect because the correct expression is $\vec{u} \cdot(\vec{v} \times \vec{w})$ .
- **Option C ( $3 \vec{u} \cdot(\vec{v} \times \vec{w})$ ):** This is incorrect because the correct expression is $\vec{u} \cdot(\vec{v} \times \vec{w})$ .
- Option D (0): This is incorrect because the correct expression is $\vec{u} \cdot(\vec{v} \times \vec{w})$ .

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

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

Quick Tips:
- Expressing the expression as a determinant simplifies the calculation.
- Remember the relationship between scalar triple product and determinant.

157 Easy

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Application +4 | -1

The volume (in cubic unit) of the tetrahedron with edges

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

and

\widehat \mathbf{i} + 2\widehat \mathbf{j} - \mathbf{\hat k}

is

View answer and explanation

Correct option: B

\frac{2}{3}

Option B (

\frac{2}{3}

) is correct because it is the volume of the tetrahedron calculated using the scalar triple product of the edge vectors.

Solution approach

Correct Answer + Core Reason: $\frac{2}{3}$ . The volume of a tetrahedron is calculated using the scalar triple product of its edge vectors.

Step-by-Step Solution:

Step 1: Define the edge vectors: $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ , $\vec{b} = \hat{i} - \hat{j} + \hat{k}$ , and $\vec{c} = \hat{i} + 2\hat{j} - \hat{k}$ .

Step 2: Calculate the scalar triple product $[\,\vec{a}\, \vec{b}\, \vec{c} = \vec{a} \cdot (\vec{b} \times \vec{c})$ . This can be found by calculating the determinant of the matrix formed by the components of the vectors.
$\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{vmatrix} = 1((-1)(-1) - (1)(2)) - 1((1)(-1) - (1)(1)) + 1((1)(2) - (-1)(1)) = 1(1 - 2) - 1(-1 - 1) + 1(2 + 1) = -1 + 2 + 3 = 4$ .

Step 3: Calculate the volume of the tetrahedron using the formula $V = \frac{1}{6} |[\,\vec{a}\, \vec{b}\, \vec{c}| = \frac{1}{6} |4| = \frac{4}{6} = \frac{2}{3}$ .

Why other options are wrong:
- Option A (4): This is incorrect because it is missing the factor of $\frac{1}{6}$ in the volume formula.
- **Option C ( $\frac{1}{6}$ ): This is incorrect because the scalar triple product was not calculated correctly.
- Option D ( $\frac{1}{3}$ ): This is incorrect because the scalar triple product was not calculated correctly.

Key Concepts:
- Scalar triple product
- Volume of a tetrahedron
- Determinant of a matrix

Memory Aids:**
- Remember the formula for the volume of a tetrahedron: $V = \frac{1}{6} |[\,\vec{a}\, \vec{b}\, \vec{c}|$ .
- The scalar triple product can be calculated using the determinant of a matrix formed by the components of the vectors.

Quick Tips:
- Double-check the determinant calculation to avoid errors.
- Remember to take the absolute value of the scalar triple product before multiplying by $\frac{1}{6}$ .

158 Easy

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Application +4 | -1

If

a=-3 i+7 j+5 k, \quad \mathbf{b}=-3 i+7 j-3 k, \quad c=7 i-5 j-3 k

are the three coterminous edges of a parallelopiped, then

View answer and explanation

Correct option: C

272

Option C (272) is correct because the volume of the parallelopiped is calculated by the absolute value of the determinant of the matrix formed by the vectors, resulting in 272.

Solution approach

Correct Answer + Core Reason: 272. The volume of a parallelepiped is the absolute value of the scalar triple product.

Step-by-Step Solution:

Step 1: The volume of the parallelepiped is given by $|[\mathbf{a} \mathbf{b} \mathbf{c}]|$ , where $\mathbf{a} = -3\mathbf{i} + 7\mathbf{j} + 5\mathbf{k}$ , $\mathbf{b} = -3\mathbf{i} + 7\mathbf{j} - 3\mathbf{k}$ , and $\mathbf{c} = 7\mathbf{i} - 5\mathbf{j} - 3\mathbf{k}$ .

Step 2: Calculate the determinant:
$V = \left| \begin{array}{ccc} -3 & 7 & 5 \\ -3 & 7 & -3 \\ 7 & -5 & -3 \end{array} \right|$

Step 3: Expand the determinant:\nV = -3(7*(-3) - (-3)*(-5)) - 7((-3)*(-3) - (-3)*7) + 5((-3)*(-5) - 7*7)\nV = -3(-21 - 15) - 7(9 + 21) + 5(15 - 49)\nV = -3(-36) - 7(30) + 5(-34)\nV = 108 - 210 - 170\nV = -272 $\n\n**Step 4:** Take the absolute value:\n$ |V| = |-272| = 272 $\n\n**Why other options are wrong:**\n- **Option A (108):** This value is incorrect. It might arise from an error in calculating the determinant.\n- **Option B (210):** This value is incorrect. It might arise from an error in calculating the determinant.\n- **Option D (308):** This value is incorrect. It might arise from an error in calculating the determinant.\n\n**Key Concepts:**\n- Scalar triple product\n- Volume of a parallelepiped\n- Determinants\n\n**Memory Aids:**\n- The scalar triple product gives the volume of the parallelepiped formed by the three vectors.\n\n**Quick Tips:**\n- Remember the determinant representation of the scalar triple product.\n- Double-check the signs when expanding the determinant.$

159 Easy

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Analyzing +4 | -1

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

is equal to

View answer and explanation

Correct option: C

0

Option C (0) is correct because the scalar triple product of linearly dependent vectors is zero.

Solution approach

Correct Answer + Core Reason: 0. The expression simplifies to the difference of two identical scalar triple products, resulting in zero.

Step-by-Step Solution:

Step 1: Expand the cross product: $(\vec{b} - \vec{c}) \times (\vec{c} - \vec{a}) = \vec{b} \times \vec{c} - \vec{b} \times \vec{a} + \vec{c} \times \vec{a}$

Step 2: Expand the dot product: $(\vec{a} - \vec{b}) \cdot (\vec{b} \times \vec{c} - \vec{b} \times \vec{a} + \vec{c} \times \vec{a}) = \vec{a} \cdot (\vec{b} \times \vec{c}) - \vec{a} \cdot (\vec{b} \times \vec{a}) + \vec{a} \cdot (\vec{c} \times \vec{a}) - \vec{b} \cdot (\vec{b} \times \vec{c}) + \vec{b} \cdot (\vec{b} \times \vec{a}) - \vec{b} \cdot (\vec{c} \times \vec{a})$

Step 3: Simplify using scalar triple product properties: Terms like $\vec{a} \cdot (\vec{b} \times \vec{a})$ and $\vec{b} \cdot (\vec{b} \times \vec{c})$ are zero because they involve the scalar triple product of coplanar vectors.\nSo, the expression becomes: $\vec{a} \cdot (\vec{b} \times \vec{c}) - \vec{b} \cdot (\vec{c} \times \vec{a})$

Step 4: Use the cyclic property of the scalar triple product: $\vec{b} \cdot (\vec{c} \times \vec{a}) = \vec{a} \cdot (\vec{b} \times \vec{c})$

Step 5: Final simplification: $\vec{a} \cdot (\vec{b} \times \vec{c}) - \vec{a} \cdot (\vec{b} \times \vec{c}) = 0$

Why other options are wrong:
- Option A: This is incorrect because it doesn't account for the cancellations and simplifications arising from the properties of scalar triple products.
- Option B: This is incorrect because it doesn't account for the cancellations and simplifications arising from the properties of scalar triple products.
- Option D: This is incorrect because it doesn't account for the vector nature of the expression.

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

Memory Aids:
- Scalar triple product represents the volume of a parallelepiped.
- Cyclic order: a -> b -> c -> a

Quick Tips:
- Remember the cyclic property of the scalar triple product: [a b c] = [b c a] = [c a b]
- If any two vectors in a cross product are the same, the result is zero.

160 Easy

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Understanding +4 | -1

The equation

{\mathbf{r}^{\vec 2}} - 2\mathrm{\;}\mathbf{\vec r}\cdot\mathbf{\vec c} + h = 0,\left| {\mathbf{\vec c}} \right| > \sqrt h

, represent

View answer and explanation

Correct option: D

Sphere

Option D (Sphere) is correct because the given equation is a form of the sphere equation in vector form.

Solution approach

Correct Answer + Core Reason: Sphere. The equation represents a sphere in vector form.

Step-by-Step Solution:

Step 1: Analyze the given equation: ${\mathbf{r}^{\vec 2}} - 2\mathrm{\;}\mathbf{\vec r}\cdot\mathbf{\vec c} + h = 0$ , where $\left| {\mathbf{\vec c}} \right| > \sqrt h$ .

Step 2: Recognize the form of the equation as related to the equation of a sphere.

Step 3: Recall the diameter form of the sphere equation: $\left( {\mathbf{\vec r} - \mathbf{\vec a}} \right)\cdot\left( {\mathbf{\vec r} - \mathbf{\vec b}} \right) = 0$

Step 4: Relate the given equation to the general form of a sphere equation.

Step 5: Conclude that the given equation represents a sphere.

Why other options are wrong:
- Option A (Circle): This is incorrect because the equation is in 3D space, so it represents a sphere, not a circle.
- Option B (Ellipse): This is incorrect because the equation does not have the form of an ellipse.
- Option C (Cone): This is incorrect because the equation does not have the form of a cone.

Key Concepts:
- Vector equation of a sphere
- Diameter form of the sphere equation

Memory Aids:
- Remember the general form of a sphere equation in vector form.

Quick Tips:
- Recognize the general forms of equations for different geometric shapes.

161 Easy

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Analyzing +4 | -1

Let a, b, c are three non-coplanar vectors and d be a non-zero vector, which is perpendicular to

a+b+c

Now, If

d = (\sin x)(a \times b) + (\cos y)(b \times c) + 2(c \times a)

then the minimum value of

\left( {{x^2} + {y^2}} \right)

is

View answer and explanation

Correct option: D

\frac{{5{\pi ^2}}}{4}

The correct answer is D (

\frac{{5{\pi ^2}}}{{4}}

) because it is obtained by finding the values of x and y that satisfy the given conditions and then calculating the minimum value of x² + y².

Solution approach

Correct Answer + Core Reason: $\frac{{5{\pi ^2}}}{{4}}$ . The minimum value of (x² + y²) is found by solving the vector equation and minimizing the expression.

Step-by-Step Solution:

Step 1: Given $d \cdot (a + b + c) = 0$ and $d = (\sin x)(a \times b) + (\cos y)(b \times c) + 2(c \times a)$ .

Step 2: Take the dot product of d with a, b, and c:
$a \cdot d = (\cos y)\,[a\,b\,c]$
$b \cdot d = 2[bca]$
$c \cdot d = (\sin x)[abc]$

Step 3: Add the equations from Step 2:
$a \cdot d + b \cdot d + c \cdot d = (\cos y + 2 + \sin x)[abc]$ \nSince $d \cdot (a + b + c) = 0$ , then $(a \cdot d + b \cdot d + c \cdot d) = 0$ .\nThus, $(\cos y + 2 + \sin x)[abc] = 0$ .\nSince a, b, and c are non-coplanar, $[abc] \neq 0$ . Therefore, $\sin x + \cos y + 2 = 0$ .

Step 4: Solve for \sin x and \cos y:
$\Rightarrow \sin x + \cos y = -2$ \nSince $-1 \leq \sin x \leq 1$ and $-1 \leq \cos y \leq 1$ , the only solution is $\sin x = -1$ and $\cos y = -1$ .

Step 5: Find the values of x and y:
$x = -\frac{\pi}{2}$ and $y = \pi$ .

Step 6: Calculate the minimum value of $x^2 + y^2$ :
$x^2 + y^2 = \left(-\frac{\pi}{2}\right)^2 + (\pi)^2 = \frac{{\pi^2}}{4} + \pi^2 = \frac{{5{\pi ^2}}}{4}$

Why other options are wrong:
- Option A: This value ( $\pi ^2$ ) is incorrect because it only considers the y value and not the x value when minimizing x² + y².
- Option B: This value ( $\frac{{{\pi ^2}}}{{2}}$ ) is incorrect because it doesn't correctly combine the contributions from x and y.
- Option C: This value ( $\frac{{{\pi ^2}}}{{4}}$ ) is incorrect because it only considers the x value and not the y value when minimizing x² + y².

Key Concepts:
- Scalar triple product
- Dot product of vectors
- Solving trigonometric equations
- Minimization of expressions

Memory Aids:
- Remember the range of sine and cosine functions: [-1, 1].
- Recall that $a \cdot (b \times c) = [abc]$ .

Quick Tips:
- When dealing with non-coplanar vectors, their scalar triple product is non-zero.
- Always consider the range of trigonometric functions when solving equations.

162 Easy

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Analyzing +4 | -1

If the foot of the perpendicular from point

(4,3,8)

on the line

L_1: \frac{x-a}{2}=\frac{y-2}{m}=\frac{z-b}{4}, m \neq 0

is

(3,5,7)

then the shortest distance between the line

L_1

and the line

L_2 \Rightarrow \frac{x-2}{3}=\frac{y-4}{4}=\frac{z-5}{5}

is equal to

View answer and explanation

Correct option: B

\frac{1}{\sqrt{6}}

Option B (

\frac{1}{\sqrt{6}}

) is correct because it represents the shortest distance between the two lines, calculated using the formula involving the vector difference and cross product of direction vectors.

Solution approach

Correct Answer + Core Reason: $\frac{1}{\sqrt{6}}$ . The shortest distance is calculated using the formula involving the vector difference and cross product of direction vectors.

Step-by-Step Solution:

Step 1: Identify the given information: Line $L_1: \frac{x-a}{2} = \frac{y-2}{m} = \frac{z-b}{4}$ , point $A(4,3,8)$ , foot of perpendicular $B(3,5,7)$ , and line $L_2: \frac{x-2}{3} = \frac{y-4}{4} = \frac{z-5}{5}$ .

Step 2: Find the direction ratios of $AB$ : $(-1, 2, -1)$ .

Step 3: Since $L_1$ is perpendicular to $AB$ , the dot product of their direction ratios is zero: $2(-1) + m(2) + 4(-1) = 0 \Rightarrow -2 + 2m - 4 = 0 \Rightarrow m = 3$ .

Step 4: Since $B(3,5,7)$ lies on $L_1$ , substitute the coordinates into the equation of $L_1$ : $\frac{3-a}{2} = \frac{5-2}{3} = \frac{7-b}{4} \Rightarrow \frac{3-a}{2} = 1 \Rightarrow a = 1$ and $\frac{7-b}{4} = 1 \Rightarrow b = 3$ .

Step 5: So, $L_1: \frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4}$ . The direction vector of $L_1$ is $\vec{b} = (2, 3, 4)$ and a point on $L_1$ is $\vec{a} = (1, 2, 3)$ .

Step 6: The direction vector of $L_2$ is $\vec{d} = (3, 4, 5)$ and a point on $L_2$ is $\vec{c} = (2, 4, 5)$ .

Step 7: Calculate $\vec{a} - \vec{c} = (1-2, 2-4, 3-5) = (-1, -2, -2)$ .

Step 8: Calculate the cross product $\vec{b} \times \vec{d} = (2, 3, 4) \times (3, 4, 5) = (15-16, 12-10, 8-9) = (-1, 2, -1)$ .

Step 9: Calculate the shortest distance using the formula: $SD = \frac{|(\vec{a} - \vec{c}) \cdot (\vec{b} \times \vec{d})|}{|\vec{b} \times \vec{d}|} = \frac{|(-1)(-1) + (-2)(2) + (-2)(-1)|}{\sqrt{(-1)^2 + 2^2 + (-1)^2}} = \frac{|1 - 4 + 2|}{\sqrt{1 + 4 + 1}} = \frac{|-1|}{\sqrt{6}} = \frac{1}{\sqrt{6}}$ .

Why other options are wrong:
- **Option A ( $\frac{1}{2}$ ): This is incorrect as it doesn't correctly apply the shortest distance formula between two skew lines.
- Option C ( $\sqrt{\frac{2}{3}}$ ): This is incorrect as it doesn't correctly apply the shortest distance formula between two skew lines.
- Option D ( $\frac{1}{\sqrt{3}}$ ): This is incorrect as it doesn't correctly apply the shortest distance formula between two skew lines.

Key Concepts:
- Shortest distance between two skew lines
- Direction ratios and direction vectors of a line
- Dot product and cross product of vectors
- Condition for perpendicularity of lines

Memory Aids:**
- Remember the formula for the shortest distance between two skew lines: $SD = \frac{|(\vec{a} - \vec{c}) \cdot (\vec{b} \times \vec{d})|}{|\vec{b} \times \vec{d}|}$

Quick Tips:
- Always find the direction vectors and a point on each line.
- Remember the formula for the shortest distance and apply it carefully.

163 Easy

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Analyzing +4 | -1

If V is the volume of the parallelopiped having three coterminous edges as a, b and c, then the volume of parallelopiped having three coterminous edges as

\alpha = (a \cdot a)a + (a \cdot b)b + (a \cdot c)c,\,\,\,\,\,\beta = (a \cdot b)a + (b \cdot b)b + (b \cdot c)c\,\,

and

\gamma = (a \cdot c)a + (b \cdot c)b + (c \cdot c)c

is

View answer and explanation

Correct option: A

{V^3}

Option A (

{V^3}

) is correct because the volume of the new parallelepiped is the cube of the original volume.

Solution approach

Correct Answer + Core Reason: ${V^3}$ . The volume of the new parallelopiped is the cube of the original volume.

Step-by-Step Solution:

Step 1: Define V as the volume of the parallelepiped with edges a, b, and c: $V = |[a \, b \, c]|$

Step 2: Define the new vectors: $\alpha = (a \cdot a)a + (a \cdot b)b + (a \cdot c)c$ , $\beta = (a \cdot b)a + (b \cdot b)b + (b \cdot c)c$ , and $\gamma = (a \cdot c)a + (b \cdot c)b + (c \cdot c)c$ .

Step 3: Calculate the volume of the new parallelepiped, $V_1 = |[\alpha \, \beta \, \gamma]|$ .

Step 4: Express the scalar triple product as a determinant: $[\alpha \, \beta \, \gamma] = \begin{{vmatrix}} a \cdot a & a \cdot b & a \cdot c \\ a \cdot b & b \cdot b & b \cdot c \\ a \cdot c & b \cdot c & c \cdot c \end{{vmatrix}} [a \, b \, c]$

Step 5: Recognize that the determinant is the square of the scalar triple product: $\begin{vmatrix} a \cdot a & a \cdot b & a \cdot c \\ a \cdot b & b \cdot b & b \cdot c \\ a \cdot c & b \cdot c & c \cdot c \end{{vmatrix}} = [a \, b \, c]^2$

Step 6: Therefore, $V_1 = |[\alpha \, \beta \, \gamma]| = |[a \, b \, c]^3| = V^3$

Why other options are wrong:
- Option B: $3V$ : This is incorrect because the volume is not simply three times the original volume.
- Option C: ${V^2}$ : This is incorrect because the volume is not the square of the original volume.
- Option D: $2V$ : This is incorrect because the volume is not twice the original volume.

Key Concepts:
- Volume of a parallelepiped
- Scalar triple product
- Determinants

Memory Aids:
- Remember the determinant form for the scalar triple product.

Quick Tips:
- When dealing with volumes of parallelepipeds, use the scalar triple product.
- Be careful with determinant calculations.

164 Easy

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Application +4 | -1

If

\mathbf{a}=2 \mathbf{i}+\mathbf{j}-\mathbf{k}, \mathbf{b}=\mathbf{i}+2 \mathbf{j}+\mathbf{k}

and

\mathbf{c}=\mathbf{i}-\mathbf{j}+2 \mathbf{k}

, then a.

(b \times c)=

View answer and explanation

Correct option: C

12

Option C (12) is correct because the scalar triple product a.(b x c) is calculated correctly using the determinant method.

Solution approach

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

Step-by-Step Solution:

Step 1: The scalar triple product is given by $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ .

Step 2: Calculate the cross product $\mathbf{b} \times \mathbf{c}$ :
$\mathbf{b} \times \mathbf{c} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 2 & 1 \\ 1 & -1 & 2 \end{vmatrix} = (4 - (-1))\mathbf{i} - (2 - 1)\mathbf{j} + (-1 - 2)\mathbf{k} = 5\mathbf{i} - \mathbf{j} - 3\mathbf{k}$

Step 3: Calculate the dot product $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ :
$\mathbf{a} \cdot (5\mathbf{i} - \mathbf{j} - 3\mathbf{k}) = (2\mathbf{i} + \mathbf{j} - \mathbf{k}) \cdot (5\mathbf{i} - \mathbf{j} - 3\mathbf{k}) = 2(5) + 1(-1) + (-1)(-3) = 10 - 1 + 3 = 12$

Why other options are wrong:
- Option A (6): This value is incorrect. It might arise from an error in calculating the cross product or the dot product.
- Option B (10): This value is incorrect. It might arise from an error in calculating the cross product or the dot product.
- Option D (24): This value is incorrect. It might arise from an error in calculating the cross product or the dot product.

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

Memory Aids:
- Remember the determinant representation of the cross product.
- Remember the formula for the dot product.

Quick Tips:
- Double-check the signs when calculating the cross product and the dot product.

165 Easy

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Application +4 | -1

Given unit vectors

\mathbf{m}, \mathbf{n}

and

\mathbf{p}

such that angle between

\mathbf{m}

and

\mathbf{n}=

Angle between

\mathbf{p}

and

(\mathbf{m} \times \mathbf{n})=\frac{\pi}{6}

, then [npm] is equal to

View answer and explanation

Correct option: A

\sqrt{3} / 4

Option A (

\sqrt{3} / 4

) is correct because it is the result of calculating the scalar triple product [npm] using the given angles between the vectors.

Solution approach

Correct Answer + Core Reason: $\sqrt{3} / 4$ . The scalar triple product is calculated using the sine and cosine of the given angles.

Step-by-Step Solution:

Step 1: Recognize that [npm] represents the scalar triple product of vectors n, p, and m.

Step 2: Recall that $[\mathbf{n} \mathbf{p} \mathbf{m}] = (\mathbf{n} \times \mathbf{p}) \cdot \mathbf{m} = |\mathbf{n}| |\mathbf{p}| |\mathbf{m}| \sin \theta \cos \phi$ , where $\theta$ is the angle between $\mathbf{n}$ and $\mathbf{p}$ , and $\phi$ is the angle between $\mathbf{m}$ and the normal to the plane containing $\mathbf{n}$ and $\mathbf{p}$ . In this case, $\theta$ is the angle between $\mathbf{m}$ and $\mathbf{n}$ , and $\phi$ is the angle between $\mathbf{p}$ and $(\mathbf{m} \times \mathbf{n})$ .

Step 3: Substitute the given angles: $\theta = \phi = \frac{\pi}{6}$

Step 4: Since $\mathbf{m}, \mathbf{n}, \mathbf{p}$ are unit vectors, their magnitudes are 1. Therefore, $[\mathrm{npm}]=\sin \theta \cos \phi=\sin \frac{\pi}{6} \cdot \cos \frac{\pi}{6}$

Step 5: Evaluate the trigonometric functions: $\sin \frac{\pi}{6} = \frac{1}{2}$ and $\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}$

Step 6: Calculate the scalar triple product: $[\mathrm{npm}] = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}$

Why other options are wrong:
- Option B: This option is incorrect because it uses an incorrect combination of sine and cosine values or makes a calculation error.
- Option C: This option is incorrect because it uses an incorrect combination of sine and cosine values or makes a calculation error.
- Option D: This option is incorrect because one of the other options is correct.

Key Concepts:
- Scalar triple product
- Unit vectors
- Trigonometric functions

Memory Aids:
- Scalar triple product: $[\mathbf{a} \mathbf{b} \mathbf{c}] = (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}$

Quick Tips:
- Remember that the scalar triple product can be interpreted as the volume of a parallelepiped formed by the three vectors.

166 Easy

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Application +4 | -1

The volume of the parallelopiped whose edges are represented by

-12 i+a k, 3 j-k

and

2 i+j-15 k

is 546 . Then

\alpha=

View answer and explanation

Correct option: C

-3

Option C (-3) is correct because the volume of the parallelepiped is given by the absolute value of the scalar triple product of the vectors representing its edges, and solving for alpha gives -3.

Solution approach

Correct Answer + Core Reason: -3. The value of $\alpha$ is found by calculating the scalar triple product and equating it to the volume.

Step-by-Step Solution:

Step 1: The volume of the parallelepiped is given by the absolute value of the scalar triple product of the vectors representing its edges.

Step 2: The vectors are $\mathbf{a} = -12\mathbf{i} + \alpha \mathbf{k}$ , $\mathbf{b} = 3\mathbf{j} - \mathbf{k}$ , and $\mathbf{c} = 2\mathbf{i} + \mathbf{j} - 15\mathbf{k}$ .

Step 3: The scalar triple product is given by the determinant:
$\left|\begin{{array}}{{ccc}} -12 & 0 & \alpha \\ 0 & 3 & -1 \\ 2 & 1 & -15 \end{{array}}\right| = 546$

Step 4: Calculate the determinant:
$-12(3(-15) - (-1)(1)) - 0 + \alpha(0(1) - 3(2)) = 546$
$-12(-45 + 1) - 6\alpha = 546$
$-12(-44) - 6\alpha = 546$
$528 - 6\alpha = 546$

Step 5: Solve for $\alpha$ :
$-6\alpha = 546 - 528$
$-6\alpha = 18$
$\alpha = -\frac{{18}}{{6}} = -3$

Why other options are wrong:
- Option A: This is incorrect because it has the wrong sign. The correct value is negative.
- Option B: This is incorrect because it is not the correct solution to the equation.
- Option D: This is incorrect because it is not the correct solution to the equation.

Key Concepts:
- Scalar triple product
- Volume of a parallelepiped
- Determinants

Memory Aids:
- The volume of a parallelepiped is given by $|[\mathbf{a} \mathbf{b} \mathbf{c}]|$ .

Quick Tips:
- Be careful with signs when calculating determinants.
- Double-check your arithmetic to avoid errors.

167 Easy

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Understanding +4 | -1

If

\mathbf{a}, \mathbf{b}

and

\mathbf{c}

are three non-coplanar vectors, that

\frac{\mathbf{a} \cdot \mathbf{b} \times \mathbf{c}}{\mathbf{c} \times \mathbf{a} \cdot \mathbf{b}}+\frac{\mathbf{b} \cdot \mathbf{a} \times \mathbf{c}}{\mathbf{c} \cdot \mathbf{a} \times \mathbf{b}}

is equal to

View answer and explanation

Correct option: A

0

Option A (0) is correct because the expression simplifies to

\frac{{[abc]}}{{[abc]}} - \frac{{[abc]}}{{[abc]}} = 1 - 1 = 0

.

Solution approach

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

Step-by-Step Solution:

Step 1: Rewrite the expression using scalar triple product notation: $\frac{\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})}{\mathbf{c} \cdot (\mathbf{a} \times \mathbf{b})} + \frac{\mathbf{b} \cdot (\mathbf{a} \times \mathbf{c})}{\mathbf{c} \cdot (\mathbf{a} \times \mathbf{b})}$ .

Step 2: Express the scalar triple products using brackets: $\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]}{[\mathbf{c} \mathbf{a} \mathbf{b}]} + \frac{[\mathbf{b} \mathbf{a} \mathbf{c}]}{[\mathbf{c} \mathbf{a} \mathbf{b}]}$ .

Step 3: Recall that $[\mathbf{c} \mathbf{a} \mathbf{b}] = [\mathbf{a} \mathbf{b} \mathbf{c}]$ and $[\mathbf{b} \mathbf{a} \mathbf{c}] = -[\mathbf{a} \mathbf{b} \mathbf{c}]$ .

Step 4: Substitute these identities into the expression: $\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]}{[\mathbf{a} \mathbf{b} \mathbf{c}]} + \frac{-[\mathbf{a} \mathbf{b} \mathbf{c}]}{[\mathbf{a} \mathbf{b} \mathbf{c}]}$ .

Step 5: Simplify: $1 - 1 = 0$ .

Why other options are wrong:
- Option B: This is incorrect because the expression simplifies to 0, not 2.
- Option C: This is incorrect because the expression simplifies to 0, not -2.
- Option D: This is incorrect because the expression simplifies to a specific value (0).

Key Concepts:
- Scalar triple product
- Properties of scalar triple product

Memory Aids:
- Remember that $[\mathbf{a} \mathbf{b} \mathbf{c}] = [\mathbf{b} \mathbf{c} \mathbf{a}] = [\mathbf{c} \mathbf{a} \mathbf{b}]$ and $[\mathbf{a} \mathbf{c} \mathbf{b}] = -[\mathbf{a} \mathbf{b} \mathbf{c}]$

168 Easy

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Application +4 | -1

The value of '

a

' so that the volume of the parallelopiped formed by

\hat{i}+a \hat{j}+\hat{k}, \hat{j}+a \hat{k}

and

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

becomes minimum, is

View answer and explanation

Correct option: C

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

Option C (

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

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

Solution approach

Correct Answer + Core Reason: $\frac{{1}}{{\sqrt{{3}}}}$ . This value minimizes the volume of the parallelepiped.

Step-by-Step Solution:

Step 1: Let $\vec{{\alpha}} = \hat{{i}} + a\hat{{j}} + \hat{{k}}$ , $\vec{{\beta}} = \hat{{j}} + a\hat{{k}}$ , and $\vec{{\gamma}} = a\hat{{i}} + \hat{{k}}$ .

Step 2: The volume $V$ of the parallelepiped is given by the absolute value of the scalar triple product: $V = |[\vec{{\alpha}} \,\vec{{\beta}} \,\vec{{\gamma}}]| = |\vec{{\alpha}} \cdot (\vec{{\beta}} \times \vec{{\gamma}})|$ .

Step 3: Calculate the scalar triple product: $\left[\begin{{array}}{{ccc}} \vec{{\alpha}} & \vec{{\beta}} & \vec{{\gamma}} \end{{array}}\right]$

169 Easy

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Analyzing +4 | -1

If

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

are non-coplanar vectors and

p, q

are real numbers, then the equality

\left[ {\begin{array}{llllllllllllllllllll} {3\,\vec u}&{p\,\vec v}&{p\,\overrightarrow w } \end{array}} \right] - \left[ {\begin{array}{llllllllllllllllllll} {p\,\vec v}&{\vec w}&{q\,\vec u} \end{array}} \right] - \left[ {\begin{array}{llllllllllllllllllll} {2\vec w}&{q\vec v}&{q\vec u} \end{array}} \right] = 0

holds for

View answer and explanation

Correct option: A

exactly one value of

(p, q)

Option A (exactly one value of (p, q)) is correct because the given equality holds only when p = 0 and q = 0, which is the only solution derived from the equation.

Solution approach

Correct Answer + Core Reason: exactly one value of $(p, q)$ . The given equation holds only for $p=0$ and $q=0$ .

Step-by-Step Solution:

Step 1: Given $\vec{u}, \vec{v}, \vec{w}$ are non-coplanar vectors and $\left[ {\begin{array}{llllllllllllllllllll} {3\,\vec u}&{p\,\vec v}&{p\,\overrightarrow w } \end{array}} \right] - \left[ {\begin{array}{llllllllllllllllllll} {p\,\vec v}&{\vec w}&{q\vec u} \end{array}} \right] - \left[ {\begin{array}{llllllllllllllllllll} {2\vec w}&{q\vec v}&{q\vec u} \end{array}} \right] = 0$ .

Step 2: Simplify the scalar triple products using their properties:
$3p^2 [\vec{u} \vec{v} \vec{w}] - pq [\vec{v} \vec{w} \vec{u}] - 2q^2 [\vec{w} \vec{v} \vec{u}] = 0$ .

Step 3: Since $[\begin{array}{lll}\vec{u} & \vec{v} & \vec{w}\end{array}] = [\begin{array}{lll}\vec{v} & \vec{w} & \vec{u}\end{array}] = -[\begin{array}{lll}\vec{w} & \vec{v} & \vec{u}\end{array}]$ , we have:
$3p^2 [\vec{u} \vec{v} \vec{w}] - pq [\vec{u} \vec{v} \vec{w}] + 2q^2 [\vec{u} \vec{v} \vec{w}] = 0$ .

Step 4: Factor out $[\begin{array}{lll}\vec{u} & \vec{v} & \vec{w}\end{array}]$ : $(3p^2 - pq + 2q^2) [\begin{array}{lll}\vec{u} & \vec{v} & \vec{w}\end{array}] = 0$ .

Step 5: Since $\vec{u}, \vec{v}, \vec{w}$ are non-coplanar, $[\begin{array}{lll}\vec{u} & \vec{v} & \vec{w}\end{array}] \neq 0$ . Therefore, $3p^2 - pq + 2q^2 = 0$ .

Step 6: Rewrite the equation: $3p^2 - pq + 2q^2 = 0 \Rightarrow p^2 - \frac{1}{3}pq + \frac{2}{3}q^2 = 0 \Rightarrow (p - \frac{q}{6})^2 + \frac{23}{36}q^2 = 0$ .

Step 7: The only real solution to this equation is $p - \frac{q}{6} = 0$ and $q = 0$ , which implies $p = 0$ and $q = 0$ .

Step 8: Hence, there is exactly one value of $(p, q)$ , which is $(0, 0)$ .

Why other options are wrong:
- Option B: exactly two values of $(p, q)$ : This is incorrect as there is only one solution.
- Option C: more than two but not all values of $(p, q)$ : This is incorrect as there is only one solution.
- Option D: all values of $(p, q)$ : This is incorrect as the equation is not satisfied for all values of $(p, q)$ .

Key Concepts:
- Scalar triple product
- Properties of scalar triple product

Memory Aids:
- Remember the properties of scalar triple product and how to manipulate them.

Quick Tips:
- Simplify the equation using the properties of scalar triple product.
- Solve the resulting equation to find the values of $p$ and $q$ .

170 Easy

170 of 185

Application +4 | -1

If

a

and

{b}

are two non-parallel vectors satisfying

|{a}|=|{b}|

, then the vector

({a}+{b}) \times({a} \times {b})

is parallel to

View answer and explanation

Correct option: B

{a}-{b}

Option B (

{a}-{b}

) is correct because the vector

({a}+{b}) \times({a} \times {b})

simplifies to a vector parallel to

{a}-{b}

.

Solution approach

Correct Answer + Core Reason: ${a}-{b}$ . The vector is parallel to ${a}-{b}$ after simplification.

Step-by-Step Solution:

Step 1: $({a} + {b}) \times ({a} \times {b}) = {a} \times ({a} \times {b}) + {b} \times ({a} \times {b})$ .

Step 2: Using the vector triple product formula, ${a} \times ({a} \times {b}) = ({a} \cdot {b}){a} - ({a} \cdot {a}){b}$ and ${b} \times ({a} \times {b}) = ({b} \cdot {b}){a} - ({b} \cdot {a}){b}$ .

Step 3: So, $({a} + {b}) \times ({a} \times {b}) = ({a} \cdot {b}){a} - |{a}|^2{b} + |{b}|^2{a} - ({b} \cdot {a}){b}$ .

Step 4: Since $|{a}| = |{b}|$ , we have $({a} + {b}) \times ({a} \times {b}) = ({a} \cdot {b}){a} - |{a}|^2{b} + |{a}|^2{a} - ({a} \cdot {b}){b} = ({a} \cdot {b} + |{a}|^2)({a} - {b})$ .

Step 5: Therefore, $({a} + {b}) \times ({a} \times {b})$ is parallel to ${a} - {b}$ .

Why other options are wrong:
- Option A: ${a}$ is incorrect because the vector is not parallel to ${a}$ .
- Option C: ${a}+{b}$ is incorrect because the vector is not parallel to ${a}+{b}$ .
- Option D: ${b}$ is incorrect because the vector is not parallel to ${b}$ .

Key Concepts:
- Vector triple product
- Parallel vectors

Memory Aids:
- Remember the vector triple product formula.

Quick Tips:
- Always use the vector triple product formula to simplify the expression.

171 Easy

171 of 185

Application +4 | -1

If the volume of parallelopiped formed by the vectors a, b, c as coterminous edges is 27 cu units, then the volume of the parallelopiped have

View answer and explanation

Correct option: C

81

Option C (81) is correct because the volume of the new parallelopiped is 3 times the volume of the original parallelopiped.

Solution approach

Correct Answer + Core Reason: 81. The volume of the new parallelopiped is 3 times the original volume.

Step-by-Step Solution:

Step 1: Given $|[a \, b \, c]| = 27$ cubic units.

Step 2: Let $\alpha = a + 2b - c, \beta = a - b, \gamma = a - b - c$

Step 3: $[\alpha \, \beta \, \gamma] = \left| {\begin{{array}}{{{*{{20}}{c}}}} 1&2&{ - 1} \\ 1&{ - 1}&0 \\ 1&{ - 1}&{ - 1} \end{{array}}} \right|[a \, b \, c]$

Step 4: Evaluate the determinant: $[1(1 - 0) - 2( - 1 - 0) + ( - 1)( - 1 + 1)][a \, b \, c] = [1 + 2 + 0][a \, b \, c] = 3[a \, b \, c]$

Step 5: Required volume $= |[\alpha \, \beta \, \gamma]| = |3[abc]$

172 Easy

172 of 185

Application +4 | -1

If

\vec{a}, \vec{b}

and

\vec{c}

are unit coplanar vectors, then the scalar triple product

\left[\begin{array}{ll}2 \vec{a}-\vec{b} & 2 \vec{b}-\vec{c} \quad 2 \vec{c}-\vec{a}\end{array}\right]=

View answer and explanation

Correct option: A

0

Option A (0) is correct because if the vectors are coplanar, their scalar triple product is zero, and the linear combination also results in zero.

Solution approach

Correct Answer + Core Reason: 0. Since $\vec{a}, \vec{b}, \vec{c}$ are coplanar, any linear combination of them will also have a scalar triple product of 0.

Step-by-Step Solution:

Step 1: State the coplanarity condition: Since $\vec{a}, \vec{b}, \vec{c}$ are coplanar vectors, $\left[\begin{array}{ll} {\vec a}&{\vec b}&{\vec c} \end{array}\right] = 0$ .

Step 2: Define the new vectors: $\vec{\alpha}=2 \vec{a}-\vec{b}, \vec{\beta}=2 \vec{b}-\vec{c}$ and $\vec{\gamma}=2 \vec{c}-\vec{a}$ .

Step 3: Express the new vectors in terms of $\vec{a}, \vec{b}, \vec{c}$ : $\vec{\alpha}=2 \vec{a}-\vec{b}+0\vec{c}, \vec{\beta}=0 \vec{a}+2\vec{b}-\vec{c}$ and $\vec{\gamma}=-\vec{a}+0 \vec{b}+2\vec{c}$ .

Step 4: Calculate the scalar triple product: $\left[\begin{array}{ll} {\vec \alpha }&{\vec \beta }&{\vec \gamma } \end{array}\right] = \left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr} 2&{ - 1}&0 \\ 0&2&{ - 1} \\ { - 1}&0&2 \end{array}} \right|\left[\begin{array}{ll} {\vec a}&{\vec b}&{\vec c} \end{array}\right]$

Step 5: Evaluate the determinant: $\left| {\begin{array}{rrrrrrrrrrrrrrrrrrrr} 2&{ - 1}&0 \\ 0&2&{ - 1} \\ { - 1}&0&2 \end{array}} \right| = 2(4-0) - (-1)(0-1) + 0 = 8 - 1 = 7$ .

Step 6: Final Answer: $\left[\begin{array}{ll} {\vec \alpha }&{\vec \beta }&{\vec \gamma } \end{array}\right] = 7\left[\begin{array}{ll} {\vec a}&{\vec b}&{\vec c} \end{array}\right] = 7 \times 0 = 0$ .

Why other options are wrong:
- Option B: This is incorrect because the scalar triple product is zero due to coplanarity.
- Option C: This is incorrect because the scalar triple product is zero due to coplanarity.
- Option D: This is incorrect because the scalar triple product is zero due to coplanarity.

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

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

Quick Tips:
- If vectors are coplanar, the scalar triple product is always zero.

173 Easy

173 of 185

Analyzing +4 | -1

Let

\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{k}}, \mathbf{b}=x \hat{\mathbf{i}}+\hat{\mathbf{j}}+(1-x) \hat{\mathbf{k}}

and

\mathbf{c}=y \hat{\mathbf{i}}+x \hat{\mathbf{j}}+(1+x-y) \hat{\mathbf{k}}

. Then,

[\mathbf{a\,} \mathbf{b\,} \mathbf{c}]

depends on

View answer and explanation

Correct option: A

Neither

x

nor

y

Option A (Neither

x

nor

y

) is correct because the scalar triple product simplifies to 1, which is independent of both

x

and

y

.

Solution approach

Correct Answer + Core Reason: Neither $x$ nor $y$ . The scalar triple product simplifies to a constant value.

Step-by-Step Solution:

Step 1: Given $\mathbf{a}=\hat{\mathbf{i}}-\hat{\mathbf{k}}$ , $\mathbf{b}=x \hat{\mathbf{i}}+\hat{\mathbf{j}}+(1-x) \hat{\mathbf{k}}$ , and $\mathbf{c}=y \hat{\mathbf{i}}+x \hat{\mathbf{j}}+(1+x-y) \hat{\mathbf{k}}$ .

Step 2: Calculate the scalar triple product $[\mathbf{a} \, \mathbf{b} \, \mathbf{c}] = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ .
$\left|\begin{{array}}{{ccc}}1 & 0 & -1 \x & 1 & 1-x \y & x & 1+x-y\end{{array}}\right|$

Step 3: Expand the determinant:
$1 \cdot \left|\begin{{array}}{{cc}}1 & 1-x \x & 1+x-y\end{{array}}\right| - 0 + (-1) \cdot \left|\begin{{array}}{{cc}}x & 1 \y & x\end{{array}}\right| = (1+x-y - x(1-x)) - (x^2 - y) = 1+x-y - x + x^2 - x^2 + y = 1$ .

Step 4: The scalar triple product is equal to 1, which is independent of both $x$ and $y$ .

Final Answer: Neither $x$ nor $y$ .

Why other options are wrong:
- Option B: Incorrect because the scalar triple product does not depend on both $x$ and $y$ .
- Option C: Incorrect because the scalar triple product does not depend only on $x$ .
- Option D: Incorrect because the scalar triple product does not depend only on $y$ .

Key Concepts:
- Scalar triple product
- Determinants

Memory Aids:
- Remember how to calculate the determinant of a 3x3 matrix.

Quick Tips:
- Calculate the determinant carefully.
- If the result is a constant, it is independent of the variables.

174 Easy

174 of 185

Analyzing +4 | -1

Let

{x_0}

be the point of local maximum of

f(x) = a.(b \times c),

where

a = x\widehat i - 2\widehat j + 3\widehat k,b = - 2\widehat i + x\widehat j - \widehat k\mathrm{ }\,\mathrm{and }\,\,c = 7\widehat i - 2\widehat j + x\widehat k

Then the value

a.\,b + b.\,c + c.\,a

at

x = {x_0}

is

View answer and explanation

Correct option: C

-22

Option C (-22) is correct because calculating the scalar triple product, finding the local maximum, and substituting it into the expression yields -22.

Solution approach

Correct Answer + Core Reason: -22. The value is obtained by finding the local maximum of f(x) and substituting it into the expression a.b + b.c + c.a.

Step-by-Step Solution:

Step 1: Calculate $f(x) = a.(b \times c) = \begin{vmatrix} x & -2 & 3 \\ -2 & x & -1 \\ 7 & -2 & x \end{vmatrix}$ .

Step 2: Expand the determinant: $f(x) = x({x^2} - 2) + 2(-2x + 7) + 3(4 - 7x) = {x^3} - 2x - 4x + 14 + 12 - 21x = {x^3} - 27x + 26$ .

Step 3: Find the derivative: $f'(x) = 3{x^2} - 27$ .

Step 4: Set $f'(x) = 0$ to find critical points: $3{x^2} - 27 = 0 \Rightarrow {x^2} = 9 \Rightarrow x = \pm 3$ .

Step 5: Find the second derivative: $f''(x) = 6x$ .

Step 6: Determine the local maximum: $f''(-3) = -18 < 0$ , so $x_0 = -3$ is a local maximum.

Step 7: Calculate $a.b + b.c + c.a = (x)(-2) + (-2)(x) + (3)(-1) + (-2)(7) + (x)(-2) + (-1)(x) + (7)(x) + (-2)(-2) + (x)(3) = -2x - 2x - 3 - 14 - 2x - x + 7x + 4 + 3x = 3x - 13$ .

Step 8: Substitute $x_0 = -3$ : $3(-3) - 13 = -9 - 13 = -22$ .

Why other options are wrong:
- Option A (14): This is incorrect because it doesn't account for the correct value of x at the local maximum and the correct calculation of the expression.
- Option B (-4): This is incorrect because it doesn't account for the correct value of x at the local maximum and the correct calculation of the expression.
- Option D (-30): This is incorrect because it doesn't account for the correct value of x at the local maximum and the correct calculation of the expression.

Key Concepts:
- Scalar triple product
- Local maxima and minima
- Derivatives
- Dot product of vectors

Memory Aids:
- Remember the determinant formula for scalar triple product.
- Recall the conditions for local maxima and minima using first and second derivatives.

Quick Tips:
- Always double-check your calculations, especially when dealing with multiple terms.
- Remember to find the correct x value for the local maximum before substituting into the expression.

175 Easy

175 of 185

Analyzing +4 | -1

If the lines

\frac{x-1}{2}=\frac{y-2}{x_1}=\frac{z-3}{x_2}

and

\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}

lie in the same plane, then for the equation

x_1 t^2+\left(x_2+2\right) t+a=0

, sum of roots of equation is

View answer and explanation

Correct option: D

-2

Option D (

-2

) is correct because the condition for the lines to lie in the same plane leads to a relationship between

x_1

and

x_2

, allowing us to find the sum of the roots of the given quadratic equation.

Solution approach

Correct Answer + Core Reason: $-2$ . The condition for coplanarity gives a relation between $x_1$ and $x_2$ , which leads to the sum of roots.

Step-by-Step Solution:

Step 1: The given lines are $\frac{{x-1}}{{2}} = \frac{{y-2}}{{x_1}} = \frac{{z-3}}{{x_2}}$ and $\frac{{x-2}}{{3}} = \frac{{y-3}}{{4}} = \frac{{z-4}}{{5}}$ .

Step 2: The direction vectors of the lines are $\vec{{b}} = (2, x_1, x_2)$ and $\vec{{d}} = (3, 4, 5)$ .

Step 3: The vector joining a point on each line is $\vec{{a}} - \vec{{c}} = (1-2, 2-3, 3-4) = (-1, -1, -1)$ .

Step 4: For the lines to be coplanar, the scalar triple product of these vectors must be zero: $[(\vec{{a}} - \vec{{c}}) \vec{{b}} \vec{{d}}] = 0$ .\nThis means $\begin{{vmatrix}} -1 & -1 & -1 \\ 2 & x_1 & x_2 \\ 3 & 4 & 5 \end{{vmatrix}} = 0$ .

Step 5: Expanding the determinant, we get $-1(5x_1 - 4x_2) + 1(10 - 3x_2) - 1(8 - 3x_1) = 0$ .\nSimplifying, $-5x_1 + 4x_2 + 10 - 3x_2 - 8 + 3x_1 = 0$ , which gives $-2x_1 + x_2 + 2 = 0$ , or $x_2 = 2x_1 - 2$ .

Step 6: The given equation is $x_1 t^2 + (x_2 + 2)t + a = 0$ . Substituting $x_2 = 2x_1 - 2$ , we get $x_1 t^2 + (2x_1 - 2 + 2)t + a = 0$ , which simplifies to $x_1 t^2 + 2x_1 t + a = 0$ .

Step 7: Dividing by $x_1$ , we get $t^2 + 2t + \frac{{a}}{{x_1}} = 0$ .

Step 8: The sum of the roots of this equation is given by $-\frac{{\text{{coefficient of }} t}}{{\text{{coefficient of }} t^2}} = -\frac{{2}}{{1}} = -2$ .

Final Answer: $-2$

Why other options are wrong:
- Option A (1): Incorrect calculation of the sum of the roots or the coplanarity condition.
- Option B (2): Incorrect calculation of the sum of the roots or the coplanarity condition.
- Option C (-1): Incorrect calculation of the sum of the roots or the coplanarity condition.

Key Concepts:
- Coplanarity of Lines
- Scalar Triple Product
- Sum of Roots of a Quadratic Equation

Memory Aids:
- Remember the determinant condition for coplanarity.
- Sum of roots formula: $-\frac{{b}}{{a}}$ for $ax^2 + bx + c = 0$ .

Quick Tips:
- Always write down the condition for coplanarity in terms of a determinant.
- Simplify the resulting equation before calculating the sum of roots.

176 Easy

176 of 185

Understanding +4 | -1

If

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

are three vectors such that

\mathbf{a}=\mathbf{b}+\mathbf{c}

and the angle between

\mathbf{b}

and

\mathbf{c}

is

\pi / 2

, then

View answer and explanation

Correct option: A

a^2=b^2+c^2

Option A (

a^2=b^2+c^2

) is correct because it is derived from the Pythagorean theorem, given that

\mathbf{a}=\mathbf{b}+ \mathbf{c}

and the angle between

\mathbf{b}

and

\mathbf{c}

is

\pi / 2

.

Solution approach

Correct Answer + Core Reason: $a^2=b^2+c^2$ . This follows from the Pythagorean theorem since the angle between b and c is 90 degrees.

Step-by-Step Solution:

Step 1: Given $\mathbf{a} = \mathbf{b} + \mathbf{c}$ .

Step 2: Square both sides: $|\mathbf{a}|^2 = |\mathbf{b} + \mathbf{c}|^2$ .

Step 3: Expand the right side: $a^2 = (\mathbf{b} + \mathbf{c}) \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{b} \cdot \mathbf{b} + 2(\mathbf{b} \cdot \mathbf{c}) + \mathbf{c} \cdot \mathbf{c}$ .

Step 4: Since the angle between $\mathbf{b}$ and $\mathbf{c}$ is $\pi/2$ , $\mathbf{b} \cdot \mathbf{c} = |\mathbf{b}||\mathbf{c}|\cos{(\pi/2)} = 0$ .

Step 5: Therefore, $a^2 = b^2 + 2(0) + c^2 = b^2 + c^2$ .

Why other options are wrong:
- Option B: This is incorrect because it implies $b^2 = c^2 + a^2$ , which is not true given the conditions.
- Option C: This is incorrect because it implies $c^2 = a^2 + b^2$ , which is not true given the conditions.
- Option D: This is incorrect because it implies $2a^2 - b^2 = c^2$ , which is not true given the conditions.

Key Concepts:
- Vector addition
- Dot product of vectors
- Pythagorean theorem

Memory Aids:
- Remember that $\mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos{\theta}$ .

Quick Tips:
- When squaring a vector sum, remember to use the dot product to expand it.

177 Easy

177 of 185

Analyzing +4 | -1

If

\mathbf{a}, \mathbf{b}

and

\mathbf{c}

are any three vectors forming a linearly independent system, then

\forall \theta \in R

\left[\mathbf{a} \cos \theta+\mathbf{b} \sin \theta+\mathbf{c o s} 2 \theta, \mathbf{a} \cos \left(\frac{2 \pi}{3}+\theta\right)\right.

+ b\sin \left( {\frac{{2\pi }}{3} + \theta } \right) + c\cos 2\left( {\frac{{2\pi }}{3} + \theta } \right){\text{, }}

acos\left. {\left( {\theta - \frac{{2\pi }}{3}} \right) + b\sin \left( {\theta - \frac{{2\pi }}{3}} \right) + c\cos 2\left( {\theta - \frac{{2\pi }}{3}} \right)} \right]

equals

View answer and explanation

Correct option: D

None of the above

Option D (

\mathbf{a} \times(\mathbf{b} \times \mathbf{c})+\frac{(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{b} \times \mathbf{c})}{|b|^2} \mathbf{b}

) is correct because it represents the component of

\mathbf{a} \times(\mathbf{b} \times \mathbf{c})

perpendicular to

\mathbf{b}

.

Solution approach

Correct Answer + Core Reason: $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})+\frac{(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{b} \times \mathbf{c})}{|b|^2} \mathbf{b}$ . The component of a vector perpendicular to another vector is found by subtracting the projection of the vector along the other vector from the original vector.

Step-by-Step Solution:

Step 1: We want to find the component of $\mathbf{a} \times (\mathbf{b} \times \mathbf{c})$ perpendicular to $\mathbf{b}$ .

Step 2: First, find the component of $\mathbf{a} \times (\mathbf{b} \times \mathbf{c})$ along $\mathbf{b}$ . This is given by the projection of $\mathbf{a} \times (\mathbf{b} \times \mathbf{c})$ onto $\mathbf{b}$ :
$\text{proj}_{\mathbf{b}} (\mathbf{a} \times (\mathbf{b} \times \mathbf{c})) = \frac{(\mathbf{a} \times (\mathbf{b} \times \mathbf{c})) \cdot \mathbf{b}}{|\mathbf{b}|^2} \mathbf{b}$ .

Step 3: We know that $\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$ .

Step 4: So, $(\mathbf{a} \times (\mathbf{b} \times \mathbf{c})) \cdot \mathbf{b} = ((\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}) \cdot \mathbf{b} = (\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{b}) - (\mathbf{a} \cdot \mathbf{b})(\mathbf{c} \cdot \mathbf{b})$ .

Step 5: The component of $\mathbf{a} \times (\mathbf{b} \times \mathbf{c})$ along $\mathbf{b}$ is $\frac{(\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{b}) - (\mathbf{a} \cdot \mathbf{b})(\mathbf{c} \cdot \mathbf{b})}{|\mathbf{b}|^2} \mathbf{b}$ .

Step 6: The component of $\mathbf{a} \times (\mathbf{b} \times \mathbf{c})$ perpendicular to $\mathbf{b}$ is obtained by subtracting the component along $\mathbf{b}$ from the original vector:
$\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) - \frac{(\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{b}) - (\mathbf{a} \cdot \mathbf{b})(\mathbf{c} \cdot \mathbf{b})}{|\mathbf{b}|^2} \mathbf{b} = \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) + \frac{(\mathbf{a} \cdot \mathbf{b})(\mathbf{c} \cdot \mathbf{b}) - (\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{b})}{|\mathbf{b}|^2} \mathbf{b}$ .

Step 7: Using the vector identity $(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{b})(\mathbf{c} \cdot \mathbf{b}) - (\mathbf{a} \cdot \mathbf{c})(\mathbf{b} \cdot \mathbf{b})$ , we can rewrite the expression as:
$\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) + \frac{(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{b} \times \mathbf{c})}{|\mathbf{b}|^2} \mathbf{b}$ .

Why other options are wrong:
- **Option A ( $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})+\frac{(\mathbf{a} \times \mathbf{b}) \cdot(\mathbf{c} \times \mathbf{a})}{|b|^2} \mathbf{b}$ ): Incorrect because the dot product in the numerator is wrong.
- Option B ( $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})+\frac{(\mathbf{a} \times \mathbf{c}) \cdot(\mathbf{a} \times \mathbf{b})}{|b|^2} \mathbf{b}$ ): Incorrect because the dot product in the numerator is wrong.
- Option C ( $\mathbf{a} \times(\mathbf{b} \times \mathbf{c})+\frac{(\mathbf{b} \times \mathbf{c}) \cdot(\mathbf{b} \times \mathbf{a})}{|b|^2} \mathbf{b}$ ): Incorrect because the dot product in the numerator is wrong.

Key Concepts:
- Vector triple product
- Projection of a vector
- Component of a vector perpendicular to another vector

Memory Aids:**
- Remember the vector triple product identity: $\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$ .

Quick Tips:
- Use the vector triple product identity to simplify expressions.
- Remember the formula for the projection of one vector onto another.

178 Easy

178 of 185

Application +4 | -1

Let

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

are three vectors along the adjacent edges of a tetrahedron, if

|\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=2

and

\mathbf{a} \cdot \mathbf{b}=\mathbf{b} \cdot \mathbf{c}=\mathbf{c} \cdot \mathbf{a}=2

, then volume of tetrahedron is

View answer and explanation

Correct option: D

\frac{2 \sqrt{2}}{3}

Option D (

\frac{2 \sqrt{2}}{3}

) is correct because it is the volume calculated using the scalar triple product of the vectors.

Solution approach

Correct Answer + Core Reason: $\frac{2 \sqrt{2}}{3}$ . The volume of the tetrahedron is calculated using the scalar triple product of the vectors representing its adjacent edges.

Step-by-Step Solution:

Step 1: The volume of a tetrahedron is given by $V = \frac{1}{6} |[\mathbf{a} \mathbf{b} \mathbf{c}]|$ .

Step 2: We know that $[\mathbf{a} \mathbf{b} \mathbf{c}]^2 = \begin{vmatrix} \mathbf{a} \cdot \mathbf{a} & \mathbf{a} \cdot \mathbf{b} & \mathbf{a} \cdot \mathbf{c} \\ \mathbf{b} \cdot \mathbf{a} & \mathbf{b} \cdot \mathbf{b} & \mathbf{b} \cdot \mathbf{c} \\ \mathbf{c} \cdot \mathbf{a} & \mathbf{c} \cdot \mathbf{b} & \mathbf{c} \cdot \mathbf{c} \end{vmatrix}$ .

Step 3: Given $|\mathbf{a}|=|\mathbf{b}|=|\mathbf{c}|=2$ and $\mathbf{a} \cdot \mathbf{b}=\mathbf{b} \cdot \mathbf{c}=\mathbf{c} \cdot \mathbf{a}=2$ , we have $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2 = 4$ , $\mathbf{b} \cdot \mathbf{b} = |\mathbf{b}|^2 = 4$ , and $\mathbf{c} \cdot \mathbf{c} = |\mathbf{c}|^2 = 4$ .

Step 4: Substitute the values into the determinant: $[\mathbf{a} \mathbf{b} \mathbf{c}]^2 = \begin{vmatrix} 4 & 2 & 2 \\ 2 & 4 & 2 \\ 2 & 2 & 4 \end{vmatrix} = 4(16-4) - 2(8-4) + 2(4-8) = 4(12) - 2(4) + 2(-4) = 48 - 8 - 8 = 32$ .

Step 5: Therefore, $|[\mathbf{a} \mathbf{b} \mathbf{c}]| = \sqrt{32} = 4\sqrt{2}$ .

Step 6: The volume of the tetrahedron is $V = \frac{1}{6} |[\mathbf{a} \mathbf{b} \mathbf{c}]| = \frac{1}{6} (4\sqrt{2}) = \frac{2\sqrt{2}}{3}$ .

Why other options are wrong:
- **Option A ( $\frac{1}{\sqrt{2}}$ ): This is incorrect because it is not the correct volume calculated using the scalar triple product.
- Option B ( $\frac{2}{\sqrt{3}}$ ): This is incorrect because it is not the correct volume calculated using the scalar triple product.
- Option C ( $\frac{\sqrt{3}}{2}$ ): This is incorrect because it is not the correct volume calculated using the scalar triple product.

Key Concepts:
- Volume of a tetrahedron
- Scalar triple product
- Determinant of a matrix

Memory Aids:**
- Volume of tetrahedron: $V = \frac{1}{6} |[\mathbf{a} \mathbf{b} \mathbf{c}]|$

Quick Tips:
- Remember the formula for the volume of a tetrahedron in terms of the scalar triple product.
- Be careful when calculating the determinant of the matrix.

179 Easy

179 of 185

Application +4 | -1

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

equals

View answer and explanation

Correct option: B

\left[ {\vec a\vec b\vec c} \right]\left( {\vec a\cdot\vec d} \right)

Option B (

\left[ {\vec a\vec b\vec c} \right]\left( {\vec a\cdot\vec d} \right)

) is correct because the triple vector product and scalar triple product properties are correctly applied.

Solution approach

Correct Answer + Core Reason: $\left[ {\vec a\vec b\vec c} \right]\left( {\vec a\cdot\vec d} \right)$ . This result is obtained by applying the vector triple product and scalar triple product identities.

Step-by-Step Solution:

Step 1: Start with the given expression: $\left( {\vec a \times \vec b} \right) \times \left( {\vec a \times \vec c} \right)\cdot\vec d$

Step 2: Apply the vector triple product identity: $\vec A \times (\vec B \times \vec C) = (\vec A \cdot \vec C)\vec B - (\vec A \cdot \vec B)\vec C$ . Let $\vec A = (\vec a \times \vec b)$ , $\vec B = \vec a$ , and $\vec C = \vec c$ . Then, $(\vec a \times \vec b) \times (\vec a \times \vec c) = [(\vec a \times \vec b) \cdot \vec c]\vec a - [(\vec a \times \vec b) \cdot \vec a]\vec c$

Step 3: Simplify the expression. Note that $(\vec a \times \vec b) \cdot \vec a = 0$ because the scalar triple product of vectors where two are the same is zero. Thus, the expression becomes: $[(\vec a \times \vec b) \cdot \vec c]\vec a - 0 = [\vec a \vec b \vec c]\vec a$

Step 4: Take the dot product with $\vec d$ : $[\vec a \vec b \vec c]\vec a \cdot \vec d = [\vec a \vec b \vec c](\vec a \cdot \vec d)$

Step 5: Final Answer: $\left[ {\vec a\vec b\vec c} \right]\left( {\vec a\cdot\vec d} \right)$

Why other options are wrong:
- Option A: This is incorrect because it has $(\vec b \cdot \vec d)$ instead of $(\vec a \cdot \vec d)$ .
- Option C: This is incorrect because it has $(\vec c \cdot \vec d)$ instead of $(\vec a \cdot \vec d)$ .
- Option D: This is incorrect because the correct simplification leads to option B.

Key Concepts:
- Vector triple product identity
- Scalar triple product
- Properties of dot and cross products

Memory Aids:
- Remember the vector triple product formula: $\vec A \times (\vec B \times \vec C) = (\vec A \cdot \vec C)\vec B - (\vec A \cdot \vec B)\vec C$
- Scalar triple product: $[\,\vec a \,\vec b \,\vec c] = \vec a \cdot (\vec b \times \vec c)$

Quick Tips:
- Always remember the vector triple product identity.
- Simplify step by step to avoid errors.

180 Easy

180 of 185

Application +4 | -1

If

\mathbf{a}

and

\mathbf{b}

are orthogonal unit vectors, then for any non-zero vector

\mathbf{r}

, the vector

(\mathbf{r} \times \mathbf{a})

is equal to

View answer and explanation

Correct option: C

[\mathbf{r} \hat{\mathbf{a}} \hat{\mathbf{b}}] \hat{\mathbf{b}}+(\mathbf{r} \cdot \hat{\mathbf{b}})(\hat{\mathbf{b}} \times \hat{\mathbf{a}})

Option D (

[\mathbf{r} \hat{\mathbf{a}} \hat{\mathbf{b}}] \hat{\mathbf{b}}+(\mathbf{r} \cdot \hat{\mathbf{a}})(\hat{\mathbf{a}} \times \hat{\mathbf{b}})

) is correct because it correctly expresses the vector

(\mathbf{r} \times \mathbf{a})

in terms of the given orthogonal unit vectors and their scalar triple product.

Solution approach

Correct Answer + Core Reason: $[\mathbf{r} \hat{\mathbf{a}} \hat{\mathbf{b}}] \hat{\mathbf{b}}+(\mathbf{r} \cdot \hat{\mathbf{b}})(\hat{\mathbf{b}} \times \hat{\mathbf{a}})$ . The vector $(\mathbf{r} \times \mathbf{a})$ is expressed in terms of the given orthogonal unit vectors.

Step-by-Step Solution:

Step 1: Recognize that since $\mathbf{a}$ and $\mathbf{b}$ are orthogonal unit vectors, $\mathbf{a}, \mathbf{b}$ and $\mathbf{a} \times \mathbf{b}$ are non-coplanar.

Step 2: Express the vector $(\mathbf{r} \times \mathbf{a})$ as a linear combination of $\mathbf{a}, \mathbf{b}$ and $(\mathbf{a} \times \mathbf{b})$ : $\mathbf{r} \times \mathbf{a} = x \mathbf{a}+y \mathbf{b}+z(\mathbf{a} \times \mathbf{b})$

Step 3: Take the dot product of the equation with $\mathbf{a}$ : $(\mathbf{r} \times \mathbf{a}) \cdot \mathbf{a} = x \mathbf{a} \cdot \mathbf{a}+y \mathbf{b} \cdot \mathbf{a}+z(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{a}$ . Since $(\mathbf{r} \times \mathbf{a}) \cdot \mathbf{a} = 0$ , $\mathbf{a} \cdot \mathbf{a} = 1$ , and $\mathbf{b} \cdot \mathbf{a} = 0$ , we get $0 = x$ , so $x = 0$ .

Step 4: Take the dot product of the equation with $\mathbf{b}$ : $(\mathbf{r} \times \mathbf{a}) \cdot \mathbf{b} = x \mathbf{a} \cdot \mathbf{b}+y \mathbf{b} \cdot \mathbf{b}+z(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b}$ . Since $(\mathbf{r} \times \mathbf{a}) \cdot \mathbf{b} = [\mathbf{r} \mathbf{a} \mathbf{b}]$ , $\mathbf{b} \cdot \mathbf{b} = 1$ , and $(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{b} = 0$ , we get $[\mathbf{r} \mathbf{a} \mathbf{b}] = y$ , so $y = [\mathbf{r} \mathbf{a} \mathbf{b}]$ .

Step 5: Take the dot product of the equation with $(\mathbf{a} \times \mathbf{b})$ : $(\mathbf{r} \times \mathbf{a}) \cdot (\mathbf{a} \times \mathbf{b}) = x \mathbf{a} \cdot (\mathbf{a} \times \mathbf{b})+y \mathbf{b} \cdot (\mathbf{a} \times \mathbf{b})+z(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{a} \times \mathbf{b})$ . Since $(\mathbf{r} \times \mathbf{a}) \cdot (\mathbf{a} \times \mathbf{b}) = (\mathbf{r} \cdot \mathbf{a})(\mathbf{a} \cdot \mathbf{b}) - (\mathbf{r} \cdot \mathbf{b})(\mathbf{a} \cdot \mathbf{a}) = -(\mathbf{r} \cdot \mathbf{b})$ , and $(\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{a} \times \mathbf{b}) = 1$ , we get $-(\mathbf{r} \cdot \mathbf{b}) = z$ , so $z = -(\mathbf{r} \cdot \mathbf{b})$ .

Step 6: Substitute the values of x, y, and z back into the equation: $\mathbf{r} \times \mathbf{a} = 0 \mathbf{a}+[\mathbf{r} \mathbf{a} \mathbf{b}] \mathbf{b}-(\mathbf{r} \cdot \mathbf{b})(\mathbf{a} \times \mathbf{b}) = [\mathbf{r} \mathbf{a} \mathbf{b}] \mathbf{b}+(\mathbf{r} \cdot \mathbf{b})(\mathbf{b} \times \mathbf{a})$

Why other options are wrong:
- Option A: This option is incorrect because it does not correctly express the vector in terms of the given orthogonal unit vectors.
- Option B: This option is incorrect because it does not correctly express the vector in terms of the given orthogonal unit vectors.
- Option C: This option is incorrect because it does not correctly express the vector in terms of the given orthogonal unit vectors.

Key Concepts:
- Orthogonal unit vectors
- Scalar triple product
- Vector decomposition

Memory Aids:
- Any vector can be expressed as a linear combination of three non-coplanar vectors.

Quick Tips:
- Remember the properties of dot and cross products with orthogonal vectors.

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