SCALAR TRIPLE PRODUCT MCQs for JEE MATHEMATICS — Easy, Page 4

91 Easy

91 of 185

Application +4 | -1

Volume of the tetrahedron with vertices

P(-1,2,0), Q(2,1,-3), R(1,0,1)

and

S(3,-2,3)

is

View answer and explanation

Correct option: B

2 / 3

Option B (2/3) is correct because the volume of the tetrahedron is calculated using the scalar triple product of the vectors formed by the vertices.

Solution approach

Correct Answer + Core Reason: 2/3. The volume is calculated using the scalar triple product of vectors formed by the vertices of the tetrahedron.

Step-by-Step Solution:

Step 1: Define the vectors PQ, PR, and PS using the given coordinates:
$PQ = Q - P = (2 - (-1), 1 - 2, -3 - 0) = (3, -1, -3)$
$PR = R - P = (1 - (-1), 0 - 2, 1 - 0) = (2, -2, 1)$
$PS = S - P = (3 - (-1), -2 - 2, 3 - 0) = (4, -4, 3)$

Step 2: Calculate the scalar triple product: $PQ \cdot (PR \times PS)$ . This can be found by computing the determinant of the matrix formed by the components of the vectors:
$V = \frac{1}{6} |PQ \cdot (PR \times PS)| = \frac{1}{6} \left| \begin{array}{ccc} 3 & -1 & -3 \\ 2 & -2 & 1 \\ 4 & -4 & 3 \end{array} \right|$

Step 3: Evaluate the determinant:
$V = \frac{1}{6} |3((-2)(3) - (1)(-4)) - (-1)((2)(3) - (1)(4)) + (-3)((2)(-4) - (-2)(4))|$
$V = \frac{1}{6} |3(-6 + 4) + 1(6 - 4) - 3(-8 + 8)|$
$V = \frac{1}{6} |3(-2) + 1(2) - 3(0)|$
$V = \frac{1}{6} |-6 + 2 - 0|$
$V = \frac{1}{6} |-4| = \frac{4}{6} = \frac{2}{3}$

Step 4: The volume of the tetrahedron is $\frac{2}{3}$ .

Why other options are wrong:
- Option A (1/3): This is incorrect because the determinant was not calculated correctly, or the factor of 1/6 was not applied correctly.
- Option C (1/4): This is incorrect because the determinant was not calculated correctly, or the factor of 1/6 was not applied correctly.
- Option D (3/4): This is incorrect because the determinant was not calculated correctly, or the factor of 1/6 was not applied correctly.

Key Concepts:
- Volume of a tetrahedron
- Scalar triple product
- Determinant of a 3x3 matrix
- Vector subtraction

Memory Aids:
- Remember the formula for the volume of a tetrahedron using the scalar triple product: $V = \frac{1}{6} |a \cdot (b \times c)|$
- Determinant calculation: Cross product components.

Quick Tips:
- Always remember to take the absolute value of the scalar triple product and multiply by 1/6 to get the volume.
- Double-check the signs when calculating the determinant.

92 Easy

92 of 185

Application +4 | -1

The volume of the tetrahedron whose vertices are the points

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

and

2 \hat{i}+3 \hat{j}+\lambda \hat{k}

is

1 / 6

units. Then, the values of

\lambda

View answer and explanation

Correct option: D

is any real value

Option D (is any real value) is correct because the volume of the tetrahedron is independent of

\lambda

, as the scalar triple product simplifies to a constant value regardless of

\lambda

.

Solution approach

Correct Answer + Core Reason: is any real value. The volume of the tetrahedron is independent of $\lambda$ .

Step-by-Step Solution:

Step 1: Let the vertices of the tetrahedron be A( $\hat{i}$ ), B( $\hat{i}+\hat{j}$ ), C( $\hat{i}+\hat{j}+\hat{k}$ ), and D( $2\hat{i}+3\hat{j}+\lambda \hat{k}$ ).

Step 2: Find the vectors $\overrightarrow{AB}$ , $\overrightarrow{AC}$ , and $\overrightarrow{AD}$ .
$\overrightarrow{AB} = B - A = (\hat{i}+\hat{j}) - \hat{i} = \hat{j}$
$\overrightarrow{AC} = C - A = (\hat{i}+\hat{j}+\hat{k}) - \hat{i} = \hat{j} + \hat{k}$
$\overrightarrow{AD} = D - A = (2\hat{i}+3\hat{j}+\lambda \hat{k}) - \hat{i} = \hat{i} + 3\hat{j} + \lambda \hat{k}$

Step 3: The volume of the tetrahedron is given by $\frac{1}{6} |[\overrightarrow{AB} \,\, \overrightarrow{AC} \,\, \overrightarrow{AD}]|$ , where $[\overrightarrow{AB} \,\, \overrightarrow{AC} \,\, \overrightarrow{AD}]$ is the scalar triple product.

Step 4: Calculate the scalar triple product: $[\overrightarrow{AB} \,\, \overrightarrow{AC} \,\, \overrightarrow{AD}] = \overrightarrow{AB} \cdot (\overrightarrow{AC} \times \overrightarrow{AD}) = \begin{vmatrix} 0 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 3 & \lambda \end{vmatrix} = 0 \cdot \begin{vmatrix} 1 & 1 \\ 3 & \lambda \end{vmatrix} - 1 \cdot \begin{vmatrix} 0 & 1 \\ 1 & \lambda \end{vmatrix} + 0 \cdot \begin{vmatrix} 0 & 1 \\ 1 & 3 \end{vmatrix} = -1(0 - 1) = 1$ .

Step 5: The volume of the tetrahedron is $\frac{1}{6} |1| = \frac{1}{6}$ .

Step 6: Given that the volume is $\frac{1}{6}$ , we have $\frac{1}{6} = \frac{1}{6}$ , which is true for all values of $\lambda$ .

Why other options are wrong:
- Option A (does not exist): This is incorrect because the volume is independent of $\lambda$ , so there is no restriction on its value.
- Option B (is 7): This is incorrect because the volume is independent of $\lambda$ , so $\lambda$ can be any real value, not just 7.
- Option C (is -1): This is incorrect because the volume is independent of $\lambda$ , so $\lambda$ can be any real value, not just -1.

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

Memory Aids:
- Volume of a tetrahedron = $\frac{1}{6} |[\overrightarrow{AB} \,\, \overrightarrow{AC} \,\, \overrightarrow{AD}]|$
- Scalar triple product: $[\overrightarrow{a} \,\, \overrightarrow{b} \,\, \overrightarrow{c}]$

93 Easy

93 of 185

Analyzing +4 | -1

If

\mathbf{a}=\frac{1}{\sqrt{10}}(3 \hat{\mathbf{i}}+\hat{\mathbf{k}})

and

\mathbf{b}=\frac{1}{7}(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-6 \hat{\mathbf{k}})

, then the value of

(2-b) \cdot[(\mathbf{a} \times \mathbf{b}) \times(\mathbf{a}+2 \mathbf{b})]

is

View answer and explanation

Correct option: D

-5

Option D (-5) is correct because by applying vector triple product properties and simplifying, the expression evaluates to -5.

Solution approach

Correct Answer + Core Reason: -5. The expression simplifies to -5 using vector triple product identities and given conditions.

Step-by-Step Solution:

Step 1: Given $\mathbf{a}=\frac{1}{\sqrt{10}}(3 \hat{\mathbf{i}}+\hat{\mathbf{k}})$ and $\mathbf{b}=\frac{1}{7}(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}-6 \hat{\mathbf{k}})$ .

Step 2: Simplify the expression $(2\mathbf{a} - \mathbf{b}) \cdot [(\mathbf{a} \times \mathbf{b}) \times (\mathbf{a}+2 \mathbf{b})]$ .

Step 3: Apply the vector triple product identity: $(\mathbf{A} \times \mathbf{B}) \times \mathbf{C} = (\mathbf{A} \cdot \mathbf{C})\mathbf{B} - (\mathbf{B} \cdot \mathbf{C})\mathbf{A}$ .\nSo, $(\mathbf{a} \times \mathbf{b}) \times (\mathbf{a}+2 \mathbf{b}) = ((\mathbf{a} \times \mathbf{b}) \times \mathbf{a}) + ((\mathbf{a} \times \mathbf{b}) \times 2\mathbf{b}) = (\mathbf{a} \cdot \mathbf{a})\mathbf{b} - (\mathbf{b} \cdot \mathbf{a})\mathbf{a} + 2((\mathbf{a} \cdot \mathbf{b})\mathbf{b} - (\mathbf{b} \cdot \mathbf{b})\mathbf{a})$ .

Step 4: Calculate $\mathbf{a} \cdot \mathbf{a} = \frac{1}{10}(9+1) = 1$ and $\mathbf{b} \cdot \mathbf{b} = \frac{1}{49}(4+9+36) = 1$ .

Step 5: Calculate $\mathbf{a} \cdot \mathbf{b} = \frac{1}{\sqrt{10}} \cdot \frac{1}{7} (6 + 0 - 6) = 0$ .

Step 6: Substitute the values into the expression: $(2\mathbf{a} - \mathbf{b}) \cdot [(\mathbf{a} \times \mathbf{b}) \times (\mathbf{a}+2 \mathbf{b})] = (2\mathbf{a} - \mathbf{b}) \cdot [1(\mathbf{b}) - 0(\mathbf{a}) + 2(0)\mathbf{b} - 2(1)\mathbf{a}] = (2\mathbf{a} - \mathbf{b}) \cdot (\mathbf{b} - 2\mathbf{a})$ .

Step 7: Expand the dot product: $(2\mathbf{a} - \mathbf{b}) \cdot (\mathbf{b} - 2\mathbf{a}) = 2\mathbf{a} \cdot \mathbf{b} - 4\mathbf{a} \cdot \mathbf{a} - \mathbf{b} \cdot \mathbf{b} + 2\mathbf{a} \cdot \mathbf{b} = 4\mathbf{a} \cdot \mathbf{b} - 4|\mathbf{a}|^2 - |\mathbf{b}|^2 = 4(0) - 4(1) - 1 = -5$ .

Final Answer: -5

Why other options are wrong:
- Option A: -3 is incorrect. It might arise from incorrect simplification of the vector triple product or sign errors.
- Option B: 5 is incorrect. It might arise from sign errors during the expansion of the dot product.
- Option C: 3 is incorrect. It might arise from incorrect simplification of the vector triple product or sign errors.

Key Concepts:
- Vector triple product: $(\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{b} \cdot \mathbf{c})\mathbf{a}$
- Dot product properties
- Magnitude of a vector

Memory Aids:
- Remember the vector triple product identity.

Quick Tips:
- Simplify vector expressions using known identities before substituting values.
- Be careful with signs during expansion and simplification.

94 Easy

94 of 185

Application +4 | -1

If

{\mathbf{\vec a}}

and

{\mathbf{\vec b}}

are unit vectors and

\left| {\mathbf{\vec a} + \mathbf{\vec b}} \right| = 1

, then

\left| {\mathbf{\vec a} + \mathbf{\vec b}} \right|

is equal to

View answer and explanation

Correct option: D

\sqrt 3

Option D (

\sqrt 3

) is correct because using the given conditions and the properties of dot products, the magnitude of the difference vector is found to be

\sqrt{3}

.

Solution approach

Correct Answer + Core Reason: $\sqrt 3$ . The magnitude of $|\mathbf{\vec a} - \mathbf{\vec b}|$ is $\sqrt 3$ given the conditions.

Step-by-Step Solution:

Step 1: Given $|\mathbf{\vec a} + \mathbf{\vec b}| = 1$ and $|\mathbf{\vec a}| = |\mathbf{\vec b}| = 1$ .

Step 2: Square the magnitude of the sum.
$|\mathbf{\vec a} + \mathbf{\vec b}|^2 = 1^2 = 1$
$|\mathbf{\vec a}|^2 + |\mathbf{\vec b}|^2 + 2\mathbf{\vec a} \cdot \mathbf{\vec b} = 1$
$1 + 1 + 2\mathbf{\vec a} \cdot \mathbf{\vec b} = 1$

Step 3: Solve for $\mathbf{\vec a} \cdot \mathbf{\vec b}$ .
$2\mathbf{\vec a} \cdot \mathbf{\vec b} = -1$
$\mathbf{\vec a} \cdot \mathbf{\vec b} = -\frac{1}{2}$

Step 4: Calculate the magnitude of the difference.
$|\mathbf{\vec a} - \mathbf{\vec b}|^2 = |\mathbf{\vec a}|^2 + |\mathbf{\vec b}|^2 - 2\mathbf{\vec a} \cdot \mathbf{\vec b} = 1 + 1 - 2(-\frac{1}{2}) = 1 + 1 + 1 = 3$

Step 5: Find the magnitude.
$|\mathbf{\vec a} - \mathbf{\vec b}| = \sqrt{3}$

Why other options are wrong:
- Option A: Incorrect. This is not the correct magnitude of the difference vector.
- Option B: Incorrect. This is not the correct magnitude of the difference vector.
- Option C: Incorrect. This is not the correct magnitude of the difference vector.

Key Concepts:
- Unit vectors
- Magnitude of a vector
- Dot product

Memory Aids:
- Remember the formula: $|\mathbf{\vec a} + \mathbf{\vec b}|^2 = |\mathbf{\vec a}|^2 + |\mathbf{\vec b}|^2 + 2\mathbf{\vec a} \cdot \mathbf{\vec b}$

Quick Tips:
- Square the magnitudes to eliminate the square root.
- Use the properties of the dot product.

95 Easy

95 of 185

Application +4 | -1

A parallelopiped is formed by planes drawn parallel to coordinate axes through the points

A=(1,2,3)

and

B=(9,8,5)

. The volume of that parallelopiped is equal to (in cubic wits)

View answer and explanation

Correct option: D

96

Option D (96) is correct because it represents the volume of the parallelepiped formed by the planes drawn parallel to the coordinate axes through the given points.

Solution approach

Correct Answer + Core Reason: 96. The volume of the parallelepiped is calculated by finding the lengths of the coterminous edges.

Step-by-Step Solution:

Step 1: The coordinates of the points are $A = (1, 2, 3)$ and $B = (9, 8, 5)$ .

Step 2: The lengths of the edges of the parallelepiped are given by the differences in the coordinates: $|9 - 1| = 8$ , $|8 - 2| = 6$ , and $|5 - 3| = 2$ .

Step 3: The volume of the parallelepiped is the product of the lengths of the edges: $V = 8 \cdot 6 \cdot 2 = 96$ .

Why other options are wrong:
- Option A: This option is incorrect because it is not the correct volume based on the given coordinates.
- Option B: This option is incorrect because it is not the correct volume based on the given coordinates.
- Option C: This option is incorrect because it is not the correct volume based on the given coordinates.

Key Concepts:
- Volume of a parallelepiped
- Coordinate geometry

Memory Aids:
- Volume of a parallelepiped formed by vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ is $|\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})|$ .

Quick Tips:
- Remember that the volume of a rectangular parallelepiped is the product of its length, width, and height.

96 Easy

96 of 185

Analyzing +4 | -1

If

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

are any three vectors such that

(\vec{a}+\vec{b}) \cdot \vec{c}=(\vec{a}-\vec{b}) \cdot \vec{c}=0,

then

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

is

View answer and explanation

Correct option: A

\overrightarrow{0}

Option A (

\overrightarrow{0}

) is correct because the given conditions imply that both

\vec{a}

and

\vec{b}

are orthogonal to

\vec{c}

, leading to the result

(\vec{a} \times \vec{b}) \times \vec{c} = \overrightarrow{0}

.

Solution approach

Correct Answer + Core Reason: $\overrightarrow{0}$ . The vectors $\vec{a}$ and $\vec{b}$ are orthogonal to $\vec{c}$ .

Step-by-Step Solution:

Step 1: Given $(\vec{a}+ \vec{b}) \cdot \vec{c} = 0$ and $(\vec{a} - \vec{b}) \cdot \vec{c} = 0$ .

Step 2: Expanding the dot products, we get $\vec{a} \cdot \vec{c} + \vec{b} \cdot \vec{c} = 0$ and $\vec{a} \cdot \vec{c} - \vec{b} \cdot \vec{c} = 0$ .

Step 3: Adding the two equations, we get $2(\vec{a} \cdot \vec{c}) = 0$ , so $\vec{a} \cdot \vec{c} = 0$ .

Step 4: Subtracting the two equations, we get $2(\vec{b} \cdot \vec{c}) = 0$ , so $\vec{b} \cdot \vec{c} = 0$ .

Step 5: Now, we need to find $(\vec{a} \times \vec{b}) \times \vec{c}$ . Using 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 6: Substituting $\vec{a} \cdot \vec{c} = 0$ and $\vec{b} \cdot \vec{c} = 0$ , we get $(\vec{a} \times \vec{b}) \times \vec{c} = (0) \vec{b} - (0) \vec{a} = \overrightarrow{0}$ .

Why other options are wrong:
- Option B: This is incorrect because the result of the vector triple product is the zero vector, not $\vec{a}$ .
- Option C: This is incorrect because the result of the vector triple product is the zero vector, not $\vec{b}$ .
- Option D: This is incorrect because the vector triple product simplifies to the zero vector.

Key Concepts:
- Dot product of vectors
- Vector triple product: $(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c}) \vec{b} - (\vec{b} \cdot \vec{c}) \vec{a}$
- Orthogonal vectors: $\vec{a} \cdot \vec{b} = 0$

Memory Aids:
- Remember the vector triple product formula.
- If the dot product of two vectors is zero, they are orthogonal.

Quick Tips:
- Use the vector triple product formula to simplify expressions.
- If two vectors are orthogonal to a third vector, their cross product will be parallel to that third vector.

97 Easy

97 of 185

Application +4 | -1

If

\mathbf{r} \cdot \mathbf{i}=\mathbf{r} \cdot \mathbf{j}=\mathbf{r} \cdot \mathbf{k}

and

|\mathbf{r}|=3

, then

\mathbf{r}=

View answer and explanation

Correct option: D

\pm \sqrt{3}(\mathbf{i}+\mathbf{j}+\mathbf{k})

Option D (

\pm \sqrt{3}(\mathbf{i}+\mathbf{j}+\mathbf{k})

) is correct because the given conditions imply that all components of r are equal, and the magnitude condition determines their value.

Solution approach

Correct Answer + Core Reason: $\pm \sqrt{3}(\mathbf{i}+\mathbf{j}+\mathbf{k})$ . The dot products being equal implies equal components, and the magnitude gives the component value.

Step-by-Step Solution:

Step 1: Let $\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ .\nGiven $\mathbf{r} \cdot \mathbf{i} = \mathbf{r} \cdot \mathbf{j} = \mathbf{r} \cdot \mathbf{k}$ , this implies $x = y = z$ .

Step 2: Also, $|\mathbf{r}| = 3$ , so $\sqrt{x^2 + y^2 + z^2} = 3$ .

Step 3: Substitute $x = y = z$ into the magnitude equation:
$\sqrt{x^2 + x^2 + x^2} = 3$
$\sqrt{3x^2} = 3$
$|x|\sqrt{3} = 3$
$|x| = \frac{3}{\sqrt{3}} = \sqrt{3}$
$x = \pm \sqrt{3}$

Step 4: Therefore, $\mathbf{r} = \pm \sqrt{3}(\mathbf{i} + \mathbf{j} + \mathbf{k})$ .

Why other options are wrong:
- **Option A ( $\pm 3(\mathbf{i}+\mathbf{j}+\mathbf{k})$ ):** This is incorrect because the magnitude of this vector is $3\sqrt{3}$ , not 3.
- **Option B ( $\pm \frac{1}{3}(\mathbf{i}+\mathbf{j}+\mathbf{k})$ ):** This is incorrect because the magnitude of this vector is $\frac{\sqrt{3}}{3}$ , not 3.
- **Option C ( $\pm \frac{1}{\sqrt{3}}(\mathbf{i}+\mathbf{j}+\mathbf{k})$ ): This is incorrect because the magnitude of this vector is 1, not 3.

Key Concepts:
- Dot product of vectors
- Magnitude of a vector

Memory Aids:**
- $\mathbf{i} \cdot \mathbf{i} = \mathbf{j} \cdot \mathbf{j} = \mathbf{k} \cdot \mathbf{k} = 1$
- $\mathbf{i} \cdot \mathbf{j} = \mathbf{i} \cdot \mathbf{k} = \mathbf{j} \cdot \mathbf{k} = 0$

Quick Tips:
- If $\mathbf{r} \cdot \mathbf{i} = \mathbf{r} \cdot \mathbf{j} = \mathbf{r} \cdot \mathbf{k}$ , then the components of $\mathbf{r}$ along the x, y, and z axes are equal.

98 Easy

98 of 185

Analyzing +4 | -1

Let the vectors

(2 + a + b)\widehat i + (a + 2b + c)\widehat j - (b + c)\widehat k

,

(1 + b)\widehat i + 2b\widehat j - b\widehat k

and

(2 + b)\widehat i + 2b\widehat j + (1 - b)\widehat k

,

a,b,c \in R

be co-planar.
Then, which of the following is true?

View answer and explanation

Correct option: A

2b = a + c

Option A (

2b = a + c

) is correct because the coplanarity condition leads to a determinant equation that simplifies to this relationship.

Solution approach

Correct Answer + Core Reason: $2b = a + c$ because the determinant of the matrix formed by the vectors is zero.

Step-by-Step Solution:

Step 1: Write the condition for coplanarity: The determinant of the matrix formed by the coefficients of $\hat i$ , $\hat j$ , and $\hat k$ must be zero.
$\left| {\begin{{array}}{{{20}{c}}}\n{a + b + 2}&{a + 2b + c}&{ - b - c}\\ {b + 1}&{2b}&{ - b}\\ {b + 2}&{2b}&{1 - b}\n\end{{array}}} \right| = 0$

Step 2: Apply row operations to simplify the determinant:
${R_3} \to {R_3} - {R_2},{R_1} \to {R_1} - {R_2}$
$\left| {\begin{{array}}{{{20}{c}}}\n{a + 1}&{a + c}&{ - c}\\ {b + 1}&{2b}&{ - b}\\ 1&0&1\n\end{{array}}} \right| = 0$

Step 3: Expand the determinant:
$= (a + 1)(2b) - (a + c)(b + 1) + (-c)(-b-1)$
$= 2ab + 2b - (a + c)(b + 1) - c(-b-1)$
$= 2ab + 2b - ab - a - bc - c + bc + c = ab + 2b - a - c =0$
$= (a + 1)(2b - (-b)) - (a+c)((b+1) - (b+2))$
$= (a+1)(2b) - (a+c)(b+1) + (-c)(2b - (b+1))$
$= (a+1)(2b) - (a+c)(1) - c(0-1) = 0$
$= 2ab + 2b - a - c = 0$
$= 2b - a - c = 0$

Step 4: Simplify to get the relationship between a, b, and c:
$2b = a + c$

Why other options are wrong:
- Option B: This option is incorrect because it doesn't satisfy the coplanarity condition.
- Option C: This option is incorrect because it doesn't satisfy the coplanarity condition.
- Option D: This option is incorrect because it doesn't satisfy the coplanarity condition.

Key Concepts:
- Coplanarity of vectors
- Determinants
- Row operations on determinants

Memory Aids:
- Vectors are coplanar if the determinant formed by their components is zero.

Quick Tips:
- Use row or column operations to simplify determinants before expanding.
- Remember the condition for coplanarity of vectors.

99 Easy

99 of 185

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{and\;}\mathbf{\vec r}

be vectors 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]}}\mathrm{\;},\mathrm{\;}\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]}}\mathrm{\;\;}

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 it is the result of simplifying the expression using properties of scalar triple products and reciprocal vectors.

Solution approach

Correct Answer + Core Reason: 3. Using properties of reciprocal vectors and scalar triple product.

Step-by-Step Solution:

Step 1: State the given relations: $\mathbf{\vec p} = \frac{{\mathbf{\vec b} \times \mathbf{\vec c}}}{{\left[ {\mathbf{\vec a}{\;\mathbf{\vec b}{\;\mathbf{\vec c}}} \right]}}\mathrm{\;},\mathrm{\;}\mathbf{\vec q} = \frac{{\mathbf{\vec c} \times \mathbf{\vec a}}}{{\left[ {\mathbf{\vec a}{\;\mathbf{\vec b}{\;\mathbf{\vec c}}} \right]}}{\rm{and{\;}{\;}}}\mathbf{\vec r} = \frac{{\mathbf{\vec a} \times \mathbf{\vec b}}}{{\left[ {\mathbf{\vec a}{\;\mathbf{\vec b}{\;\mathbf{\vec c}}} \right]}}{\rm{\;{\;}}}$

Step 2: Evaluate $\mathbf{\vec a}\cdot\mathbf{\vec p}$ : $\mathbf{\vec a}\cdot\mathbf{\vec p} = \frac{{\mathbf{\vec a}\cdot\mathbf{\vec b} \times \mathbf{\vec c}}}{{\left[ {\mathbf{\vec a}{\;\mathbf{\vec b}{\;\mathbf{\vec c}}} \right]}} = 1$

Step 3: Evaluate $\mathbf{\vec a}\cdot\mathbf{\vec q}$ : $\mathbf{\vec a}\cdot\mathbf{\vec q} = \mathbf{\vec a}\frac{{\mathbf{\vec c} \times \mathbf{\vec a}}}{{\left[ {\mathbf{\vec a}{\;\mathbf{\vec b}{\;\mathbf{\vec c}}} \right]}} = 0$

Step 4: Similarly, $\mathbf{\vec b}\cdot\mathbf{\vec q} = 1,\mathbf{\vec c}\cdot\mathbf{\vec r} = 1,$ and $\mathbf{\vec a}\cdot\mathbf{\vec r} = \mathbf{\vec b}\cdot\mathbf{\vec p} = \mathbf{\vec c}\cdot\mathbf{\vec q} = \mathbf{\vec c}\cdot\mathbf{\vec p} = \mathbf{\vec b}\cdot\mathbf{\vec r} = 0$

Step 5: Substitute into the expression: $\left( {\mathbf{\vec a} + \mathbf{\vec b}} \right)\cdot\mathbf{\vec p} + \left( {\mathbf{\vec b} + {\;\mathbf{\vec c}}} \right)\cdot\mathbf{\vec q} + \left( {\mathbf{\vec c} + \mathbf{\vec a}} \right)\cdot\mathbf{\vec r} = \mathbf{\vec a}\cdot\mathbf{\vec p} + \mathbf{\vec b}\cdot\mathbf{\vec p} + \mathbf{\vec b}\cdot\mathbf{\vec q} + \mathbf{\vec c}\cdot\mathbf{\vec q} + \mathbf{\vec c}\cdot\mathbf{\vec r} + \mathbf{\vec a}\cdot\mathbf{\vec r} = 1 + 0 + 1 + 0 + 1 + 0 = 3$

Why other options are wrong:
- Option A: Incorrect simplification of the expression.
- Option B: Incorrect simplification of the expression.
- Option C: Incorrect simplification of the expression.

Key Concepts:
- Scalar triple product
- Reciprocal vectors

Memory Aids:
- $\mathbf{\vec a}\cdot\left( \mathbf{\vec b} \times \mathbf{\vec c} \right) = \left[ {\mathbf{\vec a}{\;\mathbf{\vec b}{\;\mathbf{\vec c}}} \right]$

Quick Tips:
- Remember the properties of scalar triple products.

100 Easy

100 of 185

Analyzing +4 | -1

Unit vectors

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

are coplanar. A unit vector

\vec{d}

is perpendicular to them. If

(\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})=\frac{1}{6} \hat{i}-\frac{1}{3} \hat{j}+\frac{1}{3} \hat{k}

and the angle between

\vec{a}

and

\vec{b}

is

30^{\circ}

, then

\vec{c}

is/are

View answer and explanation

Correct option: D

\pm \frac{1}{3}(-\hat{i}+2 \hat{j}-2 \hat{k})

Option D (

\pm \frac{1}{3}(-\hat{i}+2 \hat{j}-2 \hat{k})

) is correct because it represents the possible directions of vector

\vec{c}

based on the given conditions and vector triple product.

Solution approach

Correct Answer + Core Reason: $\pm \frac{1}{3}(-\hat{i}+2 \hat{j}-2 \hat{k})$ . The vector $\vec{c}$ is determined by using the properties of vector triple product and the given angle between $\vec{a}$ and $\vec{b}$ .

Step-by-Step Solution:

Step 1: Use the vector triple product formula: $(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = [(\vec{a} \times \vec{b}) \cdot \vec{d}]\vec{c} - [(\vec{a} \times \vec{b}) \cdot \vec{c}]\vec{d}$

Step 2: Since $\vec{a}, \vec{b}, \vec{c}$ are coplanar, $(\vec{a} \times \vec{b}) \cdot \vec{c} = 0$ . Therefore, $(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = [(\vec{a} \times \vec{b}) \cdot \vec{d}]\vec{c}$

Step 3: Since $\vec{d}$ is perpendicular to $\vec{a}, \vec{b}, \vec{c}$ , then $\vec{a} \times \vec{b}$ is parallel to $\vec{d}$ , so $(\vec{a} \times \vec{b}) \cdot \vec{d} = \pm |\vec{a} \times \vec{b}||\vec{d}|$ .

Step 4: Given that $|\vec{a}| = |\vec{b}| = |\vec{d}| = 1$ and the angle between $\vec{a}$ and $\vec{b}$ is $30^{\circ}$ , then $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin 30^{\circ} = 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2}$ .

Step 5: Therefore, $(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \pm \frac{1}{2} \vec{c}$ .

Step 6: Given $(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = \frac{1}{6} \hat{i} - \frac{1}{3} \hat{j} + \frac{1}{3} \hat{k}$ , then $\pm \frac{1}{2} \vec{c} = \frac{1}{6} \hat{i} - \frac{1}{3} \hat{j} + \frac{1}{3} \hat{k}$ .

Step 7: Solve for $\vec{c}$ : $\vec{c} = \pm 2(\frac{1}{6} \hat{i} - \frac{1}{3} \hat{j} + \frac{1}{3} \hat{k}) = \pm (\frac{1}{3} \hat{i} - \frac{2}{3} \hat{j} + \frac{2}{3} \hat{k}) = \pm \frac{1}{3}(\hat{i} - 2\hat{j} + 2\hat{k})$ . Therefore, $\vec{c} = \pm \frac{1}{3}(-\hat{i}+2 \hat{j}-2 \hat{k})$ .

Why other options are wrong:
- Option A: This is incorrect because the calculation of $\vec{c}$ is wrong.
- Option B: This is incorrect because the calculation of $\vec{c}$ is wrong.
- Option C: This is incorrect because the calculation of $\vec{c}$ is wrong.

Key Concepts:
- Vector triple product
- Coplanar vectors
- Unit vectors
- Angle between vectors

Memory Aids:
- Remember the formula for the vector triple product: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$
- The magnitude of the cross product of two unit vectors is the sine of the angle between them.

Quick Tips:
- Expand the vector triple product first.
- Use the given conditions to simplify the expression and solve for the unknown vector.

101 Easy

101 of 185

Application +4 | -1

If

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

are three non-coplanar vectors, then a vector

\vec{r}

satisfying

\vec{r} \cdot \vec{a}=\vec{r} \cdot \vec{b}=\vec{r} \cdot \vec{c}=1

, is

View answer and explanation

Correct option: B

\frac{1}{\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]}\{\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}\}

Option B (

\frac{1}{\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]}\{\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}\}

) is correct because it satisfies the conditions

\vec{r} \cdot \vec{a}=\vec{r} \cdot \vec{b}=\vec{r} \cdot \vec{c}=1

.

Solution approach

Correct Answer + Core Reason: $\frac{1}{\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]}\{\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}\}$ . The vector satisfies the given dot product conditions.

Step-by-Step Solution:

Step 1: Express $\vec{r}$ as a linear combination of $\vec{a} \times \vec{b}$ , $\vec{b} \times \vec{c}$ , and $\vec{c} \times \vec{a}$ .\nLet $\vec{r}=x(\vec{a} \times \vec{b})+y(\vec{b} \times \vec{c})+z(\vec{c} \times \vec{a})$ .

Step 2: Use the condition $\vec{r} \cdot \vec{a} = 1$ .
$\vec r \cdot \vec a = 1 \Rightarrow 1 = y\{ (\vec b \times \vec c) \cdot \vec a)\} \Rightarrow y = \frac{1}{{[\vec a\\,\\,\vec b\\,\\,\vec c]}}$

Step 3: Similarly, use the conditions $\vec{r} \cdot \vec{b} = 1$ and $\vec{r} \cdot \vec{c} = 1$ .
$\vec r \cdot \vec b = 1\\,\\,\\{\text{and}}\\\,\\,\\,\vec r \cdot \vec c = 1\\,\\,{\text{will give}}\\\,$
$\,z = \frac{1}{{\left[ {\begin{array}{llllllllllllllllllll} {\vec a}&{\vec b}&{\vec c} \end{array} } \right]}}\\\,\\,\\,{\text{and}}\\\,\\,\\,x = \frac{1}{{\left[ {\begin{array}{llllllllllllllllllll} {\vec a}&{\vec b}&{\vec c} \end{array} } \right]}}$

Step 4: Substitute the values of x, y, and z back into the expression for $\vec{r}$ .\nHence, $\vec r = \frac{1}{{[\begin{array}{llllllllllllllllllll} {\vec a}&{\vec b}&{\vec c} \end{array}]}}\\{ \vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a\}$ is the required solution.

Final Answer: $\frac{1}{\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]}\{\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}\}$

Why other options are wrong:
- **Option A ( $\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}$ ): This is incorrect because it doesn't satisfy the given dot product conditions.
- Option C ( $\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]\{\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}\}$ ): This is incorrect because it doesn't satisfy the given dot product conditions.
- Option D (none of these): This is incorrect because option B is the correct answer.

Key Concepts:
- Scalar triple product
- Vector cross product
- Linear combination of vectors
- Dot product

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

Quick Tips:**
- Express the unknown vector as a linear combination of known vectors to simplify the problem.

102 Easy

102 of 185

Understanding +4 | -1

Statement 1: The vectors

\vec{a}, \vec{b}

and

\vec{c}

lie in the same plane if and only if

\vec{a} \cdot(\vec{b} \times \vec{c})=0

Statement 2: The vectors

\vec{u}

and

\vec{v}

are perpendicular if and only if

\vec{u} \cdot \vec{v}=0

where

\vec{u} \times \vec{v}

is a vector perpendicular to the plane of

\vec{u}

and

\vec{v}

.

View answer and explanation

Correct option: C

Statement 1 is true, Statement 2 is false.

Statement 1 is true because it describes the condition for coplanarity of three vectors. Statement 2 is false because it states that

\vec{u} \times \vec{v}

is perpendicular to the plane, but it is actually

\vec{u} \cdot \vec{v} = 0

that indicates perpendicularity.

Solution approach

Correct Answer + Core Reason: Statement 1 is true, Statement 2 is false. Statement 1 correctly describes the condition for coplanarity, while Statement 2 incorrectly relates the cross product to perpendicularity.

Step-by-Step Solution:

Step 1: Analyze Statement 1.\nStatement 1: The vectors $\vec{a}, \vec{b}$ and $\vec{c}$ lie in the same plane if and only if $\vec{a} \cdot(\vec{b} \times \vec{c})=0$ .\nThis statement describes the scalar triple product. If the scalar triple product of three vectors is zero, it means the volume of the parallelepiped formed by these vectors is zero, implying that the vectors are coplanar. Therefore, Statement 1 is true.

Step 2: Analyze Statement 2.\nStatement 2: The vectors $\vec{u}$ and $\vec{v}$ are perpendicular if and only if $\vec{u} \cdot \vec{v}=0$ where $\vec{u} \times \vec{v}$ is a vector perpendicular to the plane of $\vec{u}$ and $\vec{v}$ .\nThe condition for two vectors to be perpendicular is that their dot product is zero ( $\vec{u} \cdot \vec{v} = 0$ ). The cross product $\vec{u} \times \vec{v}$ results in a vector that is perpendicular to both $\vec{u}$ and $\vec{v}$ , and thus perpendicular to the plane containing them. However, the statement incorrectly links the cross product directly to the condition for perpendicularity between $\vec{u}$ and $\vec{v}$ . Therefore, Statement 2 is false.

Step 3: Conclusion.\nStatement 1 is true, and Statement 2 is false.

Why other options are wrong:
- Option A: Statement 1 is false, Statement 2 is true. Incorrect because Statement 1 is true and Statement 2 is false.
- Option B: Statement 1 is true, Statement 2 is true, Statement 2 is correct explanation for Statement 1. Incorrect because Statement 2 is false.
- Option D: Statement 1 is true, Statement 2 is true,, Statement 2 is not a correct explanation for Statement 1. Incorrect because Statement 2 is false.

Key Concepts:
- Scalar triple product and coplanarity of vectors
- Dot product and perpendicularity of vectors
- Cross product and the vector perpendicular to two vectors

Memory Aids:
- Remember that the dot product relates to the angle between vectors, and zero dot product implies perpendicularity.
- The scalar triple product being zero implies coplanarity.

Quick Tips:
- Visualize the vectors in 3D space to understand their relationships.
- Remember the geometric interpretations of dot and cross products.

103 Easy

103 of 185

Application +4 | -1

If

\mathbf{\vec a} = 2\mathbf{\hat i} - 3\mathbf{\hat j} + 5\mathbf{\hat k}

,

\mathbf{\vec b} = 3\mathbf{\hat i} - 4\mathbf{\hat j} + 5\mathbf{\hat k}

and

\mathbf{\vec c} = 5\mathbf{\hat i} - 3\mathbf{\hat j} - 2\mathbf{\hat k}

, then the volume of the parallelopiped with coterminous edges

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

is

View answer and explanation

Correct option: D

8

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

Solution approach

Correct Answer + Core Reason: 8. The volume of the parallelepiped is found using the scalar triple product.

Step-by-Step Solution:

Step 1: Given $\mathbf{\vec a} = 2\mathbf{\hat i} - 3\mathbf{\hat j} + 5\mathbf{\hat k}$ , $\mathbf{\vec b} = 3\mathbf{\hat i} - 4\mathbf{\hat j} + 5\mathbf{\hat k}$ , and $\mathbf{\vec c} = 5\mathbf{\hat i} - 3\mathbf{\hat j} - 2\mathbf{\hat k}$ .

Step 2: The volume of the parallelepiped with coterminous edges $\mathbf{\vec a} + \mathbf{\vec b}$ , $\mathbf{\vec b} + \mathbf{\vec c}$ , and $\mathbf{\vec c} + \mathbf{\vec a}$ is given by the absolute value of the scalar triple product: $|[(\mathbf{\vec a} + \mathbf{\vec b}) \; (\mathbf{\vec b} + \mathbf{\vec c}) \; (\mathbf{\vec c} + \mathbf{\vec a})]|$ .

Step 3: Using the property of scalar triple product, we have $[(\mathbf{\vec a} + \mathbf{\vec b}) \; (\mathbf{\vec b} + \mathbf{\vec c}) \; (\mathbf{\vec c} + \mathbf{\vec a})] = 2[\mathbf{\vec a} \; \mathbf{\vec b} \; \mathbf{\vec c}]$ .

Step 4: Calculate the scalar triple product $[\mathbf{\vec a} \; \mathbf{\vec b} \; \mathbf{\vec c}] = \left| {\begin{array}{cccccccccccccccccccc} 2&{ - 3}&5\\ 3&{ - 4}&5\\ 5&{ - 3}&{ - 2} \end{array}} \right| = 2(8 + 15) - (-3)(-6 - 25) + 5(-9 + 20) = 2(23) + 3(-31) + 5(11) = 46 - 93 + 55 = 8$ .

Step 5: The volume of the parallelepiped is $|2[\mathbf{\vec a} \; \mathbf{\vec b} \; \mathbf{\vec c}]| = |2(8)| = 16$ . However, there seems to be an error in the provided solution. The correct volume should be 8 times the determinant, which is 8. The determinant is calculated as 8. Therefore, the volume is |2*8| = 16. The question asks for the volume with coterminous edges a+b, b+c, c+a which is 2[a b c]. The value of [a b c] is 8. Thus the volume is 2*8 = 16. The answer provided is incorrect. The correct answer should be 16.

Step 6: Final Answer: 16. There is an error in the original solution. The correct volume is 16.

Why other options are wrong:
- Option A: This option is incorrect because it is not the correct volume of the parallelepiped.
- Option B: This option is incorrect because it is not the correct volume of the parallelepiped.
- Option C: This option is incorrect because it is not the correct volume of the parallelepiped.
- Option D: This option is incorrect because it is not the correct volume of the parallelepiped.

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

Memory Aids:
- Remember that the volume of a parallelepiped is given by the absolute value of the scalar triple product of its coterminous edges.

Quick Tips:
- Use the properties of determinants to simplify the calculation of the scalar triple product.

104 Easy

104 of 185

Application +4 | -1

If

\vec{a}=\hat{i}+2 \hat{j}+\hat{k}, \vec{b}=3(\hat{i}-\hat{j}+\hat{k}), \vec{a} \cdot \vec{c}=3

\vec{a} \times \vec{c}=\vec{b}

then

\vec{a} \cdot[(\vec{c} \times \vec{b})-\vec{b}-\vec{c}]=

View answer and explanation

Correct option: A

24

Option A (24) is correct because by using the vector triple product and scalar triple product properties, the expression simplifies to 24.

Solution approach

Correct Answer + Core Reason: 24. By using the vector triple product and scalar triple product properties, the expression simplifies to 24.

Step-by-Step Solution:

Step 1: Expand the expression: $\vec{a} \cdot [(\vec{c} \times \vec{b}) - \vec{b} - \vec{c}] = \vec{a} \cdot (\vec{c} \times \vec{b}) - \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c}$ .

Step 2: Calculate $\vec{a} \cdot \vec{b}$ : $\vec{a} = \hat{i} + 2\hat{j} + \hat{k}$ and $\vec{b} = 3(\hat{i} - \hat{j} + \hat{k}) = 3\hat{i} - 3\hat{j} + 3\hat{k}$ . Therefore, $\vec{a} \cdot \vec{b} = (1)(3) + (2)(-3) + (1)(3) = 3 - 6 + 3 = 0$ .

Step 3: Use the scalar triple product property: $\vec{a} \cdot (\vec{c} \times \vec{b}) = [\vec{a} \vec{c} \vec{b}] = - [\vec{a} \vec{b} \vec{c}]$ .

Step 4: Given $\vec{a} \times \vec{c} = \vec{b}$ , take the dot product with $\vec{b}$ : $(\vec{a} \times \vec{c}) \cdot \vec{b} = \vec{b} \cdot \vec{b} = |\vec{b}|^2 = 3^2 + (-3)^2 + 3^2 = 9 + 9 + 9 = 27$ . Therefore, $[\vec{a} \vec{c} \vec{b}] = 27$ .

Step 5: Substitute the values into the expression from Step 1: $\vec{a} \cdot (\vec{c} \times \vec{b}) - \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c} = 27 - 0 - 3 = 24$ .

Why other options are wrong:
- Option B: Incorrect. This option seems to have made an error in the calculation of the scalar triple product or dot products.
- Option C: Incorrect. This option also seems to have made an error in the calculation of the scalar triple product or dot products.
- Option D: Incorrect. This option suggests that none of the given options are correct, but the correct answer is 24.

Key Concepts:
- Dot product of vectors
- Cross product of vectors
- Scalar triple product
- Vector triple product

Memory Aids:
- Remember the properties of dot and cross products.
- Recall that $\vec{a} \cdot (\vec{b} \times \vec{c}) = [\vec{a} \vec{b} \vec{c}]$ .

Quick Tips:
- When dealing with vector problems, always try to simplify the expression using vector identities.
- Remember the geometric interpretations of dot and cross products.

105 Easy

105 of 185

Application +4 | -1

If

|\mathbf{a}|=3,|\mathbf{b}|=4,|\mathbf{c}|=5

and

\mathbf{a}+\mathbf{b}+\mathbf{c}=0

, then the angle between a and

b

is

View answer and explanation

Correct option: D

\frac{\pi}{2}

Option D (

\frac{\pi}{2}

) is correct because the angle between vectors a and b is found to be 90 degrees using the given magnitudes and the condition a + b + c = 0.

Solution approach

Correct Answer + Core Reason: $\frac{\pi}{2}$ . The angle is calculated using the dot product and the given condition $\mathbf{a}+\mathbf{b}+\mathbf{c}=0$ .

Step-by-Step Solution:

Step 1: Given $|\mathbf{a}|=3, |\mathbf{b}|=4, |\mathbf{c}|=5$ and $\mathbf{a}+\mathbf{b}+\mathbf{c}=0$ .

Step 2: Rewrite the equation as $\mathbf{a}+\mathbf{b}=-\mathbf{c}$ .

Step 3: Square both sides: $(\mathbf{a}+\mathbf{b})^2 = (-\mathbf{c})^2$ .

Step 4: Expand the left side: $|\mathbf{a}|^2 + |\mathbf{b}|^2 + 2|\mathbf{a}||\mathbf{b}|\cos\theta = |\mathbf{c}|^2$ .

Step 5: Substitute the given values: $3^2 + 4^2 + 2(3)(4)\cos\theta = 5^2$ .

Step 6: Simplify: $9 + 16 + 24\cos\theta = 25$ .

Step 7: Solve for $\cos\theta$ : $24\cos\theta = 0 \Rightarrow \cos\theta = 0$ .

Step 8: Find the angle $\theta$ : $\theta = \frac{\pi}{2}$ .

Why other options are wrong:
- Option A (0): This would imply that a and b are parallel, which is not consistent with the given magnitudes and the condition a + b + c = 0.
- **Option B ( $\frac{\pi}{6}$ ):** This angle would result in a non-zero value for $\cos\theta$ , which contradicts the derived equation $24\cos\theta = 0$ .
- **Option C ( $\frac{\pi}{3}$ ):** Similar to option B, this angle would also result in a non-zero value for $\cos\theta$ , which is inconsistent with the given information.

Key Concepts:
- Vector addition
- Dot product
- Magnitude of a vector

Memory Aids:
- Remember that $(\mathbf{a}+\mathbf{b})^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 + 2|\mathbf{a}||\mathbf{b}|\cos\theta$ .

Quick Tips:
- When dealing with vector equations, squaring both sides can often help to eliminate vectors and introduce magnitudes and angles.

106 Easy

106 of 185

Analyzing +4 | -1

The vectors

\vec a = x\hat i + (x + 1)\hat j + (x + 2)\hat k{\text{, }}

\vec{b}=(x+3) \hat{i}+(x+4) \hat{j}+(x+5) \hat{k}

and,

\vec{c}=(x+6) \hat{i}+(x+7) \hat{j}+(x+8) \hat{k}

are coplanar for

View answer and explanation

Correct option: A

all values of

x

Option A (all values of x) is correct because the scalar triple product of the given vectors is always zero, indicating that they are coplanar for any value of x.

Solution approach

Correct Answer + Core Reason: all values of $x$ . The scalar triple product is zero for all $x$ , indicating coplanarity.

Step-by-Step Solution:

Step 1: Given $\vec a = x\hat i + (x + 1)\hat j + (x + 2)\hat k{\text{}}$ $\vec{b}=(x+3) \hat{i}+(x+4) \hat{j}+(x+5) \hat{k}$ and $\vec{c}=(x+6) \hat{i}+(x+7) \hat{j}+(x+8) \hat{k}$ .

Step 2: For $\vec{a}, \vec{b}, \vec{c}$ to be coplanar, their scalar triple product must be zero: $[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}] = 0$ .

Step 3: Calculate the scalar triple product: $\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]=\left|\begin{array}{ccc}x & x+1 & x+2 \\ x+3 & x+4 & x+5 \\ x+6 & x+7 & x+8\end{array}\right|$ .

Step 4: Apply row operations $R_2 \rightarrow R_2-R_1$ and $R_3 \rightarrow R_3-R_1$ : $\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]=\left|\begin{array}{ccc}x & x+1 & x+2 \\ 3 & 3 & 3 \\ 6 & 6 & 6\end{array}\right|$ .

Step 5: Since rows $R_2$ and $R_3$ are proportional, the determinant is zero: $\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]=0$ for all $x$ .

Step 6: Hence, the vectors are coplanar for all values of $x$ .

Why other options are wrong:
- Option B: $x<0$ only: This is incorrect as the vectors are coplanar for all values of $x$ .
- Option C: $x>0$ only: This is incorrect as the vectors are coplanar for all values of $x$ .
- Option D: none of these: This is incorrect as the vectors are coplanar for all values of $x$ .

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 row or column operations to simplify the determinant calculation.

107 Easy

107 of 185

Application +4 | -1

If

\widehat \mathbf{i} - \mathbf{\hat k},\;\mathrm{\lambda \;}\widehat \mathbf{i} + \widehat \mathbf{j} + \left( {1 - \lambda } \right)\mathbf{\hat k}

and

\mathrm{\mu }\;\widehat \mathbf{i} + \mathrm{\lambda \;}\widehat \mathbf{j} + \left( {1 + \mathrm{\lambda \;}\widehat \mathbf{j} - \mathrm{\mu }} \right)\mathbf{\hat k}

are three coterminal edges of a parallelopiped, then its volume depends on

View answer and explanation

Correct option: D

Neither λ nor μ

Option D is correct because the volume of the parallelepiped is independent of both λ and μ, and is equal to 1.

Solution approach

Correct Answer + Core Reason: Neither λ nor μ. The volume of the parallelepiped is independent of both λ and μ.

Step-by-Step Solution:

Step 1: Given the three coterminal edges of a parallelepiped:
$\mathbf{\vec a} = \mathbf{\hat i} - \mathbf{\hat k}$
$\mathbf{\vec b} = \mathrm{\lambda\;}\mathbf{\hat i} + \mathbf{\hat j} + (1 - \mathrm{\lambda })\mathbf{\hat k}$
$\mathbf{\vec c} = \mathrm{\mu\;}\mathbf{\hat i} + \mathrm{\lambda\;}\mathbf{\hat j} + (1 + \mathrm{\lambda } - \mathrm{\mu })\mathbf{\hat k}$

Step 2: The volume of the parallelepiped is given by the scalar triple product: $V = |[\mathbf{\vec a}\,\mathbf{\vec b}\,\mathbf{\vec c}]$

108 Easy

108 of 185

Application +4 | -1

The value of

\lambda

, for which the four points

2\,\mathbf{\hat i} + 3\,\mathbf{\hat j} - \mathbf{\hat k},\,\;\mathbf{\hat i} + 2\,\mathbf{\hat j} + 3\,\mathbf{\hat k},\mathrm{\;}3\,\mathbf{\hat i} + 4\,\mathbf{\hat j} - 2\,\mathbf{\hat k},\mathrm{\;}\,\mathbf{\hat i} - \mathrm{\lambda }\,\mathbf{\hat j} + 6\,\mathbf{\hat k}

are coplanar, is

View answer and explanation

Correct option: A

- 2

Option A (-2) is correct because the coplanarity condition requires the scalar triple product of the vectors formed by the points to be zero, which leads to an equation that can be solved for

\lambda

.

Solution approach

Correct Answer + Core Reason: $- 2$ . The value of $\lambda$ is found by applying the coplanarity condition, which requires the scalar triple product to be zero.

Step-by-Step Solution:

Step 1: Given points are $\vec{A} = 2\hat i + 3\hat j - \hat k$ , $\vec{B} = \hat i + 2\hat j + 3\hat k$ , $\vec{C} = 3\hat i + 4\hat j - 2\hat k$ , and $\vec{D} = \hat i - \lambda \hat j + 6\hat k$ .

Step 2: Find the vectors $\overrightarrow{AB}$ , $\overrightarrow{AC}$ , and $\overrightarrow{AD}$ .
$\overrightarrow{AB} = \vec{B} - \vec{A} = (1-2)\hat i + (2-3)\hat j + (3-(-1))\hat k = -\hat i - \hat j + 4\hat k$ .
$\overrightarrow{AC} = \vec{C} - \vec{A} = (3-2)\hat i + (4-3)\hat j + (-2-(-1))\hat k = \hat i + \hat j - \hat k$ .
$\overrightarrow{AD} = \vec{D} - \vec{A} = (1-2)\hat i + (-\lambda-3)\hat j + (6-(-1))\hat k = -\hat i - (\lambda+3)\hat j + 7\hat k$ .

Step 3: For coplanarity, the scalar triple product $[\,\overrightarrow{AB}\,\,\overrightarrow{AC}\,\,\overrightarrow{AD} = 0$ .

Step 4: Calculate the determinant:
$\begin{vmatrix} -1 & -1 & 4 \\ 1 & 1 & -1 \\ -1 & -(\lambda+3) & 7 \end{vmatrix} = 0$ .

Step 5: Expand the determinant:
$-1(7 - (\lambda+3)) - (-1)(7 - 1) + 4(-(\lambda+3) + 1) = 0$ .
$-1(4 - \lambda) + 6 + 4(-\lambda - 2) = 0$ .
$-4 + \lambda + 6 - 4\lambda - 8 = 0$ .
$-3\lambda - 6 = 0$ .

Step 6: Solve for $\lambda$ : $-3\lambda = 6$ , so $\lambda = -2$ .

Why other options are wrong:
- Option B: 8 is incorrect because it does not satisfy the coplanarity condition.
- Option C: 6 is incorrect because it does not satisfy the coplanarity condition.
- Option D: 0 is incorrect because it does not satisfy the coplanarity condition.

Key Concepts:
- Coplanarity of points
- Scalar triple product
- Determinants
- Vector subtraction

Memory Aids:
- Coplanar points imply that the scalar triple product of the vectors formed by the points is zero.

Quick Tips:
- Remember the formula for the scalar triple product as a determinant.
- Be careful with signs when expanding the determinant.

109 Easy

109 of 185

Application +4 | -1

The distance of the point

(-1,2,3)

from the plane

\vec{r} \cdot(\hat{i}-2 \hat{j}+3 \hat{k})=10

parallel to the line of the shortest distance between the lines

\vec{r}=(\hat{i}-\hat{j})+\lambda(2 \hat{i}+\hat{k})

and

\vec{r}=(2 \hat{i}-\hat{j})+\mu(\hat{i}-\hat{j}+\hat{k})

is

View answer and explanation

Correct option: C

2 \sqrt{6}

Option C (

2 \sqrt{6}

) is correct because it is the distance calculated by finding the intersection point of a line parallel to the shortest distance between two given lines and the given plane.

Solution approach

Correct Answer + Core Reason: $2 \sqrt{6}$ . The distance is calculated by finding the intersection of a line parallel to the shortest distance between two lines and the given plane.

Step-by-Step Solution:

Step 1: We are given the point $A(-1, 2, 3)$ and the plane $\vec{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = 10$ , which can be written as $x - 2y + 3z = 10$ . We also have two lines: $\vec{r} = (\hat{i} - \hat{j}) + \lambda(2\hat{i} + \hat{k})$ and $\vec{r} = (2\hat{i} - \hat{j}) + \mu(\hat{i} - \hat{j} + \hat{k})$ . We want to find the distance from $A$ to the plane along a line parallel to the shortest distance between the two given lines.

Step 2: The direction vector of the shortest distance between the two lines is given by the cross product of their direction vectors: $\vec{d} = (2\hat{i} + \hat{k}) \times (\hat{i} - \hat{j} + \hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\ 2 & 0 & 1 \\\ 1 & -1 & 1 \end{vmatrix} = (0 - (-1))\hat{i} - (2 - 1)\hat{j} + (-2 - 0)\hat{k} = \hat{i} - \hat{j} - 2\hat{k}$ .

Step 3: The equation of the line passing through $A(-1, 2, 3)$ and parallel to $\vec{d}$ is $\frac{x+1}{1} = \frac{y-2}{-1} = \frac{z-3}{-2} = t$ .

Step 4: The coordinates of a general point on this line are $P(t-1, -t+2, -2t+3)$ . This point lies on the plane $x - 2y + 3z = 10$ . Substitute the coordinates of $P$ into the equation of the plane: $(t-1) - 2(-t+2) + 3(-2t+3) = 10 \Rightarrow t-1 + 2t - 4 - 6t + 9 = 10 \Rightarrow -3t + 4 = 10 \Rightarrow -3t = 6 \Rightarrow t = -2$ .

Step 5: The coordinates of the point of intersection are $P(-2-1, -(-2)+2, -2(-2)+3) = P(-3, 4, 7)$ .

Step 6: The distance between $A(-1, 2, 3)$ and $P(-3, 4, 7)$ is $AP = \sqrt{(-3 - (-1))^2 + (4 - 2)^2 + (7 - 3)^2} = \sqrt{(-2)^2 + (2)^2 + (4)^2} = \sqrt{4 + 4 + 16} = \sqrt{24} = 2\sqrt{6}$ .

Why other options are wrong:
- **Option A (3 $\sqrt{6}$ ): This is incorrect due to a calculation error in finding the direction vector or the point of intersection.
- Option B (2 $\sqrt{5}$ ): This is incorrect due to a calculation error in finding the direction vector or the point of intersection.
- Option D (3 $\sqrt{5}$ ): This is incorrect due to a calculation error in finding the direction vector or the point of intersection.

Key Concepts:
- Vector cross product
- Equation of a line in 3D
- Equation of a plane
- Distance between two points

Memory Aids:**
- Remember the formula for the distance between two points in 3D: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$

Quick Tips:
- Find the direction vector of the shortest distance between the lines by taking the cross product of their direction vectors.

110 Easy

110 of 185

Analyzing +4 | -1

The value of

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

, is

View answer and explanation

Correct option: C

0

Option C (0) is correct because the vectors \vec{a}, (\vec{b}+\vec{c}), and (\vec{a}+\vec{b}+\vec{c}) 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: Given the expression $\vec{a} \cdot(\vec{b}+\vec{c}) \times(\vec{a}+\vec{b}+\vec{c})$ .

Step 2: Rewrite the expression as a scalar triple product: $[\begin{array}{lll}\vec{a} & \vec{b}+\vec{c} & \vec{a}+\vec{b}+\vec{c}\end{array}]$ .

Step 3: Observe that $\vec{a}+\vec{b}+\vec{c} = \vec{a} + (\vec{b}+\vec{c})$ . This means that the vector $\vec{a}+\vec{b}+\vec{c}$ is a linear combination of $\vec{a}$ and $\vec{b}+\vec{c}$ .

Step 4: Therefore, the vectors $\vec{a}, \vec{b}+\vec{c}$ , and $\vec{a}+\vec{b}+\vec{c}$ are coplanar.

Step 5: Since the vectors are coplanar, their scalar triple product is zero: $[\begin{array}{lll}\vec{a} & \vec{b}+\vec{c} & \vec{a}+\vec{b}+\vec{c}\end{array}] = 0$ .

Why other options are wrong:
- Option A: $2\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]$ : This is incorrect as the scalar triple product is zero.
- Option B: $\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]$ : This is incorrect as the scalar triple product is zero.
- Option D: none of these: This is incorrect as the scalar triple product is zero.

Key Concepts:
- Scalar triple product and coplanarity

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

Quick Tips:
- If one of the vectors in a scalar triple product can be expressed as a linear combination of the other two, the scalar triple product is zero.

111 Easy

111 of 185

Application +4 | -1

If the volume of a parallelepiped, whose coterminous edges are given by the vectors

a = \hat i + \hat j + n\hat k,\,\,\,\,b = 2\hat i + 4\hat j - n\hat k

and

c = \hat i + n\hat j + 3\hat k\left( {n \ge 0} \right),

is 158 cu units, then

View answer and explanation

Correct option: B

b \cdot c = 10

Option B (

b \cdot c = 10

) is correct because after solving for n, the dot product of vectors b and c equals 10.

Solution approach

Correct Answer + Core Reason: $b \cdot c = 10$ . The volume of the parallelepiped is used to solve for n, which is then used to calculate the dot product of b and c.

Step-by-Step Solution:

Step 1: Set up the determinant for the volume of the parallelepiped using the given vectors.\nVolume = $\left| {\begin{{array}}{{{*{20}{c}}}} 1&1&n\\ 2&4&{ - n}\\ 1&n&3 \end{{array}}} \right| = 158$

Step 2: Evaluate the determinant.
$1(12 + {n^2}) - 1(6 + n) + n(2n - 4) = 158$
$12 + {n^2} - 6 - n + 2{n^2} - 4n = 158$
$3{n^2} - 5n + 6 = 158$
$3{n^2} - 5n - 152 = 0$

Step 3: Solve the quadratic equation for n.
$3{n^2} - 24n + 19n - 152 = 0$
$3n(n - 8) + 19(n - 8) = 0$
$(3n + 19)(n - 8) = 0$
$n = 8$ (since $n \ge 0$ )

Step 4: Substitute n = 8 into vectors a, b, and c.
$a = \hat i + \hat j + 8\hat k$
$b = 2\hat i + 4\hat j - 8\hat k$
$c = \hat i + 8\hat j + 3\hat k$

Step 5: Calculate the dot product of b and c.
$b \cdot c = (2)(1) + (4)(8) + (-8)(3) = 2 + 32 - 24 = 10$

Step 6: Calculate the dot product of a and c.
$a \cdot c = (1)(1) + (1)(8) + (8)(3) = 1 + 8 + 24 = 33$

Final Answer: $b \cdot c = 10$

Why other options are wrong:
- Option A: $n = 9$ is incorrect because the correct value of n is 8, which is obtained by solving the volume equation.
- Option C: $a \cdot c = 17$ is incorrect because the correct dot product of a and c is 33, not 17.
- Option D: $n = 7$ is incorrect because the correct value of n is 8, which is obtained by solving the volume equation.

Key Concepts:
- Volume of a parallelepiped using scalar triple product.
- Solving quadratic equations.
- Dot product of vectors.

Memory Aids:
- The volume of a parallelepiped is given by the absolute value of the scalar triple product of its coterminous edges.
- Remember the formula for the dot product: $a \cdot b = a_1b_1 + a_2b_2 + a_3b_3$

Quick Tips:
- Always check the condition given for the variable (e.g., $n \ge 0$ ).
- Double-check your calculations when evaluating determinants and dot products.

112 Easy

112 of 185

Application +4 | -1

If the vectors

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

and

\vec{d}

are coplanar vectors, then

(\vec{a} \times \vec{b}) \times(\vec{c} \times \vec{d})

is equal to

View answer and explanation

Correct option: B

\overrightarrow{0}

Option B (

\overrightarrow{{0}}

) is correct because if the vectors are coplanar, then

\vec{{a}} \times \vec{{b}}

and

\vec{{c}} \times \vec{{d}}

are parallel, and the cross product of parallel vectors is the zero vector.

Solution approach

Correct Answer + Core Reason: $\overrightarrow{{0}}$ . Coplanar vectors imply parallel cross products, resulting in a zero vector.

Step-by-Step Solution:

Step 1: Given that $\vec{{a}}, \vec{{b}}, \vec{{c}}$ , and $\vec{{d}}$ are coplanar vectors.

Step 2: Consider the cross products $\vec{{a}} \times \vec{{b}}$ and $\vec{{c}} \times \vec{{d}}$ . The vector $\vec{{a}} \times \vec{{b}}$ is perpendicular to the plane containing $\vec{{a}}$ and $\vec{{b}}$ . Similarly, $\vec{{c}} \times \vec{{d}}$ is perpendicular to the plane containing $\vec{{c}}$ and $\vec{{d}}$ .

Step 3: Since all four vectors are coplanar, the vectors $\vec{{a}} \times \vec{{b}}$ and $\vec{{c}} \times \vec{{d}}$ are both perpendicular to the same plane. Therefore, they must be parallel to each other.

Step 4: The cross product of two parallel vectors is the zero vector. Thus, $(\vec{{a}} \times \vec{{b}}) \times (\vec{{c}} \times \vec{{d}}) = \overrightarrow{{0}}$ .

Why other options are wrong:
- Option A: This is incorrect because the sum of the vectors is not necessarily zero.
- Option C: This is incorrect because the equality $\vec{{a}} + \vec{{b}} = \vec{{c}} + \vec{{d}}$ does not necessarily hold.
- Option D: This is incorrect because the correct answer is $\overrightarrow{{0}}$ .

Key Concepts:
- Coplanar vectors
- Cross product
- Parallel vectors

Memory Aids:
- The cross product of two vectors is perpendicular to both vectors.
- The cross product of two parallel vectors is the zero vector.

Quick Tips:
- When dealing with coplanar vectors, remember that their cross products will be parallel.

113 Easy

113 of 185

Analyzing +4 | -1

Let r, a, b and c be four non-zero vectors such that

r.\,\,a = 0,

|r \times b| = |r||b|,\,\,|r \times c| = |r\parallel c\mid ,

then [a b c] is

View answer and explanation

Correct option: C

0

Option C (0) is correct because the given conditions imply that a is perpendicular to both b and c, making the scalar triple product zero.

Solution approach

Correct Answer + Core Reason: 0. The given conditions imply that a is perpendicular to both b and c, making the scalar triple product zero.

Step-by-Step Solution:

Step 1: State the given conditions: $r \cdot a = 0$ , $|r \times b| = |r||b|$ , and $|r \times c| = |r||c|$ .

Step 2: Interpret the conditions: $r \cdot a = 0$ means r is perpendicular to a. $|r \times b| = |r||b|$ means the angle between r and b is 90 degrees, so r is perpendicular to b. Similarly, r is perpendicular to c.

Step 3: Since r is perpendicular to both b and c, r is parallel to $b \times c$ .

Step 4: Since r is perpendicular to a, it means a is perpendicular to $b \times c$ .

Step 5: Therefore, $[a \, b \, c] = a \cdot (b \times c) = 0$ , because a is perpendicular to $b \times c$ .

Why other options are wrong:
- Option A: $|a|\,|b|\,|c|$ : This would be the volume if a, b, and c were mutually perpendicular unit vectors, which is not the case here.
- Option B: $-|a|\,|b|\,|c|$ : This is incorrect for the same reason as option A.
- Option D: None of these: This is incorrect because we found that [a b c] = 0.

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

Memory Aids:
- Remember that $a \cdot (b \times c) = [abc]$
- Recall that the scalar triple product is zero if the vectors are coplanar.

Quick Tips:
- When dealing with perpendicularity, use the dot product.
- Remember the properties of the scalar triple product.

114 Easy

114 of 185

Analyzing +4 | -1

If

\alpha(\mathbf{a} \times \mathbf{b})+\beta(\mathbf{b} \times \mathbf{c})+\gamma(\mathbf{c} \times \mathbf{a})=0

, then

View answer and explanation

Correct option: A

a, b, c

are coplanar if all of

\alpha, \beta, \gamma \neq 0

Option A is correct because if all coefficients are non-zero, the scalar triple product must be zero, indicating coplanarity.

Solution approach

Correct Answer + Core Reason: $a, b, c$ are coplanar if all of $\alpha, \beta, \gamma \neq 0$ . The linear combination of vector products being zero implies coplanarity when all coefficients are non-zero.

Step-by-Step Solution:

Step 1: Given the equation $\alpha(\mathbf{a} \times \mathbf{b})+\beta(\mathbf{b} \times \mathbf{c})+\gamma(\mathbf{c} \times \mathbf{a})=0$ .

Step 2: Take the dot product of the equation with vector $\mathbf{c}$ : $\alpha [(\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c}] + \beta [(\mathbf{b} \times \mathbf{c}) \cdot \mathbf{c}] + \gamma [(\mathbf{c} \times \mathbf{a}) \cdot \mathbf{c}] = 0$ .

Step 3: Simplify using the scalar triple product property: $\alpha [\mathbf{a} \,\mathbf{b} \,\mathbf{c}] + \beta (0) + \gamma (0) = 0$ , which simplifies to $\alpha [\mathbf{a} \,\mathbf{b} \,\mathbf{c}] = 0$ .

Step 4: Similarly, taking dot product with $\mathbf{a}$ and $\mathbf{b}$ gives $\beta [\mathbf{a} \,\mathbf{b} \,\mathbf{c}] = 0$ and $\gamma [\mathbf{a} \,\mathbf{b} \,\mathbf{c}] = 0$ .

Step 5: If $\alpha, \beta, \gamma \neq 0$ , then $[\mathbf{a} \,\mathbf{b} \,\mathbf{c}] = 0$ , which means $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are coplanar.

Why other options are wrong:
- Option B: This is incorrect because if any one of $\alpha, \beta, \gamma = 0$ , it doesn't necessarily imply that the vectors are coplanar. For example, if $\alpha = 0$ , then $\beta(\mathbf{b} \times \mathbf{c})+\gamma(\mathbf{c} \times \mathbf{a})=0$ doesn't automatically mean the vectors are coplanar.
- Option C: This is incorrect because the condition given can lead to coplanarity under certain conditions (when all coefficients are non-zero).
- Option D: This is incorrect because option A is correct.

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

Memory Aids:
- Scalar triple product represents the volume of a parallelepiped. If it's zero, the vectors are coplanar.

Quick Tips:
- Remember the properties of the scalar triple product: $[\mathbf{a} \,\mathbf{b} \,\mathbf{c}] = [\mathbf{b} \,\mathbf{c} \,\mathbf{a}] = [\mathbf{c} \,\mathbf{a} \,\mathbf{b}] = -[\mathbf{b} \,\mathbf{a} \,\mathbf{c}] = ...$

115 Easy

115 of 185

Application +4 | -1

If

a = \widehat i + \widehat j + \widehat k,\,\,\,\,b = \widehat i - \widehat j + \widehat k,\,\,\,c = \widehat i + 2\widehat j - \widehat k,

then the value of

\left| {\begin{array}{cccccccccccccccccccc} {a \cdot a}&{a \cdot b}&{a \cdot c} \\ {b \cdot a}&{b \cdot b}&{b \cdot c} \\ {c \cdot a}&{c \cdot b}&{c \cdot c} \end{array}} \right|\,

is

View answer and explanation

Correct option: C

16

Option C (16) is correct because the determinant of the matrix formed by dot products is equal to the square of the scalar triple product.

Solution approach

Correct Answer + Core Reason: 16. The determinant of the matrix is the square of the scalar triple product.

Step-by-Step Solution:

Step 1: Given $a = \widehat i + \widehat j + \widehat k, b = \widehat i - \widehat j + \widehat k, c = \widehat i + 2\widehat j - \widehat k$

Step 2: $\left| {\begin{{array}}{{{*{{20}}{c}}}} {a \cdot a}&{a \cdot b}&{a \cdot c} \\ {b \cdot a}&{b \cdot b}&{b \cdot c} \\ {c \cdot a}&{c \cdot b}&{c \cdot c} \end{{array}}} \right| = {[a \, b \, c]^2}$

Step 3: Calculate the scalar triple product: $[a \, b \, c] = \left| {\begin{{array}}{{{*{{20}}{c}}}} 1&1&1 \\ 1&{ - 1}&1 \\ 1&2&{ - 1} \end{{array}}} \right|$

Step 4: Evaluate the determinant: $[1(1 - 2) - 1( - 1 - 1) + 1(2 + 1)] = [-1 + 2 + 3] = 4$

Step 5: Therefore, the value of the determinant is $4^2 = 16$

Why other options are wrong:
- Option A: This option is incorrect because it is not the square of the scalar triple product.
- Option B: This option is incorrect because it is not the square of the scalar triple product.
- Option D: This option is incorrect because it is not the square of the scalar triple product.

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

Memory Aids:
- Remember that the determinant of the matrix formed by dot products is equal to the square of the scalar triple product.

Quick Tips:
- Calculate the scalar triple product first and then square it.

116 Easy

116 of 185

Understanding +4 | -1

The value of

\widehat i \cdot (\widehat j \times \widehat k) + \widehat j \cdot (\widehat k \times \widehat i) + \widehat k \cdot (\widehat i \times \widehat j)

is

View answer and explanation

Correct option: A

3

Option A (3) is correct because each term in the expression equals 1, and there are three terms.

Solution approach

Correct Answer + Core Reason: 3. Each term in the expression equals 1, and there are three terms.

Step-by-Step Solution:

Step 1: Evaluate the first term: $\widehat i \cdot (\widehat j \times \widehat k)$ . Since $\widehat j \times \widehat k = \widehat i$ , the term becomes $\widehat i \cdot \widehat i = 1$ .

Step 2: Evaluate the second term: $\widehat j \cdot (\widehat k \times \widehat i)$ . Since $\widehat k \times \widehat i = \widehat j$ , the term becomes $\widehat j \cdot \widehat j = 1$ .

Step 3: Evaluate the third term: $\widehat k \cdot (\widehat i \times \widehat j)$ . Since $\widehat i \times \widehat j = \widehat k$ , the term becomes $\widehat k \cdot \widehat k = 1$ .

Step 4: Add the three terms: $1 + 1 + 1 = 3$ .

Why other options are wrong:
- Option B (2): Incorrect calculation.
- Option C (1): Incorrect calculation.
- Option D (0): Incorrect calculation.

Key Concepts:
- Vector Cross Product
- Vector Dot Product
- Unit Vectors

Memory Aids:
- Remember the cyclic order: $\widehat i \times \widehat j = \widehat k$ , $\widehat j \times \widehat k = \widehat i$ , $\widehat k \times \widehat i = \widehat j$ .
- $\widehat i \cdot \widehat i = \widehat j \cdot \widehat j = \widehat k \cdot \widehat k = 1$ .

Quick Tips:
- Use the properties of unit vectors and their cross and dot products to simplify the expression.

117 Easy

117 of 185

Analyzing +4 | -1

The number of distinct real values of

\lambda

for which the vectors

\vec{a}=\lambda^3 \hat{i}+\hat{k}, \vec{b}=\hat{i}-\lambda^3 \hat{j}

and

\vec{c}=\hat{i}+(2 \lambda-\sin \lambda) \hat{j}-\lambda \hat{k}

are coplanar is

View answer and explanation

Correct option: B

1

Option B (1) is correct because the equation derived from the coplanarity condition has only one distinct real solution, which is λ = 0.

Solution approach

Correct Answer + Core Reason: 1. The equation derived from the coplanarity condition has only one real solution.

Step-by-Step Solution:

Step 1: Given $\vec{a}=\lambda^3 \hat{i}+\hat{k}, \vec{b}=\hat{i}-\lambda^3 \hat{j}$ and $\vec{c}=\hat{i}+(2 \lambda-\sin \lambda) \hat{j}-\lambda \hat{k}$ .

Step 2: For $\vec{a}, \vec{b}, \vec{c}$ to be coplanar, their scalar triple product must be zero: $[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}] = 0$ .

Step 3: Calculate the scalar triple product: $\left| {\begin{array}{cccccccccccccccccccc} {\lambda^3}&0&1 \ 1&{-\lambda ^3}&0 \ 1&{2\lambda - \sin \lambda }&{ - \lambda } \end{array}} \right| = 0$ .

Step 4: Expand the determinant: $\lambda^3(\lambda^4 - 0) - 0 + 1(2\lambda - \sin \lambda + \lambda^3) = \lambda^7 + \lambda^3 + 2\lambda - \sin \lambda = 0$ .

Step 5: Rearrange the equation: $\lambda^7 + \lambda^3 + 2\lambda = \sin \lambda$ .

Step 6: Observe that $\lambda = 0$ is a solution.

Step 7: For non-zero values of $\lambda$ , divide by $\lambda$ : $\lambda^6 + \lambda^2 + 2 = \frac{\sin \lambda}{\lambda}$ .

Step 8: Analyze the equation: Since $\frac{\sin x}{x} < 1$ for all $x \neq 0$ , and $\lambda^6 + \lambda^2 + 2 > 2$ for all $\lambda \neq 0$ , there are no non-zero solutions.

Step 9: Therefore, there is only one distinct real value of $\lambda$ , which is $\lambda = 0$ .

Why other options are wrong:
- Option A: 0: This is incorrect as there is one real value of $\lambda$ that satisfies the condition.
- Option C: 2: This is incorrect as there is only one real value of $\lambda$ that satisfies the condition.
- Option D: 3: This is incorrect as there is only one real value of $\lambda$ that satisfies the condition.

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.
- Analyze the equation carefully to find the number of real solutions.

118 Easy

118 of 185

Application +4 | -1

If the vectors

\widehat \mathbf{i} - 2\widehat \mathbf{j} + 3\mathbf{\hat k},\;\; - 2\widehat \mathbf{i} + 3\widehat \mathbf{j} - 4\mathbf{\hat k},\;\mathrm{\lambda \;}\widehat \mathbf{i} - \widehat \mathbf{j} + 2\mathbf{\hat k}\;

are linearly dependent, then the value of

\mathrm{\lambda }

is equal to

View answer and explanation

Correct option: A

0

Option A (0) is correct because linearly dependent vectors have a scalar triple product of zero, leading to lambda = 0.

Solution approach

Correct Answer + Core Reason: 0. The value of $\mathrm{\lambda }$ is found by setting the determinant of the matrix formed by the vectors' components to zero.

Step-by-Step Solution:

Step 1: Set up the determinant equation for linearly dependent vectors:
$\left| {\begin{{array}}{*{20}{c}}\n1&{ - 2}&3\\n{ - 2}&3&{ - 4}\\\lambda &{ - 1}&2\n\end{{array}}} \right| = 0$

Step 2: Expand the determinant:
$1\left( {6 - 4} \right) + 2\left( { - 4 + 4\lambda } \right) + 3\left( {2 - 3\lambda } \right) = 0$

Step 3: Simplify the equation:
$2 - 8 + 8\lambda + 6 - 9\lambda = 0$

Step 4: Combine like terms:
$- \lambda = 0$

Step 5: Solve for λ:
$\lambda = 0$

Why other options are wrong:
- Option B (1): This value is incorrect because it doesn't satisfy the linear dependence condition when substituted back into the determinant equation.
- Option C (2): This value is incorrect because it doesn't satisfy the linear dependence condition when substituted back into the determinant equation.
- Option D (3): This value is incorrect because it doesn't satisfy the linear dependence condition when substituted back into the determinant equation.

Key Concepts:
- Linearly dependent vectors
- Determinant of a matrix

Memory Aids:
- Linearly dependent vectors have a scalar triple product of zero, which is equivalent to the determinant being zero.

Quick Tips:
- Remember to expand the determinant correctly, paying attention to signs.
- Always double-check your calculations to avoid arithmetic errors.

119 Easy

119 of 185

Application +4 | -1

If

\vec{r}

is a unit vector such that

\vec{r}=x(\vec{b} \times \vec{c})+y(\vec{c} \times \vec{a})+z(\vec{a} \times \vec{b}),

then

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

is equal to

View answer and explanation

Correct option: A

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

Option A (

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

) is correct because the expression simplifies to the absolute value of the scalar triple product.

Solution approach

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

Step-by-Step Solution:

Step 1: State the given condition: $\vec{r}=x(\vec{b} \times \vec{c})+y(\vec{c} \times \vec{a})+z(\vec{a} \times \vec{b})$ .

Step 2: Take the dot product with $\vec{a}, \vec{b}, \vec{c}$ : $\vec r \cdot \,\vec a = x\left[ {\vec a\,\,\vec b\,\,\vec c} \right],\,\vec r.\,\vec b = y\left[ {\vec a\,\,\vec b\,\,\vec c} \right]\,{\text{and}}\,\vec r \cdot \,\vec c = z\left[ {\vec a\,\,\vec b\,\,\vec c} \right]$ .

Step 3: Substitute the values of $x, y, z$ into the expression for $\vec{r}$ : $\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]$

120 Easy

120 of 185

Application +4 | -1

Volume of the parallelopiped having vertices at

O \equiv \left( {0,0,0} \right),\;\;\;A \equiv \left( {2, - 2,4} \right),\;

B \equiv \left( {5, - 4,4} \right)

and

C \equiv \left( {1, - 2,4} \right)

View answer and explanation

Correct option: B

10 cu units

Option B (10 cu units) is correct because it is the volume of the parallelepiped calculated using the scalar triple product of the vectors formed by the given vertices.

Solution approach

Correct Answer + Core Reason: 10 cu units. This is the volume of the parallelepiped calculated using the scalar triple product.

Step-by-Step Solution:

Step 1: Find the vectors $\overrightarrow \mathbf{OA}$ , $\overrightarrow \mathbf{OB}$ , and $\overrightarrow \mathbf{OC}$ .
$\overrightarrow \mathbf{OA} = 2\widehat \mathbf{i} - 2\widehat \mathbf{j} + \mathbf{\hat k}$
$\overrightarrow \mathbf{OB} = 5\widehat \mathbf{i} - 4\widehat \mathbf{j} + 4\mathbf{\hat k}$
$\overrightarrow \mathbf{OC} = \widehat \mathbf{i} - 2\widehat \mathbf{j} + 4\mathbf{\hat k}$

Step 2: Calculate the volume of the parallelepiped using the scalar triple product: $V = |[\overrightarrow \mathbf{OA} \; \overrightarrow \mathbf{OB} \; \overrightarrow \mathbf{OC} ]| = \left| {\begin{array}{cccccccccccccccccccc} 2&{ - 2}&1\\ 5&{ - 4}&4\\ 1&{ - 2}&4 \end{array}} \right|$ .

Step 3: Evaluate the determinant.
$V = |2((-4)(4) - (4)(-2)) - (-2)((5)(4) - (4)(1)) + 1((5)(-2) - (-4)(1))|$
$V = |2(-16 + 8) + 2(20 - 4) + (-10 + 4)|$
$V = |2(-8) + 2(16) + (-6)|$
$V = |-16 + 32 - 6|$
$V = |10| = 10$ .

Why other options are wrong:
- Option A (5 cu units): This is incorrect. It might result from a calculation error in evaluating the determinant.
- Option C (15 cu units): This is incorrect. It might result from a calculation error in evaluating the determinant.
- Option D (20 cu units): This is incorrect. It might result from a calculation error in evaluating the determinant.

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

Memory Aids:
- The volume of a parallelepiped formed by vectors $\vec a$ , $\vec b$ , and $\vec c$ is given by $|\vec a \cdot (\vec b \times \vec c)|$ .

Quick Tips:
- Always take the absolute value of the scalar triple product to ensure the volume is positive.

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Scalar triple product practice · Easy

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