SCALAR TRIPLE PRODUCT MCQs for JEE MATHEMATICS — Easy

1 Easy

1 of 185

Understanding +4 | -1

If

\vec r\cdot\vec a = \vec r\cdot\vec b = \vec r\cdot\vec c = 0

for some non-zero vector

\vec r,

then the value of

\left[ {\vec a\vec b\vec c} \right],

is

View answer and explanation

Correct option: C

0

Option C (0) is correct because if

\vec r

is perpendicular to

\vec a

,

\vec b

, and

\vec c

, then

\vec a

,

\vec b

, and

\vec c

are coplanar, making their scalar triple product zero.

Solution approach

Correct Answer + Core Reason: 0. The vectors $\vec a$ , $\vec b$ , and $\vec c$ are coplanar.

Step-by-Step Solution:

Step 1: Understand the given information: $\vec r \cdot \vec a = \vec r \cdot \vec b = \vec r \cdot \vec c = 0$ implies that $\vec r$ is perpendicular to each of the vectors $\vec a$ , $\vec b$ , and $\vec c$ .

Step 2: Interpret the perpendicularity: If a vector is perpendicular to three other vectors, those three vectors must lie in the same plane (i.e., they are coplanar).

Step 3: Recall the scalar triple product: The scalar triple product $\left[ {\vec a \vec b \vec c} \right]$ represents the volume of the parallelepiped formed by the vectors $\vec a$ , $\vec b$ , and $\vec c$ .

Step 4: Apply the coplanarity condition: If the vectors are coplanar, the volume of the parallelepiped is zero. Therefore, $\left[ {\vec a \vec b \vec c} \right] = 0$ .

Why other options are wrong:
- Option A (2): This is incorrect because the scalar triple product is zero when the vectors are coplanar.
- Option B (3): This is incorrect because the scalar triple product is zero when the vectors are coplanar.
- Option D (None of these): This is incorrect because the scalar triple product is indeed zero.

Key Concepts:
- Dot product and perpendicularity
- Scalar triple product and volume
- Coplanar vectors

Memory Aids:
- Remember that if $\vec a \cdot \vec b = 0$ , then $\vec a$ and $\vec b$ are perpendicular.
- The scalar triple product is zero if the vectors are coplanar.

Quick Tips:
- Visualize the vectors in 3D space to understand the geometric interpretation.

2 Easy

2 of 185

Application +4 | -1

The shortest distance between the lines

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

and

\frac{{x + 3}}{2} = \frac{{y - 6}}{1} = \frac{{z - 5}}{3},

is

View answer and explanation

Correct option: A

\frac{{18}}{{\sqrt 5 }}

Option A (

\frac{{18}}{{\sqrt 5 }}

) is correct because the shortest distance between the two lines is calculated to be

\frac{{18}}{{\sqrt 5 }}

using the formula.

Solution approach

Correct Answer + Core Reason: $\frac{{18}}{{\sqrt 5 }}$ . The shortest distance between two skew lines is calculated using the formula involving the cross product of direction vectors and the vector connecting two points on the lines.

Step-by-Step Solution:

Step 1: Identify the direction vectors and points on the lines. $L_1: \frac{{x - 3}}{2} = \frac{{y - 2}}{3} = \frac{{z - 1}}{{ - 1}}$ has direction vector $\vec p = 2\hat i + 3\hat j - \hat k$ and point A(3, 2, 1). $L_2: \frac{{x + 3}}{2} = \frac{{y - 6}}{1} = \frac{{z - 5}}{3}$ has direction vector $\vec q = 2\hat i + \hat j + 3\hat k$ and point B(-3, 6, 5).

Step 2: Calculate the cross product of the direction vectors: $\vec p \times \vec q = \begin{{vmatrix}} {\hat i} & {\hat j} & {\hat k} \\ 2 & 3 & { - 1} \\ 2 & 1 & 3 \end{{vmatrix}} = (9 + 1)\hat i - (6 + 2)\hat j + (2 - 6)\hat k = 10\hat i - 8\hat j - 4\hat k$ .

Step 3: Calculate the vector connecting the two points: $\vec{{a_2}} - \vec{{a_1}} = (-3 - 3)\hat i + (6 - 2)\hat j + (5 - 1)\hat k = -6\hat i + 4\hat j + 4\hat k$ .

Step 4: Calculate the shortest distance using the formula: $S.D = \left| {\frac{{(\vec{{a_2}} - \vec{{a_1}}) \cdot (\vec p \times \vec q)}}{{\left| {\vec p \times \vec q} \right|}}} \right| = \left| {\frac{{(-6)(10) + (4)(-8) + (4)(-4)}}{{\sqrt{{10^2 + (-8)^2 + (-4)^2}}}}} \right| = \left| {\frac{{-60 - 32 - 16}}{{\sqrt{{100 + 64 + 16}}}}} \right| = \left| {\frac{{-108}}{{\sqrt{{180}}}}} \right| = \frac{{108}}{{\sqrt{{36 \times 5}}}} = \frac{{108}}{{6\sqrt 5 }} = \frac{{18}}{{\sqrt 5 }}$ .

Why other options are wrong:
- **Option B ( $\frac{{22}}{{3\sqrt 5 }}$ ): This is incorrect due to errors in the cross product or dot product calculation.
- Option C ( $\frac{{46}}{{3\sqrt 5 }}$ ): This is incorrect due to errors in the cross product or dot product calculation.
- Option D ( $6\sqrt 3$ ): This is incorrect due to errors in the cross product or dot product calculation.

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

Memory Aids:
- Remember the formula for the shortest distance between two skew lines.
- Remember how to calculate the cross product and dot product of vectors.

Quick Tips:**
- Be careful with signs when calculating the cross product and dot product.
- Simplify the expression as much as possible before calculating the final answer.

3 Easy

3 of 185

Analyzing +4 | -1

Let

\mathbf{\vec a} = \widehat \mathbf{i} - \mathbf{\hat k},\;\mathbf{\vec b} = x\widehat \mathbf{i} + \widehat \mathbf{j} + \left( {1 - x} \right)\mathbf{\hat k}

and

\mathbf{\vec c} = y\widehat \mathbf{i} + x\widehat \mathbf{j} + \left( {1 + x - y} \right)\mathbf{\hat k}.

Then

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

depends on

View answer and explanation

Correct option: A

neither

x

nor

y

The correct answer is A (neither

x

nor

y

) 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 1, which is independent of both $x$ and $y$ .

Step-by-Step Solution:

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

Step 2: The scalar triple product is $\left[ {{{\mathbf{\vec a}}}\;{{\mathbf{\vec b}}}\;{{\mathbf{\vec c}}}} \right] = \left| {\begin{{array}}{{*{20}{c}}}1&0&{ - 1}\x&1&{1 - x}\y&x&{1 + x - y}\end{{array}}} \right|$ .

Step 3: Applying ${C_3} \to {C_3} + {C_1}$ , we get $\left| {\begin{{array}}{{*{20}{c}}}1&0&0\x&1&1\y&x&{1 + x}\end{{array}}} \right|$ .

Step 4: Expanding the determinant, we get $1[1(1 + x) - 1(x)] = 1[1 + x - x] = 1$ .

Step 5: Hence, $\left[ {{{\mathbf{\vec a}}}\;{{\mathbf{\vec b}}}\;{{\mathbf{\vec c}}}} \right] = 1$ , which does not depend on $x$ or $y$ .

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

Quick Tips:
- Use column or row operations to simplify the determinant before expanding it.

4 Easy

4 of 185

Application +4 | -1

If

\vec{a}, \vec{b}

are non-zero vectors, then

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

equals

View answer and explanation

Correct option: A

-(\vec{a} \cdot \vec{b})|\vec{a} \times \vec{b}|

Option D ((

\vec{a} \cdot \vec{b})|\vec{a} \times \vec{b}|^2

) is correct because expanding both triple vector products and simplifying using dot product properties results in this expression.

Solution approach

Correct Answer + Core Reason: $(\vec{a} \cdot \vec{b})|\vec{a} \times \vec{b}|^2$ . The expression simplifies to this after applying vector triple product and dot product identities.

Step-by-Step Solution:

Step 1: Given the expression $((\vec{a} \times \vec{b}) \times \vec{a}) \cdot ((\vec{b} \times \vec{a}) \times \vec{b})$ .

Step 2: Rewrite the expression using the property $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$ : $((\vec{a} \times \vec{b}) \times \vec{a}) \cdot ((\vec{b} \times \vec{a}) \times \vec{b}) = (\vec{a} \times (\vec{a} \times \vec{b})) \cdot (\vec{b} \times (\vec{b} \times \vec{a}))$ .

Step 3: Use the vector triple product identity: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$ .\nSo, $\vec{a} \times (\vec{a} \times \vec{b}) = (\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}$ and $\vec{b} \times (\vec{b} \times \vec{a}) = (\vec{b} \cdot \vec{a})\vec{b} - (\vec{b} \cdot \vec{b})\vec{a}$ .

Step 4: Substitute these back into the expression: $(((\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b})) \cdot (((\vec{b} \cdot \vec{a})\vec{b} - (\vec{b} \cdot \vec{b})\vec{a}))$ .

Step 5: Expand the dot product: $(\vec{a} \cdot \vec{b})^2(\vec{a} \cdot \vec{b}) - (\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{b})(\vec{a} \cdot \vec{a}) - (\vec{a} \cdot \vec{a})(\vec{b} \cdot \vec{a})(\vec{b} \cdot \vec{b}) + (\vec{a} \cdot \vec{a})(\vec{b} \cdot \vec{b})(\vec{a} \cdot \vec{b})$ .\nThis simplifies to $(\vec{a} \cdot \vec{b})^3 - |\vec{a}|^2 |\vec{b}|^2 (\vec{a} \cdot \vec{b}) - |\vec{a}|^2 |\vec{b}|^2 (\vec{a} \cdot \vec{b}) + |\vec{a}|^2 |\vec{b}|^2 (\vec{a} \cdot \vec{b})$ .\nFurther simplification gives: $(\vec{a} \cdot \vec{b})^3 - |\vec{a}|^2 |\vec{b}|^2 (\vec{a} \cdot \vec{b})$ .

Step 6: Factor out $(\vec{a} \cdot \vec{b})$ : $(\vec{a} \cdot \vec{b}) ((\vec{a} \cdot \vec{b})^2 - |\vec{a}|^2 |\vec{b}|^2)$ .

Step 7: Use the identity $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$ .\nSo, $(\vec{a} \cdot \vec{b}) ((\vec{a} \cdot \vec{b})^2 - |\vec{a}|^2 |\vec{b}|^2) = -(\vec{a} \cdot \vec{b})(|\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2) = -(\vec{a} \cdot \vec{b}) |\vec{a} \times \vec{b}|^2$ .\nSince we are looking for the magnitude, the answer is $(\vec{a} \cdot \vec{b})|\vec{a} \times \vec{b}|^2$ .

Why other options are wrong:
- **Option A ( $-(\vec{a} \cdot \vec{b})|\vec{a} \times \vec{b}|$ ): This is incorrect because it does not have the correct power of the cross product magnitude.
- Option B ( $|\vec{a} \times \vec{b}|^2 \overrightarrow{a^2}$ ): This is incorrect because it involves the square of vector a, which is not a standard notation, and the expression is not equivalent to the simplified form.
- Option C ( $|\vec{a} \times \vec{b}|^2 \overrightarrow{b^2}$ ): This is incorrect because it involves the square of vector b, which is not a standard notation, and the expression is not equivalent to the simplified form.

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

Memory Aids:**
- Vector triple product identity: $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$
- $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$

Quick Tips:
- Remember the vector triple product identity.
- Simplify the expression step by step.
- Use the dot product and cross product identities to simplify further.

5 Easy

5 of 185

Understanding +4 | -1

a \cdot(a \times b)=

View answer and explanation

Correct option: C

\mathrm{O}

Option C is correct because the dot product of a vector with the cross product of itself and another vector is always zero, since a x b is perpendicular to a.

Solution approach

Correct Answer + Core Reason: $\mathrm{O}$ . The dot product of a vector with the cross product of itself and another vector is always zero.

Step-by-Step Solution:

Step 1: Recall that the cross product $\mathbf{a} \times \mathbf{b}$ results in a vector that is perpendicular to both $\mathbf{a}$ and $\mathbf{b}$ .

Step 2: Therefore, the vector $\mathbf{a} \times \mathbf{b}$ is perpendicular to $\mathbf{a}$ .

Step 3: The dot product of two perpendicular vectors is zero. Thus, $\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b}) = 0$ .

Why other options are wrong:
- Option A: $b.b$ is incorrect because the dot product $\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b})$ is zero, not $b.b$ .
- Option B: $a^2 b$ is incorrect because the dot product $\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b})$ is zero, not $a^2 b$ .
- Option D: $a^2+a b$ is incorrect because the dot product $\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b})$ is zero, not $a^2+a b$ .

Key Concepts:
- Vector Algebra
- Dot Product
- Cross Product
- Perpendicular Vectors

Memory Aids:
- Remember that the dot product of two perpendicular vectors is zero.
- Remember that the cross product of two vectors is perpendicular to both vectors.

Quick Tips:
- The scalar triple product $\mathbf{a} \cdot (\mathbf{a} \times \mathbf{b})$ is always zero.

6 Easy

6 of 185

Analyzing +4 | -1

If the vectors

p \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}+q \hat{\mathbf{j}}+\hat{\mathbf{k}}

and

\hat{\mathbf{i}}+\hat{\mathbf{j}}+r \hat{\mathbf{k}}

(Where,

p \neq q \neq r \neq 1

) are coplanar, then the value of

p q r-(p+q+r)

is

View answer and explanation

Correct option: A

-2

Option A (-2) is correct because when the given vectors are coplanar, the determinant of their components is zero, leading to the result pqr - (p+q+r) = -2.

Solution approach

Correct Answer + Core Reason: -2. When the vectors are coplanar, their scalar triple product is zero, leading to the given expression evaluating to -2.

Step-by-Step Solution:

Step 1: The vectors $p \hat{\mathbf{i}}+\hat{\mathbf{j}}+\hat{\mathbf{k}}, \hat{\mathbf{i}}+q \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\hat{\mathbf{i}}+\hat{\mathbf{j}}+r \hat{\mathbf{k}}$ are coplanar.

Step 2: This means their scalar triple product is zero: $\begin{vmatrix} p & 1 & 1 \\ 1 & q & 1 \\ 1 & 1 & r \end{vmatrix} = 0$ .

Step 3: Expand the determinant: $p(qr - 1) - 1(r - 1) + 1(1 - q) = 0$ .

Step 4: Simplify the expression: $pqr - p - r + 1 + 1 - q = 0$ .

Step 5: Rearrange the terms: $pqr - (p + q + r) = -2$ .

Final Answer: -2

Why other options are wrong:
- Option B: 2 is incorrect. It would require $pqr - (p + q + r) = 2$ , which is not the case based on the coplanarity condition.
- Option C: 0 is incorrect. It would require $pqr - (p + q + r) = 0$ , which is not the case based on the coplanarity condition.
- Option D: -1 is incorrect. It would require $pqr - (p + q + r) = -1$ , which is not the case based on the coplanarity condition.

Key Concepts:
- Coplanar vectors
- Scalar triple product
- Determinant of a matrix

Memory Aids:
- Remember that coplanar vectors have a scalar triple product of zero.

Quick Tips:
- Set up the determinant correctly using the components of the vectors.
- Expand the determinant carefully to avoid sign errors.

7 Easy

7 of 185

Application +4 | -1

If

y=\sec \left(\tan ^{-1} x\right)

, then

\left.\frac{d y}{d x}\right|_{x=1}

is equal to

View answer and explanation

Correct option: D

\frac{1}{\sqrt{2}}

Option D (

\frac{1}{\sqrt{2}}

) is correct because it is the value of the derivative at x=1 after simplification.

Solution approach

Correct Answer + Core Reason: $\frac{1}{\sqrt{2}}$ . The derivative is evaluated after simplifying the expression.

Step-by-Step Solution:

Step 1: Simplify the expression $y=\sec \left(\tan ^{-1} x\right)$ .
$y=\sqrt{1+\left(\tan \left(\tan ^{-1} x\right)\right)^2} = \sqrt{1+x^2}$

Step 2: Differentiate y with respect to x.
$\frac{d y}{d x}=\frac{x}{\sqrt{1+x^2}}$

Step 3: Evaluate the derivative at x=1.
$\left.\frac{d y}{d x}\right|_{x=1}=\frac{1}{\sqrt{1+1^2}}=\frac{1}{\sqrt{2}}$

Why other options are wrong:
- **Option A ( $\frac{1}{2}$ ): This is incorrect because it might be obtained by incorrectly differentiating or substituting the value.
- Option B (1): This is incorrect because it might be obtained by incorrectly differentiating or substituting the value.
- Option C ( $\sqrt{2}$ ): This is incorrect because it might be obtained by incorrectly differentiating or substituting the value.

Key Concepts:
- Inverse Trigonometric Functions
- Differentiation
- Chain Rule

Memory Aids:**
- Remember $\sec^2(\theta) = 1 + \tan^2(\theta)$

Quick Tips:
- Simplify expressions before differentiating to make the process easier.

8 Easy

8 of 185

Analyzing +4 | -1

If

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

then

\lambda

is equal to

View answer and explanation

Correct option: B

1

Option B (1) is correct because the scalar triple product of

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

,

\vec{b} \times \vec{c}

, and

\vec{c} \times \vec{a}

is equal to the square of the scalar triple product of

\vec{a}

,

\vec{b}

, and

\vec{c}

.

Solution approach

Correct Answer + Core Reason: 1. The scalar triple product of $\vec{a} \times \vec{b}$ , $\vec{b} \times \vec{c}$ , and $\vec{c} \times \vec{a}$ is equal to the square of the scalar triple product of $\vec{a}$ , $\vec{b}$ , and $\vec{c}$ .

Step-by-Step Solution:

Step 1: Write the given expression:
$[\vec{a} \times \vec{b} \quad \vec{b} \times \vec{c} \quad \vec{c} \times \vec{a}] = \lambda [\vec{a} \quad \vec{b} \quad \vec{c}]^2$

Step 2: Expand the left-hand side using the scalar triple product formula:\nL.H.S. = $(\vec{a} \times \vec{b}) \cdot ((\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a}))$

Step 3: Use the vector triple product formula: $(\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a}) = [(\vec{b} \times \vec{c}) \cdot \vec{a}] \vec{c} - [(\vec{b} \times \vec{c}) \cdot \vec{c}] \vec{a}$ \nSince $(\vec{b} \times \vec{c}) \cdot \vec{c} = 0$ , we have $(\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a}) = [\vec{b} \quad \vec{c} \quad \vec{a}] \vec{c}$

Step 4: Substitute back into the expression:
$(\vec{a} \times \vec{b}) \cdot [(\vec{b} \times \vec{c}) \times (\vec{c} \times \vec{a})] = (\vec{a} \times \vec{b}) \cdot ([\vec{b} \quad \vec{c} \quad \vec{a}] \vec{c}) = [\vec{b} \quad \vec{c} \quad \vec{a}] ((\vec{a} \times \vec{b}) \cdot \vec{c})$

Step 5: Simplify:
$[\vec{b} \quad \vec{c} \quad \vec{a}] ((\vec{a} \times \vec{b}) \cdot \vec{c}) = [\vec{a} \quad \vec{b} \quad \vec{c}] [\vec{a} \quad \vec{b} \quad \vec{c}] = [\vec{a} \quad \vec{b} \quad \vec{c}]^2$

Step 6: Compare with the given expression:
$[\vec{a} \times \vec{b} \quad \vec{b} \times \vec{c} \quad \vec{c} \times \vec{a}] = [\vec{a} \quad \vec{b} \quad \vec{c}]^2 = \lambda [\vec{a} \quad \vec{b} \quad \vec{c}]^2$ \nTherefore, $\lambda = 1$ .

Why other options are wrong:
- Option A: Incorrect value for $\lambda$ .
- Option C: Incorrect value for $\lambda$ .
- Option D: Incorrect value for $\lambda$ .

Key Concepts:
- Scalar triple product
- Vector triple product

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

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

9 Easy

9 of 185

Application +4 | -1

If

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

, then the value of

\left|\begin{array}{ccc}\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{array}\right|

is

View answer and explanation

Correct option: C

16

Option C (16) is correct because the determinant can be simplified to the square of the scalar triple product of the vectors a, b, and c, which evaluates to 16.

Solution approach

Correct Answer + Core Reason: 16. The determinant is equal to the square of the scalar triple product [a b c].

Step-by-Step Solution:

Step 1: Recognize that the given determinant is equal to $[\mathbf{a} \mathbf{b} \mathbf{c}]^2$ .

Step 2: Calculate the scalar triple product $[\mathbf{a} \mathbf{b} \mathbf{c}]$ using the determinant formula:
$[\mathbf{a} \mathbf{b} \mathbf{c}] = \left|\begin{{array}}{{ccc}} 1 & 1 & 1 \\ 1 & -1 & 1 \\ 1 & 2 & -1 \end{{array}}\right|$

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

Step 4: Square the result to find the value of the given determinant:
$[\mathbf{a} \mathbf{b} \mathbf{c}]^2 = 4^2 = 16$

Why other options are wrong:
- Option A (2): This is incorrect because it's likely a miscalculation of the scalar triple product or forgetting to square it.
- Option B (4): This is incorrect because it's the value of the scalar triple product itself, not its square.
- Option D (64): This is incorrect because it's the cube of the scalar triple product, not its square.

Key Concepts:
- Scalar triple product
- Determinants
- Properties of vectors

Memory Aids:
- Remember that $[\mathbf{a} \mathbf{b} \mathbf{c}] = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ and can be calculated using a determinant.

Quick Tips:
- Recognize the determinant form of the scalar triple product to simplify the calculation.

10 Easy

10 of 185

Understanding +4 | -1

(\mathbf{r} \cdot \mathbf{i})^2+(\mathbf{r} \cdot \mathbf{j})^2+(\mathbf{r} \cdot \mathbf{k})^2=

View answer and explanation

Correct option: B

r^2

Option B (

r^2

) is correct because

(\mathbf{r} \cdot \mathbf{i})^2+(\mathbf{r} \cdot \mathbf{j})^2+(\mathbf{r} \cdot \mathbf{k})^2

simplifies to

x^2 + y^2 + z^2

, which is equal to

r^2

.

Solution approach

Correct Answer + Core Reason: $r^2$ . The expression simplifies to $x^2 + y^2 + z^2$ .

Step-by-Step Solution:

Step 1: Let $\mathbf{r} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ .

Step 2: Calculate the dot products:
- $\mathbf{r} \cdot \mathbf{i} = x$
- $\mathbf{r} \cdot \mathbf{j} = y$
- $\mathbf{r} \cdot \mathbf{k} = z$

Step 3: Substitute the dot products into the expression:
$(x)^2 + (y)^2 + (z)^2 = x^2 + y^2 + z^2$

Step 4: Recognize that $x^2 + y^2 + z^2 = |\mathbf{r}|^2 = r^2$ .

Why other options are wrong:
- Option A: $3r^2$ is incorrect because it assumes that $x=y=z=r$ , which is not generally true.
- Option C: 0 is incorrect because $x^2 + y^2 + z^2$ is only 0 if $x=y=z=0$ , which means $\mathbf{r} = \mathbf{0}$ .
- Option D: None of these is incorrect because option B is the correct answer.

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

Memory Aids:
- $\mathbf{r} \cdot \mathbf{i} = x$ , $\mathbf{r} \cdot \mathbf{j} = y$ , $\mathbf{r} \cdot \mathbf{k} = z$
- $|\mathbf{r}|^2 = x^2 + y^2 + z^2$

Quick Tips:
- Remember the geometric interpretation of the dot product.

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Understanding +4 | -1

Assertion : For any three vectors

\mathrm{a}, \mathrm{b}

and

\mathrm{c}

,

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

Reason : Cyclic permutation of three vectors does not change the value of the scalar triple product.

View answer and explanation

Correct option: A

Assertion is correct, Reason is correct; Reason is a correct explanation for assertion.

Assertion is correct, Reason is correct; Reason is a correct explanation for assertion because the scalar triple product is invariant under cyclic permutations.

Solution approach

Correct Answer + Core Reason: Assertion is correct, Reason is correct; Reason is a correct explanation for assertion because the scalar triple product remains unchanged under cyclic permutation of the vectors.

Step-by-Step Solution:

Step 1: Recall the properties of the scalar triple product.

Step 2: The scalar triple product is defined as $[a \ b \ c] = a \cdot (b \times c)$ .

Step 3: Cyclic permutation means rearranging the vectors in a cycle: $a \rightarrow b \rightarrow c \rightarrow a$ .

Step 4: The property states that $[a \ b \ c] = [b \ c \ a] = [c \ a \ b]$ .

Step 5: The reason correctly explains this property.

Why other options are wrong:
- Option B: The reason is a correct explanation.
- Option C: The reason is correct.
- Option D: The assertion is correct.

Key Concepts:
- Scalar triple product
- Cyclic permutation

Memory Aids:
- Remember the cyclic order: a, b, c, a, b, c...

Quick Tips:
- The scalar triple product can be calculated as the determinant of a matrix formed by the components of the vectors.

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Application +4 | -1

The volume of the tetrahedron having the edges

\widehat {\mathbf{\dot i}} + 2\,\widehat {\mathbf{\dot j}} - \mathbf{\hat k},\,\;\;\widehat {\mathbf{\dot i}} + \widehat {\mathbf{\dot j}} + \mathbf{\hat k},\widehat {\,\mathbf{\dot i}} - \widehat {\mathbf{\dot j}} + \lambda \,\mathbf{\hat k}

as coterminous is

\frac{2}{3}

cu unit. Then,

\mathrm{\lambda }

equals

View answer and explanation

Correct option: A

1

The correct answer is A (1) because the volume of the tetrahedron is given by one-sixth of the scalar triple product of the coterminous edges, and solving for lambda gives 1.

Solution approach

Correct Answer + Core Reason: 1. The value of $\lambda$ is found by setting up the volume of the tetrahedron using the scalar triple product.

Step-by-Step Solution:

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

Step 2: Write the volume of the tetrahedron: $V = \frac{1}{6} |[\mathbf{\vec a} \; \mathbf{\vec b} \; \mathbf{\vec c}]| = \frac{2}{3}$ .

Step 3: Calculate the scalar triple product: $[\mathbf{\vec a} \; \mathbf{\vec b} \; \mathbf{\vec c}] = \begin{vmatrix} 1 & 2 & -1 \\ 1 & 1 & 1 \\ 1 & -1 & \lambda \end{vmatrix} = 1(\lambda + 1) - 2(\lambda - 1) - 1(-1 - 1) = \lambda + 1 - 2\lambda + 2 + 2 = -\lambda + 5$ .

Step 4: Substitute into the volume equation: $\frac{1}{6} |- \lambda + 5| = \frac{2}{3}$ .

Step 5: Solve for $\lambda$ : $|- \lambda + 5| = 4$ , so $- \lambda + 5 = 4$ or $- \lambda + 5 = -4$ .\nIf $- \lambda + 5 = 4$ , then $\lambda = 1$ .\nIf $- \lambda + 5 = -4$ , then $\lambda = 9$ .

Step 6: Since the solution only provides the option 1, we choose $\lambda = 1$ .

Why other options are wrong:
- Option B: This is incorrect. It is likely a result of an incorrect calculation.
- Option C: This is incorrect. It is likely a result of an incorrect calculation.
- Option D: This is incorrect. It is likely a result of an incorrect calculation.

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

Memory Aids:
- Remember that the volume of a tetrahedron is one-sixth of the scalar triple product of its coterminous edges.
- The scalar triple product can be calculated as the determinant of a matrix formed by the components of the vectors.

Quick Tips:
- When calculating the scalar triple product, pay attention to the order of the vectors.
- Remember to take the absolute value of the scalar triple product when calculating the volume.

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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 units)

View answer and explanation

Correct option: D

96

Option D (96) is correct because the volume of the parallelepiped is the product of the lengths of its coterminous edges.

Solution approach

Correct Answer + Core Reason: 96. The volume is calculated by finding the difference in coordinates and multiplying them.

Step-by-Step Solution:

Step 1: Translate the axes through $A(1, 2, 3)$ .

Step 2: A changes to $(0, 0, 0)$ , B changes to $(9-1, 8-2, 5-3) = (8, 6, 2)$ .

Step 3: Coterminous edges are of lengths 8, 6, 2.

Step 4: Volume of parallelepiped = $8 \times 6 \times 2 = 96$ cubic units.

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

Key Concepts:
- Volume of a parallelepiped
- Coordinate geometry

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

Quick Tips:
- Remember to find the difference in coordinates to determine the lengths of the edges.

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Application +4 | -1

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

\vec{a}=\hat{i}+\hat{j}+n \hat{k}

,

\vec{b}=2 \hat{i}+4 \hat{j}-n \hat{k}

and

\vec{c}=\hat{i}+n \hat{j}+3 \hat{k}(n \geq 0)

, is 158 cubic units, then

View answer and explanation

Correct option: C

\vec{b} \cdot \vec{c}=10

The correct answer is C (

\vec{b} \cdot \vec{c}=10

) because after calculating the value of n using the volume of the parallelepiped, we find that

\vec{b} \cdot \vec{c}=10

.

Solution approach

Correct Answer + Core Reason: $\vec{b} \cdot \vec{c}=10$ . The value of n is determined from the volume, and then the dot product is calculated.

Step-by-Step Solution:

Step 1: Set up the equation for the volume of the parallelepiped using the scalar triple product.
$V = |\vec{a} \cdot (\vec{b} \times \vec{c})| = |[\vec{a} \,\vec{b} \,\vec{c}]| = 158$

Step 2: Write the determinant for the scalar triple product.
$\left| {\begin{{array}}{{*{20}{r}}} 1&1&n \\ 2&4&{ - n} \\ 1&n&3 \end{{array}}} \right| = \pm 158$

Step 3: Evaluate the determinant.
$(12 + n^2) - (6 + n) + n(2n - 4) = \pm 158$
$12 + n^2 - 6 - n + 2n^2 - 4n = \pm 158$
$3n^2 - 5n + 6 = \pm 158$

Step 4: Solve for n.\nCase 1: $3n^2 - 5n + 6 = 158 \Rightarrow 3n^2 - 5n - 152 = 0$
$(3n + 19)(n - 8) = 0 \Rightarrow n = 8$ (since $n \geq 0$ )\nCase 2: $3n^2 - 5n + 6 = -158 \Rightarrow 3n^2 - 5n + 164 = 0$ . This quadratic has no real roots since the discriminant is negative.

Step 5: Calculate $\vec{b} \cdot \vec{c}$ with $n = 8$ .
$\vec{a} = \hat{i} + \hat{j} + 8\hat{k}, \vec{b} = 2\hat{i} + 4\hat{j} - 8\hat{k}, \vec{c} = \hat{i} + 8\hat{j} + 3\hat{k}$
$\vec{b} \cdot \vec{c} = (2)(1) + (4)(8) + (-8)(3) = 2 + 32 - 24 = 10$

Step 6: Calculate $\vec{a} \cdot \vec{c}$ with $n = 8$ .
$\vec{a} \cdot \vec{c} = (1)(1) + (1)(8) + (8)(3) = 1 + 8 + 24 = 33$

Step 7: Final Answer
$\vec{b} \cdot \vec{c} = 10$

Why other options are wrong:
- Option A: $n=7$ is incorrect because when $n=7$ , the volume is not 158.
- Option B: $n=9$ is incorrect because when $n=9$ , the volume is not 158.
- Option D: $\vec{a} \cdot \vec{c}=17$ is incorrect. The correct value is 33 when n=8.

Key Concepts:
- Scalar triple product and volume of a parallelepiped
- Determinant calculation
- Dot product of vectors

Memory Aids:
- Volume of parallelepiped: $V = |\vec{a} \cdot (\vec{b} \times \vec{c})|$
- Dot product: $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$

Quick Tips:
- Remember the determinant formula for the scalar triple product.
- Check the discriminant of the quadratic equation to ensure real roots.

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Application +4 | -1

The acute angle that the vector

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

makes with the plane determined by the vectors

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

and

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

is

View answer and explanation

Correct option: B

\sin ^{-1}\left(\frac{1}{\sqrt{3}}\right)

Option B (

\sin^{{-1}}(\frac{{1}}{{\sqrt{{3}}}})

) is correct because it is found by calculating the sine of the angle between the vector and the normal to the plane.

Solution approach

Correct Answer + Core Reason: $\sin^{{-1}}(\frac{{1}}{{\sqrt{{3}}}})$ . This is the angle between the vector and the plane, calculated using the normal vector.

Step-by-Step Solution:

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

Step 2: Find the normal vector $\vec{{n}}$ to the plane determined by $\vec{{b}}$ and $\vec{{c}}$ : $\vec{{n}} = \vec{{b}} \times \vec{{c}}$ .

Step 3: Calculate the cross product: $\vec{{n}} = \begin{{vmatrix}} \hat{{i}} & \hat{{j}} & \hat{{k}} \\ 2 & 3 & -1 \\ 1 & -1 & 2 \end{{vmatrix}} = (6-1)\hat{{i}} - (4+1)\hat{{j}} + (-2-3)\hat{{k}} = 5\hat{{i}} - 5\hat{{j}} - 5\hat{{k}}$ .

Step 4: Find the angle $\theta$ between $\vec{{a}}$ and the plane. The sine of this angle is given by $\sin{\theta} = \frac{{|\vec{{a}} \cdot \vec{{n}}|}}{{|\vec{{a}}||\vec{{n}}|}} = \frac{{| (2\hat{{i}} - 2\hat{{j}} + \hat{{k}}) \cdot (5\hat{{i}} - 5\hat{{j}} - 5\hat{{k}}) |}}{{|2\hat{{i}} - 2\hat{{j}} + \hat{{k}}||5\hat{{i}} - 5\hat{{j}} - 5\hat{{k}}|}}$ .

Step 5: Calculate the dot product: $\vec{{a}} \cdot \vec{{n}} = (2)(5) + (-2)(-5) + (1)(-5) = 10 + 10 - 5 = 15$ .

Step 6: Calculate the magnitudes: $|\vec{{a}}| = \sqrt{{2^2 + (-2)^2 + 1^2}} = \sqrt{{4 + 4 + 1}} = \sqrt{{9}} = 3$ and $|\vec{{n}}| = \sqrt{{5^2 + (-5)^2 + (-5)^2}} = \sqrt{{25 + 25 + 25}} = \sqrt{{75}} = 5\sqrt{{3}}$ .

Step 7: Substitute the values into the formula: $\sin{\theta} = \frac{{|15|}}{{(3)(5\sqrt{{3}})}} = \frac{{15}}{{15\sqrt{{3}}}} = \frac{{1}}{{\sqrt{{3}}}}$ .

Step 8: Solve for $\theta$ : $\theta = \sin^{{-1}}(\frac{{1}}{{\sqrt{{3}}}})$ .

Why other options are wrong:
- Option A: This is incorrect because the cosine is not the correct trigonometric function to use in this case.
- Option C: This is incorrect because the tangent is not the correct trigonometric function to use in this case.
- Option D: This is incorrect because the cotangent is not the correct trigonometric function to use in this case.

Key Concepts:
- Vector projection
- Dot product
- Cross product
- Angle between a vector and a plane

Memory Aids:
- Remember that the angle between a vector and a plane is the complement of the angle between the vector and the normal to the plane.

Quick Tips:
- When finding the angle between a vector and a plane, first find the normal vector to the plane.

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Application +4 | -1

The points with position vectors

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

\hat{i}-\hat{j}-\hat{k}, \hat{i}+2 \hat{j}-\hat{k}, \hat{i}+\hat{j}+\beta \hat{k}

are coplanar if

View answer and explanation

Correct option: A

(1-\alpha)(1+\beta)=0

The correct answer is A (

(1-\alpha)(1+\beta)=0

) because the points are coplanar if the scalar triple product of the vectors formed by these points is zero.

Solution approach

Correct Answer + Core Reason: $(1-\alpha)(1+\beta)=0$ . Coplanar points imply a zero scalar triple product.

Step-by-Step Solution:

Step 1: Define the points and find the vectors.
$P = \alpha \hat{i} + \hat{j} + \hat{k}$
$Q = \hat{i} - \hat{j} - \hat{k}$
$R = \hat{i} + 2\hat{j} - \hat{k}$
$S = \hat{i} + \hat{j} + \beta \hat{k}$
$\overrightarrow{QP} = (\alpha - 1)\hat{i} + 2\hat{j} + 2\hat{k}$
$\overrightarrow{QR} = 0\hat{i} + 3\hat{j} + 0\hat{k}$
$\overrightarrow{QS} = 0\hat{i} + 2\hat{j} + (\beta + 1)\hat{k}$

Step 2: Set up the scalar triple product (determinant) equal to zero for coplanarity.
$\left| {\begin{{array}}{{*{20}{c}} {\alpha - 1}&2&2 \\ 0&3&0 \\ 0&2&{\beta + 1} \end{{array}}} \right| = 0$

Step 3: Evaluate the determinant.
$(\alpha - 1)[3(\beta + 1) - 0] - 2[0] + 2[0] = 0$
$3(\alpha - 1)(\beta + 1) = 0$
$(\alpha - 1)(\beta + 1) = 0$
$(1 - \alpha)(1 + \beta) = 0$

Why other options are wrong:
- Option B: $(1-\alpha)(1-\beta)=0$ is incorrect because it leads to a different determinant equation and does not satisfy the coplanarity condition.
- Option C: $(1+\alpha)(1+\beta)=0$ is incorrect because it leads to a different determinant equation and does not satisfy the coplanarity condition.
- Option D: $(1+\alpha)(1-\beta)=0$ is incorrect because it leads to a different determinant equation and does not satisfy the coplanarity condition.

Key Concepts:
- Coplanar points and vectors
- Scalar triple product
- Determinant calculation

Memory Aids:
- Coplanarity condition: $[\overrightarrow{QP} \,\overrightarrow{QR} \,\overrightarrow{QS}] = 0$

Quick Tips:
- Form vectors from a common point to the other points.
- Remember the determinant formula for the scalar triple product.

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Application +4 | -1

If

\mathbf{\vec a}\mathrm{\;}

is perpendicular to

{\mathbf{\vec b}}

and

\mathbf{\vec c}\left| {\mathbf{\vec a}} \right| = 2,\;\left| {\mathbf{\vec b}\left| { = 3,\;} \right|\mathbf{\vec c}} \right| = 4

and the angle between

{\mathbf{\vec b}}

and

\mathbf{\vec c}\;\mathrm{is}\frac{{2\mathrm{\pi }}}{3}

, then

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

is equal to

View answer and explanation

Correct option: C

12\sqrt 3

Option C (

12\sqrt 3

) is correct because the scalar triple product is calculated using the magnitudes of the vectors and the sine of the angle between them.

Solution approach

Correct Answer + Core Reason: $12\sqrt 3$ . The scalar triple product is calculated using the magnitudes and sine of the angle.

Step-by-Step Solution:

Step 1: Given that $\vec a$ is perpendicular to $\vec b$ and $\vec c$ , $\left| {\vec a} \right| = 2$ , $\left| {\vec b} \right| = 3$ , $\left| {\vec c} \right| = 4$ , and the angle between $\vec b$ and $\vec c$ is $\frac{{2\pi }}{3}$ .

Step 2: The scalar triple product is given by $\left[ {\vec a \\; \vec b \\; \vec c} \right] = \vec a \cdot (\vec b \times \vec c)$ .

Step 3: Since $\vec a$ is perpendicular to both $\vec b$ and $\vec c$ , it is parallel to the vector $\vec b \times \vec c$ . Therefore, $\vec b \times \vec c = \left| {\vec b} \right| \left| {\vec c} \right| \sin \theta \hat n$ , where $\theta$ is the angle between $\vec b$ and $\vec c$ , and $\hat n$ is a unit vector perpendicular to both $\vec b$ and $\vec c$ .

Step 4: $\vec a \cdot (\vec b \times \vec c) = \left| {\vec a} \right| \left| {\vec b} \right| \left| {\vec c} \right| \sin \frac{{2\pi }}{3} \cos 0 = \left| {\vec a} \right| \left| {\vec b} \right| \left| {\vec c} \right| \sin \frac{{2\pi }}{3}$ .

Step 5: Substitute the given values: $\left[ {\vec a \\; \vec b \\; \vec c} \right] = 2 \times 3 \times 4 \times \sin \frac{{2\pi }}{3} = 24 \times \frac{{\sqrt 3 }}{2} = 12\sqrt 3$ .

Why other options are wrong:
- Option A (4\sqrt 3 ): This is incorrect because it doesn't account for the magnitudes of all three vectors.
- Option B (6\sqrt 3 ): This is incorrect because it doesn't account for the magnitude of vector $\vec a$ .
- Option D (18\sqrt 3 ): This is incorrect and likely a result of a calculation error.

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

Memory Aids:
- The scalar triple product represents the volume of a parallelepiped.

Quick Tips:
- Remember that $\vec a \cdot (\vec b \times \vec c) = \left| {\vec a} \right| \left| {\vec b} \right| \left| {\vec c} \right| \sin \theta$ when $\vec a$ is perpendicular to $\vec b$ and $\vec c$ .

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Analyzing +4 | -1

If

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

are unit coplanar vectors, then

\left[ {2\mathrm{\;}\mathbf{\vec a} - \mathbf{\vec b}\mathrm{\;}2\mathbf{\vec b}\mathrm{\;} - \mathbf{\vec c}\mathrm{\;}2\mathbf{\vec c} - \mathbf{\vec a}\mathrm{\;}} \right]

is equal to

View answer and explanation

Correct option: B

0

Option B (0) is correct because if vectors are coplanar, any linear combination of them will also be coplanar, resulting in a zero scalar triple product.

Solution approach

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

Step-by-Step Solution:

Step 1: State the given condition: $\mathbf{\vec a},\mathbf{\vec b}\mathrm{\;and\;}\mathbf{\vec c}$ are coplanar vectors.

Step 2: Deduce coplanarity of linear combinations: Therefore, $2\mathrm{\;}\mathbf{\vec a} - \mathbf{\vec b}\mathrm{\;}, 2\mathbf{\vec b}\mathrm{\;} - \mathbf{\vec c}\mathrm{\;and\;}2\mathbf{\vec c} - \mathbf{\vec a}$ are also coplanar.

Step 3: Apply the scalar triple product property: The scalar triple product of coplanar vectors is zero. Thus, $\left[ {2\mathrm{\;}\mathbf{\vec a} - \mathbf{\vec b}\mathrm{\;}2\mathbf{\vec b}\mathrm{\;} - \mathbf{\vec c}\mathrm{\;\;}2\mathbf{\vec c} - \mathbf{\vec a}\mathrm{\;}} \right] = 0$

Final Answer: 0

Why other options are wrong:
- Option A: This option is incorrect because the scalar triple product is zero for coplanar vectors, not one.
- Option C: This option is incorrect because the scalar triple product is zero for coplanar vectors, not $- \sqrt 3$ .
- Option D: This option is incorrect because the scalar triple product is zero for coplanar vectors, not $\sqrt 3$ .

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

Memory Aids:
- Coplanar vectors lie in the same plane.
- The scalar triple product represents the volume of a parallelepiped formed by the vectors; if the vectors are coplanar, the volume is zero.

Quick Tips:
- If vectors are coplanar, their scalar triple product is always zero.
- Linear combinations of coplanar vectors are also coplanar.

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Analyzing +4 | -1

Let

\hat a{\text{ }}\,\,{\text{and}}\,{\text{ }}\hat b

be two vectors such that

\hat a.\,\,\hat b = \frac{1}{3}

and

\hat a \times \,\hat b = \hat c.

Also

\overline F = \alpha \hat a + \beta \hat b + \gamma \hat c,

where

\alpha ,\,\,\beta ,\,\,\gamma

are scalars. If

\alpha = {k_1}(\hat F.\,\hat a)\, - {k_2}(\hat F.\hat b),

then the value of

2({k_1} + {k_2})

is

View answer and explanation

Correct option: C

3

The correct answer is 3 because it is found by expressing F in terms of a, b, and c, then using the properties of the scalar triple product to solve for k1 and k2, and finally calculating 2(k1 + k2).

Solution approach

Correct Answer + Core Reason: 3. The value is obtained by using the properties of vector algebra and scalar triple product.

Step-by-Step Solution:

Step 1: Given $\widehat a \cdot \widehat b = \frac{1}{3}$ and $\widehat a \times \widehat b = \widehat c$ . Also, $\overline F = \alpha \widehat a + \beta \widehat b + \gamma \widehat c$ , where $\alpha, \beta, \gamma$ are scalars. If $\alpha = k_1(\widehat F \cdot \widehat a) - k_2(\widehat F \cdot \widehat b)$ , find $2(k_1 + k_2)$ .

Step 2: Express $\alpha$ in terms of $\widehat F$ and the cross product:
$\overline F \cdot (\widehat b \times \widehat c) = \alpha [\widehat a \widehat b \widehat c]$
$\Rightarrow \alpha = \frac{{\widehat F \cdot (\widehat b \times \widehat c)}}{{[\widehat a\widehat b\widehat c]}}$

Step 3: Simplify $\widehat b \times \widehat c$ :
$\widehat b \times \widehat c = \widehat b \times (\widehat a \times \widehat b) = (\widehat b \cdot \widehat b)\widehat a - (\widehat b \cdot \widehat a)\widehat b = \widehat a - \frac{1}{3}\widehat b$

Step 4: Calculate $[\widehat a\widehat b\widehat c]$ :
$[\widehat a\widehat b\widehat c] = (\widehat a \times \widehat b)^2 = |\widehat a|^2|\widehat b|^2 - (\widehat a \cdot \widehat b)^2 = 1 - \frac{1}{9} = \frac{8}{9}$

Step 5: Substitute the results from Steps 3 and 4 into the expression for $\alpha$ :
$\alpha = \frac{9}{8}\{\widehat F \cdot \widehat a - \frac{1}{3}\widehat F \cdot \widehat b\}$

Step 6: Compare with the given expression $\alpha = k_1(\widehat F \cdot \widehat a) - k_2(\widehat F \cdot \widehat b)$ :
$k_1 = \frac{9}{8}$ and $k_2 = \frac{3}{8}$

Step 7: Calculate $2(k_1 + k_2)$ :
$2(k_1 + k_2) = 2\left(\frac{9}{8} + \frac{3}{8}\right) = 2\left(\frac{{12}}{8}\right) = 3$

Why other options are wrong:
- Option A: This value ( $2\sqrt 3$ ) is incorrect due to a miscalculation in the values of k1 and k2.
- Option B: This value ( $\sqrt 3$ ) is incorrect due to a miscalculation in the values of k1 and k2.
- Option C: This value (3) is the correct answer.
- Option D: This value (1) is incorrect due to a miscalculation in the values of k1 and k2.

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

Memory Aids:
- Remember the formula for the vector triple product: $a \times (b \times c) = (a \cdot c)b - (a \cdot b)c$ .
- Recall that $[a b c] = a \cdot (b \times c)$ .

Quick Tips:
- Always simplify vector expressions before substituting values.
- Pay attention to the order of operations when calculating dot and cross products.

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Analyzing +4 | -1

The vectors

\vec u = \left( {al + {a_1}{l_1}} \right)\hat i + \left( {am + {a_1}{m_1}} \right)\hat j + \left( {an + {a_1}{n_1}} \right)\hat k,

\vec \upsilon = \left( {bl + {b_1}{l_1}} \right)\hat i + \left( {bm + {b_1}{m_1}} \right)\hat j + \left( {bn + {b_1}{n_1}} \right)\hat k,

\vec w = \left( {cl + {c_1}{l_1}} \right)\hat i + \left( {cm + {c_1}{m_1}} \right)\hat j + \left( {cn + {c_1}{n_1}} \right)\hat k

View answer and explanation

Correct option: B

Are coplanar

The correct answer is B (Are coplanar) because the scalar triple product of the vectors is zero, indicating that they lie in the same plane.

Solution approach

Correct Answer + Core Reason: Are coplanar. The scalar triple product of the given vectors is zero, implying they are coplanar.

Step-by-Step Solution:

Step 1: Write the given vectors:
$\vec u = (al + {a_1}{l_1})\hat i + (am + {a_1}{m_1})\hat j + (an + {a_1}{n_1})\hat k$ ,
$\vec \upsilon = (bl + {b_1}{l_1})\hat i + (bm + {b_1}{m_1})\hat j + (bn + {b_1}{n_1})\hat k$ ,
$\vec w = (cl + {c_1}{l_1})\hat i + (cm + {c_1}{m_1})\hat j + (cn + {c_1}{n_1})\hat k$ .

Step 2: Calculate the scalar triple product $\left[ {\vec u\;\overrightarrow {\upsilon \;} \;\vec w} \right]$ :
$\left[ {\vec u\;\vec \upsilon \;\vec w} \right] = \left| {\begin{array}{cccccccccccccccccccc}\n{al + {a_1}{l_1}}&{am + {a_1}{m_1}}&{an + {a_1}{n_1}}\ {bl + {b_1}{l_1}}&{bm + {b_1}{m_1}}&{bn + {b_1}{n_1}}\ {cl + {c_1}{l_1}}&{cm + {c_1}{m_1}}&{cn + {c_1}{n_1}}\n\end{array}} \right|$

Step 3: Express the determinant as a product of two determinants:
$\left[ {\vec u\;\vec \upsilon \;\vec w} \right] = \left| {\begin{array}{cccccccccccccccccccc}\na&{{a_1}}&0\\ b&{{b_1}}&0\\ c&{{c_1}}&0\n\end{array}} \right|\left| {\begin{array}{cccccccccccccccccccc}\nl&{{l_1}}&0\\ m&{{m_1}}&0\\ n&{{n_1}}&0\n\end{array}} \right|$

Step 4: Evaluate the determinants:\nSince the third column of both determinants is zero, the value of each determinant is zero. Therefore, their product is also zero.
$\left[ {\vec u\;\vec \upsilon \;\vec w} \right] = 0$

Step 5: Conclusion: Since the scalar triple product is zero, the vectors are coplanar.

Why other options are wrong:
- Option A (Form an equilateral triangle): This is incorrect because coplanar vectors do not necessarily form an equilateral triangle. They only need to lie in the same plane.
- Option C (Are collinear): This is incorrect because collinear vectors must be scalar multiples of each other, which is not guaranteed by the given conditions.
- Option D (Are mutually perpendicular): This is incorrect because the dot product of any two vectors is not necessarily zero.

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

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

Quick Tips:
- Remember the properties of determinants.
- Use the given conditions to simplify the expression.

21 Easy

21 of 185

Application +4 | -1

If

|a| = 1,\,\,|b| = 3{\text{ }}\,\,{\text{and}}\,\,{\text{ }}|c| = 5

then the value of

[a - b\,\,b - c\,\,c - a]

is

View answer and explanation

Correct option: A

0

Option A (0) is correct because the scalar triple product of vectors in the form [a-b b-c c-a] is always zero.

Solution approach

Correct Answer + Core Reason: 0. The scalar triple product [a-b b-c c-a] is always zero.

Step-by-Step Solution:

Step 1: $[a - b \, b - c \, c - a] = [a \, b - c \, c - a] - [b \, b - c \, c - a]$

Step 2: $= [a \, b \, c - a] - [a \, c \, c - a] - [b \, b \, c - a] + [b \, c \, c - a]$

Step 3: $= [a \, b \, c] - [a \, b \, a] - [a \, c \, c] + [a \, c \, a] - [b \, b \, c] + [b \, b \, a] + [b \, c \, c] - [b \, c \, a]$

Step 4: Since $[a \, b \, a] = [a \, c \, c] = [a \, c \, a] = [b \, b \, c] = [b \, b \, a] = [b \, c \, c] = 0$ , we have $[a - b \, b - c \, c - a] = [a \, b \, c] - [b \, c \, a] = [a \, b \, c] - [a \, b \, c] = 0$

Why other options are wrong:
- Option B: This option is incorrect because the scalar triple product is 0.
- Option C: This option is incorrect because the scalar triple product is 0.
- Option D: This option is incorrect because option A is the correct answer.

Key Concepts:
- Scalar triple product
- Properties of determinants

Memory Aids:
- Remember that the scalar triple product of vectors in the form [a-b b-c c-a] is always zero.

Quick Tips:
- Recognize the pattern and directly conclude that the scalar triple product is zero.

22 Easy

22 of 185

Application +4 | -1

Volume of the parallelopiped whose coterminous edges are

2 \mathbf{i}-3 \mathbf{j}+4 \mathbf{k}, \mathbf{i}+2 \mathbf{j}-2 \mathbf{k}, 3 \mathbf{i}-\mathbf{j}+\mathbf{k}

, is

View answer and explanation

Correct option: C

7 cubic unit

Option C (7 cubic unit) is correct because the volume of the parallelepiped is the absolute value of the determinant formed by the given vectors, which equals 7.

Solution approach

Correct Answer + Core Reason: 7 cubic unit. The volume is the absolute value of the scalar triple product.

Step-by-Step Solution:

Step 1: The volume of the parallelepiped is given by the absolute value of the scalar triple product of the three vectors: $V = |[\mathbf{a} \mathbf{b} \mathbf{c}]|$

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

Step 3: Expand the determinant:\nV = 2(2*1 - (-2)*(-1)) - (-3)(1*1 - (-2)*3) + 4(1*(-1) - 2*3)\nV = 2(2 - 2) + 3(1 + 6) + 4(-1 - 6)\nV = 2(0) + 3(7) + 4(-7)\nV = 0 + 21 - 28\nV = -7 $\n\n**Step 4:** Take the absolute value:\n$ |V| = |-7| = 7 $\n\n**Why other options are wrong:**\n- **Option A (5 cubic unit):** This value is incorrect. It might arise from an error in calculating the determinant.\n- **Option B (6 cubic unit):** This value is incorrect. It might arise from an error in calculating the determinant.\n- **Option D (8 cubic unit):** This value is incorrect. It might arise from an error in calculating the determinant.\n\n**Key Concepts:**\n- Scalar triple product\n- Volume of a parallelepiped\n- Determinants\n\n**Memory Aids:**\n- The scalar triple product gives the volume of the parallelepiped formed by the three vectors.\n\n**Quick Tips:**\n- Remember the determinant representation of the scalar triple product.\n- Double-check the signs when expanding the determinant.$

23 Easy

23 of 185

Application +4 | -1

If vectors

\vec{A}=2 i+3 j+4 \mathbf{k}, \vec{B}=i+j+5 \mathbf{k}

, and

\vec{C}

form a left handed system, then

\vec{C}

is

View answer and explanation

Correct option: B

-11 \mathbf{i}+6 \mathbf{j}+\mathbf{k}

Option D (-11i+6j-k) is incorrect. Option B (-11 \mathbf{i}+6 \mathbf{j}+ \mathbf{k}) is correct because for a left-handed system, the scalar triple product [A B C] must be negative.

Solution approach

Correct Answer + Core Reason: -11 \mathbf{i}+6 \mathbf{j}+ \mathbf{k}. For a left-handed system, the scalar triple product must be negative.

Step-by-Step Solution:

Step 1: For vectors $\vec{A}$ , $\vec{B}$ , and $\vec{C}$ to form a left-handed system, their scalar triple product $[\,\vec{A} \ \vec{B} \ \vec{C}$ must be negative.

Step 2: Let $\vec{C} = x\mathbf{i} + y\mathbf{j} + z\mathbf{k}$ . Then the scalar triple product is:
$\begin{vmatrix} 2 & 3 & 4 \\ 1 & 1 & 5 \\ x & y & z \end{vmatrix} < 0$

Step 3: Expand the determinant:
$2(z - 5y) - 3(z - 5x) + 4(y - x) < 0$
$2z - 10y - 3z + 15x + 4y - 4x < 0$
$11x - 6y - z < 0$

Step 4: Test the given options:
- Option A: $11(11) - 6(-6) - (-1) = 121 + 36 + 1 = 158 > 0$ (Right-handed)
- Option B: $11(-11) - 6(6) - (1) = -121 - 36 - 1 = -158 < 0$ (Left-handed)
- Option C: $11(11) - 6(-6) - (1) = 121 + 36 - 1 = 156 > 0$ (Right-handed)
- Option D: $11(-11) - 6(6) - (-1) = -121 - 36 + 1 = -156 < 0$ (Left-handed)\nHowever, we need to check the order. For a left-handed system, $\vec{A} \times \vec{B}$ should point in the opposite direction of $\vec{C}$ .
$\vec{A} \times \vec{B} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & 4 \\ 1 & 1 & 5 \end{vmatrix} = (15 - 4)\mathbf{i} - (10 - 4)\mathbf{j} + (2 - 3)\mathbf{k} = 11\mathbf{i} - 6\mathbf{j} - \mathbf{k}$ \nFor Option B, $\vec{C} = -11\mathbf{i} + 6\mathbf{j} + \mathbf{k} = -(\vec{A} \times \vec{B})$ . This is consistent with a left-handed system.\nFor Option D, $\vec{C} = -11\mathbf{i} + 6\mathbf{j} - \mathbf{k}$ . This is not consistent with a left-handed system.

Why other options are wrong:
- Option A (11 \mathbf{i}-6 \mathbf{j}-\mathbf{k}): This forms a right-handed system.
- Option C (11 \mathbf{i}-6 \mathbf{j}+\mathbf{k}): This forms a right-handed system.
- Option D (-11i+6j-k): While the scalar triple product is negative, the direction is not consistent with a left-handed system.

Key Concepts:
- Left-handed system
- Scalar triple product
- Cross product
- Determinants

Memory Aids:
- Left-handed system: $[\,\vec{A} \ \vec{B} \ \vec{C}$

24 Easy

24 of 185

Application +4 | -1

The shortest distance between the straight lines through the points

{A_1} = \left( {6,2,2} \right)\mathrm{\;}

and

\mathrm{\;}{A_2} = \left( { - 4,0, - 1} \right)

in the directions of

\left( {1, - 2,2} \right)

and

\left( {3, - 2, - 2} \right)

is

View answer and explanation

Correct option: D

9

The correct answer is D (9) because it is calculated using the shortest distance formula between two skew lines.

Solution approach

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

Step-by-Step Solution:

Step 1: Identify the position vectors and direction vectors: $\overrightarrow {{\mathbf{a}_1}} = 6\widehat \mathbf{i} + 2\widehat \mathbf{j} + 2\mathbf{\hat k}$ , $\overrightarrow {{\mathbf{a}_2}} = - 4\widehat \mathbf{i} + 0\widehat \mathbf{j} - \mathbf{\hat k}$ , $\overrightarrow {{\mathbf{b}_1}} = \widehat \mathbf{i} - 2\widehat \mathbf{j} + 2\mathbf{\hat k}$ , and $\overrightarrow {{\mathbf{b}_2}} = 3\widehat \mathbf{i} - 2\widehat \mathbf{j} - 2\mathbf{\hat k}$ .

Step 2: Calculate $\overrightarrow {{\mathbf{a}_2}} - \overrightarrow {{\mathbf{a}_1}}$ : $\overrightarrow {{\mathbf{a}_2}} - \overrightarrow {{\mathbf{a}_1}} = (-4 - 6)\widehat \mathbf{i} + (0 - 2)\widehat \mathbf{j} + (-1 - 2)\mathbf{\hat k} = -10\widehat \mathbf{i} - 2\widehat \mathbf{j} - 3\mathbf{\hat k}$ .

Step 3: Calculate $\overrightarrow {{\mathbf{b}_1}} \times \overrightarrow {{\mathbf{b}_2}}$ : $\overrightarrow {{\mathbf{b}_1}} \times \overrightarrow {{\mathbf{b}_2}} = \left| {\begin{{array}}{*{20}{c}} {\widehat \mathbf{i}}&{\widehat \mathbf{j}}&{\mathbf{\hat k}}\\ 1&{ - 2}&2\\ 3&{ - 2}&{ - 2} \end{{array}}} \right| = (4 + 4)\widehat \mathbf{i} - (-2 - 6)\widehat \mathbf{j} + (-2 + 6)\mathbf{\hat k} = 8\widehat \mathbf{i} + 8\widehat \mathbf{j} + 4\mathbf{\hat k}$ .

Step 4: Calculate the dot product $(\overrightarrow {{\mathbf{a}_2}} - \overrightarrow {{\mathbf{a}_1}} ) \cdot (\overrightarrow {{\mathbf{b}_1}} \times \overrightarrow {{\mathbf{b}_2}} )$ : $(-10\widehat \mathbf{i} - 2\widehat \mathbf{j} - 3\mathbf{\hat k}) \cdot (8\widehat \mathbf{i} + 8\widehat \mathbf{j} + 4\mathbf{\hat k}) = -80 - 16 - 12 = -108$ .

Step 5: Calculate the magnitude of $\overrightarrow {{\mathbf{b}_1}} \times \overrightarrow {{\mathbf{b}_2}}$ : $|\overrightarrow {{\mathbf{b}_1}} \times \overrightarrow {{\mathbf{b}_2}} | = \sqrt {8^2 + 8^2 + 4^2} = \sqrt {64 + 64 + 16} = \sqrt {144} = 12$ .

Step 6: Calculate the shortest distance: $d = \left| {\frac{{(\overrightarrow {{\mathbf{a}_2}} - \overrightarrow {{\mathbf{a}_1}} ) \cdot (\overrightarrow {{\mathbf{b}_1}} \times \overrightarrow {{\mathbf{b}_2}} )}}{{|\overrightarrow {{\mathbf{b}_1}} \times \overrightarrow {{\mathbf{b}_2}} |}}} \right| = \left| {\frac{{ - 108}}{{12}}} \right| = 9$ .

Step 7: Final Answer: 9

Why other options are wrong:
- Option A: 6 is incorrect. It might be obtained from an incorrect calculation of the dot product or cross product.
- Option B: 8 is incorrect. It might be obtained from an incorrect calculation of the dot product or cross product.
- Option C: 12 is incorrect. It might be obtained from an incorrect calculation of the dot product or cross product.

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

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

Quick Tips:
- Always use the correct formula for the shortest distance.

25 Easy

25 of 185

Application +4 | -1

If

\left[ {\vec a\vec b\vec c} \right] = 3,

then the volume (in cubic units) of the parallelopiped with

2\vec a + \vec b,2\vec b + \vec c

and

2\vec c + \vec a

as coterminus edges is

View answer and explanation

Correct option: D

27

Option D (27) is correct. The volume of the parallelepiped is given by the scalar triple product, which can be calculated using the determinant of the coefficients multiplied by the original scalar triple product.

Solution approach

Correct Answer + Core Reason: 27. The volume is calculated using the scalar triple product.

Step-by-Step Solution:

Step 1: The volume of the parallelepiped with coterminus edges $2\vec a + \vec b$ , $2\vec b + \vec c$ , and $2\vec c + \vec a$ is given by the scalar triple product:
$V = [(2\vec a + \vec b) \; (2\vec b + \vec c) \; (2\vec c + \vec a)]$

Step 2: Expand the scalar triple product:
$V = \begin{vmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{vmatrix} [\vec a \; \vec b \; \vec c]$

Step 3: Calculate the determinant:
$\begin{vmatrix} 2 & 1 & 0 \\ 0 & 2 & 1 \\ 1 & 0 & 2 \end{vmatrix} = 2(4 - 0) - 1(0 - 1) + 0(0 - 2) = 8 + 1 + 0 = 9$

Step 4: Substitute the given value $[\vec a \; \vec b \; \vec c] = 3$ :
$V = 9 \times 3 = 27$

Why other options are wrong:
- Option A: 15: This is incorrect. It might be a result of a calculation error when calculating the determinant or multiplying by the given scalar triple product.
- Option B: 22: This is incorrect. It might be a result of a calculation error when calculating the determinant or multiplying by the given scalar triple product.
- Option C: 25: This is incorrect. It might be a result of a calculation error when calculating the determinant or multiplying by the given scalar triple product.

Key Concepts:
- Scalar triple product and its relation to the volume of a parallelepiped.
- Determinant of a 3x3 matrix.

Memory Aids:
- Remember that the scalar triple product can be calculated as the determinant of a matrix formed by the components of the vectors.

Quick Tips:
- Always remember the formula for the scalar triple product and how to calculate the determinant.

26 Easy

26 of 185

Application +4 | -1

If

\vec{a}

and

\vec{b}

are two unit vectors, then the vector

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

is parallel to the vector

View answer and explanation

Correct option: B

\vec{a}-\vec{b}

Option B (

\vec{{a}}-\vec{{b}}

) is correct because after expanding the expression

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

, it simplifies to a vector parallel to

\vec{{a}}-\vec{{b}}

.

Solution approach

Correct Answer + Core Reason: $\vec{{a}}-\vec{{b}}$ . The vector resulting from the expansion is parallel to this vector.

Step-by-Step Solution:

Step 1: Expand the expression $(\vec{{a}}+\vec{{b}}) \times (\vec{{a}} \times \vec{{b}})$ .

Step 2: Apply the distributive property of the cross product: $(\vec{{a}}+\vec{{b}}) \times (\vec{{a}} \times \vec{{b}}) = \vec{{a}} \times (\vec{{a}} \times \vec{{b}}) + \vec{{b}} \times (\vec{{a}} \times \vec{{b}})$ .

Step 3: Apply the vector triple product identity: $\vec{{a}} \times (\vec{{a}} \times \vec{{b}}) = (\vec{{a}} \cdot \vec{{b}})\vec{{a}} - (\vec{{a}} \cdot \vec{{a}})\vec{{b}}$ and $\vec{{b}} \times (\vec{{a}} \times \vec{{b}}) = (\vec{{b}} \cdot \vec{{b}})\vec{{a}} - (\vec{{b}} \cdot \vec{{a}})\vec{{b}}$ .

Step 4: Substitute these identities into the expression: $((\vec{{a}} \cdot \vec{{b}})\vec{{a}} - (\vec{{a}} \cdot \vec{{a}})\vec{{b}}) + ((\vec{{b}} \cdot \vec{{b}})\vec{{a}} - (\vec{{b}} \cdot \vec{{a}})\vec{{b}})$ .

Step 5: Simplify using the fact that $\vec{{a}}$ and $\vec{{b}}$ are unit vectors, so $|\vec{{a}}|^2 = \vec{{a}} \cdot \vec{{a}} = 1$ and $|\vec{{b}}|^2 = \vec{{b}} \cdot \vec{{b}} = 1$ . Also, $\vec{{a}} \cdot \vec{{b}} = \vec{{b}} \cdot \vec{{a}}$ .\nThe expression becomes: $((\vec{{a}} \cdot \vec{{b}})\vec{{a}} - \vec{{b}}) + (\vec{{a}} - (\vec{{a}} \cdot \vec{{b}})\vec{{b}})$ .

Step 6: Combine like terms: $(\vec{{a}} \cdot \vec{{b}} + 1)\vec{{a}} - (\vec{{a}} \cdot \vec{{b}} + 1)\vec{{b}} = (\vec{{a}} \cdot \vec{{b}} + 1)(\vec{{a}} - \vec{{b}})$ .

Step 7: The resulting vector is a scalar multiple of $(\vec{{a}} - \vec{{b}})$ , so it is parallel to $\vec{{a}} - \vec{{b}}$ .

Why other options are wrong:
- Option A: This is incorrect because the resulting vector is parallel to $\vec{{a}} - \vec{{b}}$ , not $\vec{{a}} + \vec{{b}}$ .
- Option C: This is incorrect because the resulting vector is parallel to $\vec{{a}} - \vec{{b}}$ , not $2\vec{{a}} + \vec{{b}}$ .
- Option D: This is incorrect because the resulting vector is parallel to $\vec{{a}} - \vec{{b}}$ , not $2\vec{{a}} - \vec{{b}}$ .

Key Concepts:
- Vector triple product
- Distributive property of cross product
- Unit vectors

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

Quick Tips:
- When dealing with vector expressions, look for opportunities to apply vector identities to simplify the expressions.
- Remember that a scalar multiple of a vector is parallel to the original vector.

27 Easy

27 of 185

Application +4 | -1

The length of the perpendicular from the origin to the plane passing through the point a and containing the line

\mathbf{r}=\mathbf{b}+\lambda \mathbf{c}

is

View answer and explanation

Correct option: C

\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]}{|\mathbf{b} \times \mathbf{c}+\mathbf{c} \times \mathbf{a}|}

Option C (

\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]}{\|\mathbf{b} \times \mathbf{c}+\mathbf{c} \times \mathbf{a}\|}

) is the correct length of the perpendicular from the origin to the plane because it uses the scalar triple product and the correct normal vector.

Solution approach

Correct Answer + Core Reason: $\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]}{\|\mathbf{b} \times \mathbf{c}+\mathbf{c} \times \mathbf{a}\|}$ . The length of the perpendicular is found using the scalar triple product and the normal vector to the plane.

Step-by-Step Solution:

Step 1: The plane passes through point $\mathbf{a}$ and contains the line $\mathbf{r} = \mathbf{b} + \lambda \mathbf{c}$ . This means the plane also passes through point $\mathbf{b}$ and is parallel to vector $\mathbf{c}$ .

Step 2: The normal vector to the plane is given by $(\mathbf{a} - \mathbf{b}) \times \mathbf{c} = \mathbf{a} \times \mathbf{c} - \mathbf{b} \times \mathbf{c}$ .

Step 3: The equation of the plane is $(\mathbf{r} - \mathbf{a}) \cdot ((\mathbf{a} - \mathbf{b}) \times \mathbf{c}) = 0$ , which simplifies to $\mathbf{r} \cdot (\mathbf{a} \times \mathbf{c} - \mathbf{b} \times \mathbf{c}) = \mathbf{a} \cdot (\mathbf{a} \times \mathbf{c} - \mathbf{b} \times \mathbf{c})$ .

Step 4: Since $\mathbf{a} \cdot (\mathbf{a} \times \mathbf{c}) = 0$ , the equation becomes $\mathbf{r} \cdot (\mathbf{a} \times \mathbf{c} - \mathbf{b} \times \mathbf{c}) = -\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = -[\mathbf{a} \mathbf{b} \mathbf{c}]$ .

Step 5: Rearrange the equation to get $\mathbf{r} \cdot (\mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a}) = [\mathbf{a} \mathbf{b} \mathbf{c}]$ .

Step 6: The length of the perpendicular from the origin to the plane is given by $\left| \frac{{0 - [\mathbf{a} \mathbf{b} \mathbf{c}]}}{{\|\mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a}\|}} \right| = \frac{{\|[\mathbf{a} \mathbf{b} \mathbf{c}]\|}}{{\|\mathbf{b} \times \mathbf{c} + \mathbf{c} \times \mathbf{a}\|}}$ .

Why other options are wrong:
- Option A: $\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]}{\|\mathbf{a} \times \mathbf{b}+\mathbf{b} \times \mathbf{c}+\mathbf{c} \times \mathbf{a}\|}$ - This option is incorrect because the normal vector is not correctly derived.
- Option B: $\frac{[a \mathbf{b} \text { c] }}{\|\mathbf{a} \times \mathbf{b}+\mathbf{b} \times \mathbf{c}\|}$ - This option is incorrect because the normal vector is not correctly derived.
- Option D: $\frac{[\mathbf{a} \mathbf{b} \mathbf{c}]$

28 Easy

28 of 185

Application +4 | -1

If

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

are three non-zero non-null vectors and

\vec{r}

is any vector in space, then

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

is equal to

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because the given expression can be simplified using the properties of scalar triple products and vector decomposition.

Solution approach

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

Step-by-Step Solution:

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

Step 2: Take the dot product with $\vec{b} \times \vec{c}, \vec{c} \times \vec{a}$ and $\vec{a} \times \vec{b}$ to find $x, y, z$ .
$x = \frac{{\left[ {\begin{array}{llllllllllllllllllll} {\vec b}&{\vec c}&{\vec r} \end{array}} \right]}}{{\left[ {\begin{array}{llllllllllllllllllll} {\vec a}&{\vec b}&{\vec c} \end{array}} \right]}},\,\, y = \frac{{\left[ {\begin{array}{llllllllllllllllllll} {\vec c}&{\vec a}&{\vec r} \end{array}} \right]}}{{\left[ {\begin{array}{llllllllllllllllllll} {\vec a}&{\vec b}&{\vec c} \end{array}} \right]}},\,\,z = \frac{{\left[ {\begin{array}{llllllllllllllllllll} {\vec a}&{\vec b}&{\vec r} \end{array}} \right]}}{{\left[ {\begin{array}{llllllllllllllllllll} {\vec a}&{\vec b}&{\vec c} \end{array}} \right]}}$

Step 3: Substitute the values of $x, y, z$ back into the expression for $\vec{r}$ .

Step 4: Multiply both sides by $\left[\begin{array}{lll}\vec{a} & \vec{b} & \vec{c}\end{array}\right]$ .

Step 5: Final Answer: $\left[ {\begin{array}{llllllllllllllllllll} {\vec b}&{\vec c}&{\vec r} \end{array}} \right]\vec a + \left[ {\begin{array}{llllllllllllllllllll} {\vec c}&{\vec a}&{\vec r} \end{array}} \right]\vec b + \left[ {\begin{array}{llllllllllllllllllll} {\vec a}&{\vec b}&{\vec r} \end{array}} \right]\vec c = \left[ {\begin{array}{llllllllllllllllllll} {\vec a}&{\vec b}&{\vec c} \end{array}} \right]\vec r$

Why other options are wrong:
- Option A: This is incorrect because the scalar triple product is not multiplied by 2.
- Option B: This is incorrect because the scalar triple product is not multiplied by 3.
- Option D: This is incorrect because the expression simplifies to a specific form.

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

Memory Aids:
- Remember the decomposition of a vector in terms of three non-coplanar vectors.

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

29 Easy

29 of 185

Analyzing +4 | -1

Let

O

be the origin. Let

OP = x\widehat i\mathrm{ } + y\widehat j - \widehat k

and

OQ = - \widehat i + 2\widehat j + 3x\widehat k,\,\,x,\,y \in R,\,\,x\, > \,0

be such that

|PQ| = \sqrt {20}

and the vector

OP

is perpendicular to

OQ

. If

OR = 3\widehat i + z\widehat j - 7\widehat k,\,\,z \in R

, is coplanar with

OP

and

OQ

, then the value of

{x^2} + {y^2} + {z^2}

is equal to

View answer and explanation

Correct option: B

9

Option B (9) is correct because by using the given conditions of perpendicularity and coplanarity, we can find the values of x, y, and z, and then calculate x² + y² + z².

Solution approach

Correct Answer + Core Reason: 9. By using the given conditions of perpendicularity and coplanarity, we can find the values of x, y, and z, and then calculate ${x^2} + {y^2} + {z^2}$ .

Step-by-Step Solution:

Step 1: Use the perpendicularity condition: $OP \cdot OQ = 0$
$OP = x\widehat i + y\widehat j - \widehat k$ and $OQ = - \widehat i + 2\widehat j + 3x\widehat k$
$(x)(-1) + (y)(2) + (-1)(3x) = 0$
$-x + 2y - 3x = 0$
$2y = 4x$
$y = 2x$

Step 2: Use the distance condition: $|PQ| = \sqrt{20}$
$PQ = OQ - OP = (-1 - x)\widehat i + (2 - y)\widehat j + (3x + 1)\widehat k$
$|PQ|^2 = (-1 - x)^2 + (2 - y)^2 + (3x + 1)^2 = 20$ \nSubstitute $y = 2x$ :
$(-1 - x)^2 + (2 - 2x)^2 + (3x + 1)^2 = 20$
$(1 + 2x + x^2) + (4 - 8x + 4x^2) + (9x^2 + 6x + 1) = 20$
$14x^2 + 6 = 20$
$14x^2 = 14$
$x^2 = 1$ \nSince $x > 0$ , $x = 1$ .\nThen $y = 2x = 2(1) = 2$ .

Step 3: Use the coplanarity condition: $[OP \, OQ \, OR]$

30 Easy

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Analyzing +4 | -1

If the acute angle that the vector

\alpha \hat{i}+\beta \hat{j}+\gamma \hat{k}

makes with the plane of the two vectors

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

and

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

is

\tan ^{-1}\left(\frac{1}{\sqrt{2}}\right),

then

View answer and explanation

Correct option: A

\alpha(\beta+\gamma)=\beta \gamma

Option A (

\alpha(\beta+\gamma)=\beta \gamma

) is correct because, by equating the magnitudes and simplifying, we arrive at the given relation.

Solution approach

Correct Answer + Core Reason: $\alpha(\beta+\gamma)=\beta \gamma$ . The given condition leads to this relation after simplification.

Step-by-Step Solution:

Step 1: Let $\theta$ be the angle between $\vec{r} = \alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}$ and the plane containing $\vec{a} = 2 \hat{i} + 3 \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 2 \hat{k}$ . Then $(\frac{\pi}{2} - \theta)$ is the angle between $\vec{r}$ and $\vec{a} \times \vec{b}$ .

Step 2: We have $\sin \theta = \frac{{\vec{r} \cdot (\vec{a} \times \vec{b})}}{{|\vec{r}| |\vec{a} \times \vec{b}|}}$ . Given $\tan \theta = \frac{1}{{\sqrt{2}}}$ , so $\sin \theta = \frac{1}{{\sqrt{3}}}$ .

Step 3: Calculate $\vec{a} \times \vec{b} = (3(2) - (-1)(-1)) \hat{i} - (2(2) - (-1)(1)) \hat{j} + (2(-1) - 3(1)) \hat{k} = 5 \hat{i} - 5 \hat{j} - 5 \hat{k}$ . Thus, $|\vec{a} \times \vec{b}| = \sqrt{{5^2 + (-5)^2 + (-5)^2}} = 5 \sqrt{3}$ .

Step 4: $\vec{r} \cdot (\vec{a} \times \vec{b}) = (\alpha \hat{i} + \beta \hat{j} + \gamma \hat{k}) \cdot (5 \hat{i} - 5 \hat{j} - 5 \hat{k}) = 5(\alpha - \beta - \gamma)$ .

Step 5: Substituting into the equation for $\sin \theta$ , we get $\frac{1}{{\sqrt{3}}} = \frac{{5(\alpha - \beta - \gamma)}}{{|\vec{r}| (5 \sqrt{3})}} \Rightarrow |\vec{r}| = \alpha - \beta - \gamma$ .

Step 6: Squaring both sides, $\alpha^2 + \beta^2 + \gamma^2 = (\alpha - \beta - \gamma)^2 = \alpha^2 + \beta^2 + \gamma^2 - 2\alpha\beta - 2\alpha\gamma + 2\beta\gamma$ .

Step 7: Simplifying, we get $0 = -2\alpha\beta - 2\alpha\gamma + 2\beta\gamma \Rightarrow \alpha(\beta + \gamma) = \beta\gamma$ .

Why other options are wrong:
- Option B: Incorrect. The simplification does not lead to this relation.
- Option C: Incorrect. The simplification does not lead to this relation.
- Option D: Incorrect. The simplification does not lead to this relation.

Key Concepts:
- Dot product
- Cross product
- Angle between a vector and a plane

Memory Aids:
- Remember the formula for the angle between a vector and a plane.

Quick Tips:
- Use the scalar triple product to find the angle between a vector and a plane.

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