The vector ${\bf{\vec a}} = {\rm{\alpha }}\widehat {\bf{i}} + 2\widehat {\bf{j}} + {\rm{\beta }}{\bf{\hat k}}$ lies in the plane of the vectors ${\bf{\vec b}} = \widehat {\bf{i}} + \widehat {\bf{j}}$ and $\vec c = \widehat {\bf{j}} + {\bf{\hat k}}$ and bisects the angle between ${{\bf{\vec b}}}$ and ${{\bf{\vec c}}}$ Then, which one of the following gives possible value of ${\rm{\alpha }}$ and ${\rm{\beta }}$?

${\rm{\alpha }} = 1,{\rm{\;\beta }} = 1$

${\rm{\alpha }} = 2,{\rm{\;\beta }} = 2$

${\rm{\alpha }} = 1,{\rm{\;\beta }} = 2$

${\rm{\alpha }} = 2,{\rm{\;\beta }} = 1$

Five points given by $A,B,C,D,E$ are in plane. Three forces $\overrightarrow {{\bf{AC}}} ,\;\overrightarrow {\;{\bf{AD}}} $ and $\overrightarrow {{\bf{AE}}} $ act a $A$ and three forces $\overrightarrow {{\bf{CB}}} ,\overrightarrow {{\bf{DB}}} ,\overrightarrow {{\bf{EB}}} $ act at $B$. Then, their resultant is

$3\overrightarrow {{\bf{AB}}} $

$2\;\overrightarrow {{\bf{AC}}} $

$3\;\overrightarrow {{\bf{DB}}} $

$2\;\overrightarrow {{\bf{BC}}} $

ADDITION OF VECTORS MCQs for JEE MATHEMATICS — Easy, Page 4

91 Easy

91 of 257

Analyzing +4 | -1

The non – zero vectors a, b and c are related by a = 8b and c = - 7b angle between a and c is

View answer and explanation

Correct option: C

\pi

Option C (

\pi

) is correct because, since a and b are in the same direction and b and c are in opposite directions, a and c must be in opposite directions, resulting in an angle of

\pi

.

Solution approach

Correct Answer + Core Reason: $\pi$ . Since a and c are in opposite directions, the angle between them is $\pi$ .

Step-by-Step Solution:

Step 1: Given that $a = 8b$ and $c = -7b$ , we can see that a and b are in the same direction because the scalar multiple is positive.

Step 2: b and c are in opposite directions because the scalar multiple is negative.

Step 3: Therefore, a and c must be in opposite directions.

Step 4: The angle between two vectors in opposite directions is $\pi$ radians or 180 degrees.

Why other options are wrong:
- Option A: $\frac{\pi}{4}$ is incorrect because the vectors are in opposite directions, not at a 45-degree angle.
- Option B: $\frac{\pi}{2}$ is incorrect because the vectors are in opposite directions, not perpendicular.
- Option D: 0 is incorrect because the vectors are in opposite directions, not in the same direction.

Key Concepts:
- Direction of vectors
- Scalar multiplication of vectors
- Angle between vectors

Memory Aids:
- Positive scalar multiple: same direction
- Negative scalar multiple: opposite direction

Quick Tips:
- Visualize the vectors to determine their relative directions.
- Remember that opposite directions correspond to an angle of $\pi$ .

92 Easy

92 of 257

Application +4 | -1

If the vectors a and b are linearly independent satisfying

(\sqrt 3 \tan \theta + 1)a + (\sqrt 3 \sec \theta - 2)b = 0,

then the most general value of

\theta

are

View answer and explanation

Correct option: D

2n\pi + \frac{{11\pi }}{6},n \in Z

Option D (

2n\pi + \frac{{11\pi }}{{6}},n \in Z

) is correct because it satisfies both equations derived from the linear independence condition.

Solution approach

Correct Answer + Core Reason: $2n\pi + \frac{{11\pi }}{{6}},n \in Z$ . This value of $\theta$ satisfies both the tangent and secant equations.

Step-by-Step Solution:

Step 1: Since a and b are linearly independent, the coefficients must be zero: $\sqrt 3 \tan \theta + 1 = 0$ and $\sqrt 3 \sec \theta - 2 = 0$

Step 2: Solve the first equation for $\tan \theta$ : $\tan \theta = -\frac{1}{{\sqrt 3 }}$

Step 3: Solve the second equation for $\sec \theta$ : $\sec \theta = \frac{2}{{\sqrt 3 }}$ which means $\cos \theta = \frac{{\sqrt 3 }}{2}$

Step 4: Find the general solution for $\theta$ that satisfies both equations. $\theta = 2n\pi + \frac{{11\pi }}{6}$ , where n is an integer.

Why other options are wrong:
- Option A: $n\pi - \frac{\pi }{6},n \in Z$ : This only satisfies the tangent equation but not the secant equation.
- Option B: $2n\pi \pm \frac{{11\pi }}{6},n \in Z$ : This would imply that $\cos \theta$ can be both positive and negative, which is not true.
- Option C: $n\pi \pm \frac{\pi }{6},n \in Z$ : This does not satisfy the secant equation.

Key Concepts:
- Linear independence of vectors
- Trigonometric equations
- General solutions of trigonometric equations

Memory Aids:
- Remember the unit circle values for sine, cosine, and tangent.

Quick Tips:
- When dealing with linear independence, set the coefficients to zero.
- Check if the solution satisfies all given equations.

93 Easy

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Application +4 | -1

The position vector of midpoint lying on the line joining the points whose position vectors are

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

and

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

is

View answer and explanation

Correct option: B

\widehat \mathbf{i}

Option B (

\widehat\mathbf{i}

) is correct because the midpoint of a line segment is found by averaging the position vectors of the endpoints.

Solution approach

Correct Answer + Core Reason: $\widehat \mathbf{i}$ . The midpoint is found by averaging the position vectors.

Step-by-Step Solution:

Step 1: Identify the position vectors of the two points:
$\vec{a} = \widehat\mathbf{i} + \widehat\mathbf{j} - \widehat\mathbf{k}$
$\vec{b} = \widehat\mathbf{i} - \widehat\mathbf{j} + \widehat\mathbf{k}$

Step 2: Use the midpoint formula:
$\vec{m} = \frac{\vec{a} + \vec{b}}{2}$

Step 3: Substitute the position vectors:
$\vec{m} = \frac{(\widehat\mathbf{i} + \widehat\mathbf{j} - \widehat\mathbf{k}) + (\widehat\mathbf{i} - \widehat\mathbf{j} + \widehat\mathbf{k})}{2}$

Step 4: Simplify the expression:
$\vec{m} = \frac{2\widehat\mathbf{i}}{2} = \widehat\mathbf{i}$

Why other options are wrong:
- Option A: $\widehat\mathbf{j}$ : This is incorrect. It might be a result of incorrectly adding or subtracting the vectors.
- Option C: $\mathbf{\hat k}$ : This is incorrect. It might be a result of incorrectly adding or subtracting the vectors.
- Option D: ${\vec 0}$ : This is incorrect. It might be a result of incorrectly adding or subtracting the vectors.

Key Concepts:
- Midpoint formula for position vectors.

Memory Aids:
- Remember that the midpoint is the average of the coordinates.

Quick Tips:
- Always add the vectors component-wise before dividing by 2.

94 Easy

94 of 257

Application +4 | -1

The two vectors

\vec a = 2\hat i + \hat j + 3\hat k,\;\vec b = 4\hat i - \lambda \hat j + 6\hat k

are parallel if

\lambda =

View answer and explanation

Correct option: D

- 2

Option D (-2) is correct because parallel vectors have proportional components.

Solution approach

Correct Answer + Core Reason: $- 2$ . Parallel vectors have proportional components.

Step-by-Step Solution:

Step 1: For parallel vectors, the ratio of their components must be equal: $\frac{2}{4} = \frac{1}{{ - \lambda }} = \frac{3}{6}$

Step 2: Solve for $\lambda$ : $\frac{2}{4} = \frac{1}{{ - \lambda }} \Rightarrow 2( - \lambda ) = 4 \Rightarrow - 2\lambda = 4 \Rightarrow \lambda = -2$

Why other options are wrong:
- Option A: This is incorrect because if $\lambda = 2$ , the vectors are not parallel: $\frac{2}{4} \ne \frac{1}{{ - 2}}$
- Option B: This is incorrect because if $\lambda = -3$ , the vectors are not parallel: $\frac{1}{{ - ( - 3)}} = \frac{1}{3} \ne \frac{2}{4}$
- Option C: This is incorrect because if $\lambda = 3$ , the vectors are not parallel: $\frac{1}{{ - 3}} \ne \frac{2}{4}$

Key Concepts:
- Parallel vectors
- Proportional components

Memory Aids:
- Parallel vectors are scalar multiples of each other.

Quick Tips:
- Set up the ratios of corresponding components and solve for the unknown.

95 Easy

95 of 257

Application +4 | -1

If

|\vec{a}|=4

and

-3 \leq \lambda \leq 2

, then the range of

|\lambda \vec{a}|

is

View answer and explanation

Correct option: C

[0,12]

Option C ([0,12]) is correct because the range of the magnitude of the vector is determined by the range of the scalar multiplier.

Solution approach

Correct Answer + Core Reason: [0,12]. The range is determined by the given range of $\lambda$ and the magnitude of $\vec{a}$ .

Step-by-Step Solution:

Step 1: We are given that $|\vec{a}| = 4$ and $-3 \leq \lambda \leq 2$ .

Step 2: We want to find the range of $|\lambda \vec{a}|$ .

Step 3: We know that $|\lambda \vec{a}| = |\lambda| |\vec{a}|$ .

Step 4: Since $-3 \leq \lambda \leq 2$ , we have $0 \leq |\lambda| \leq 3$ .

Step 5: Therefore, $0 \leq |\lambda| |\vec{a}| \leq 3 \times 4$ .

Step 6: This simplifies to $0 \leq |\lambda \vec{a}| \leq 12$ .

Step 7: Hence, the range of $|\lambda \vec{a}|$ is $[0, 12]$ .

Why other options are wrong:
- Option A: $[0,8]$ : This is incorrect because it only considers the positive upper bound of $\lambda$ and ignores the larger magnitude possible with the negative bound.
- Option B: $[-12,8]$ : This is incorrect because the magnitude of a vector cannot be negative.
- Option D: $[8,12]$ : This is incorrect because it doesn't consider the possibility of $\lambda$ being zero, which would result in a magnitude of zero.

Key Concepts:
- Magnitude of a vector
- Scalar multiplication of a vector
- Absolute value

Memory Aids:
- The magnitude of a vector is always non-negative.
- $|\lambda \vec{a}| = |\lambda| |\vec{a}|$

Quick Tips:
- Remember that the magnitude of any vector is always non-negative.
- When dealing with ranges, consider both positive and negative bounds.

96 Easy

96 of 257

Application +4 | -1

If

A B C D

is a parallelogram,

\overrightarrow{A B}=2 \mathbf{i}+4 \mathbf{j}-5 \mathbf{k}

and

\overrightarrow{A D}=\mathbf{i}+2 \mathbf{j}+3 \mathbf{k}

, then the unit vector in the direction of

B D

is

View answer and explanation

Correct option: C

\frac{1}{\sqrt{69}}(-\mathbf{i}-2 \mathbf{j}+8 \mathbf{k})

Option C (

\frac{1}{\sqrt{69}}(-\mathbf{i}-2 \mathbf{j}+8 \mathbf{k})

) is correct because the unit vector in the direction of BD is calculated by subtracting the vectors and normalizing the result.

Solution approach

Correct Answer + Core Reason: $\frac{1}{\sqrt{69}}(-\mathbf{i}-2 \mathbf{j}+8 \mathbf{k})$ . The unit vector is found by subtracting the vectors and normalizing the result.

Step-by-Step Solution:

Step 1: Find $\overrightarrow{BD}$ using the parallelogram property: $\overrightarrow{AB} + \overrightarrow{BD} = \overrightarrow{AD}$ .\nTherefore, $\overrightarrow{BD} = \overrightarrow{AD} - \overrightarrow{AB}$ .

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

Step 3: Find the magnitude of $\overrightarrow{BD}$ :
$|\overrightarrow{BD}| = \sqrt{(-1)^2 + (-2)^2 + 8^2} = \sqrt{1 + 4 + 64} = \sqrt{69}$ .

Step 4: Find the unit vector in the direction of $\overrightarrow{BD}$ :\nUnit vector = $\frac{\overrightarrow{BD}}{|\overrightarrow{BD}|} = \frac{-\mathbf{i} - 2\mathbf{j} + 8\mathbf{k}}{\sqrt{69}} = \frac{1}{\sqrt{69}}(-\mathbf{i} - 2\mathbf{j} + 8\mathbf{k})$ .

Why other options are wrong:
- **Option A ( $\frac{1}{\sqrt{69}}(i+2 j-8 k)$ ):** Incorrect sign in the calculation of $\overrightarrow{BD}$ .
- **Option B ( $\frac{1}{69}(\mathbf{i}+2 \mathbf{j}-8 \mathbf{k})$ ):** Incorrect magnitude and incorrect sign in the calculation of $\overrightarrow{BD}$ .
- **Option D ( $\frac{1}{69}(-i-2 j+8 k)$ ): Incorrect magnitude in the calculation of the unit vector.

Key Concepts:
- Vector subtraction
- Magnitude of a vector
- Unit vector
- Properties of a parallelogram

Memory Aids:
- Remember that a unit vector is found by dividing a vector by its magnitude.

Quick Tips:**
- Always check the signs when subtracting vectors.

97 Easy

97 of 257

Analyzing +4 | -1

Which of the following is not a unit vector for all values of

\theta

View answer and explanation

Correct option: C

(\sin 2 \theta) \hat{i}-(\cos \theta) \hat{j}

Option C (

\sin 2 \theta \hat{i}-(\cos \theta) \hat{j}

) is correct because the sum of the squares of its components is not always equal to 1, which is a requirement for a unit vector.

Solution approach

Correct Answer + Core Reason: $(\sin 2 \theta) \hat{i}-(\cos \theta) \hat{j}$ . A unit vector must have a magnitude of 1.

Step-by-Step Solution:

Step 1: Recall that a vector $a \hat{i} + b \hat{j}$ is a unit vector if $a^2 + b^2 = 1$ .

Step 2: Check each option:
- A: $(\cos \theta)^2 + (-\sin \theta)^2 = \cos^2 \theta + \sin^2 \theta = 1$
- B: $(\sin \theta)^2 + (\cos \theta)^2 = \sin^2 \theta + \cos^2 \theta = 1$
- C: $(\sin 2 \theta)^2 + (-\cos \theta)^2 = \sin^2 2\theta + \cos^2 \theta$ . This is not always equal to 1.
- D: $(\cos 2 \theta)^2 + (-\sin 2 \theta)^2 = \cos^2 2\theta + \sin^2 2\theta = 1$

Step 3: Option C does not always result in 1, so it is not a unit vector for all values of $\theta$ .

Why other options are wrong:
- **Option A ( $(\cos \theta) \hat{i}-(\sin \theta) \hat{j}$ ):** $(\cos \theta)^2 + (-\sin \theta)^2 = 1$ , so it is a unit vector.
- **Option B ( $(\sin \theta) \hat{i}+(\cos \theta) \hat{j}$ ):** $(\sin \theta)^2 + (\cos \theta)^2 = 1$ , so it is a unit vector.
- **Option D ( $(\cos 2 \theta) \hat{i}-(\sin 2 \theta) \hat{j}$ ):** $(\cos 2 \theta)^2 + (-\sin 2 \theta)^2 = 1$ , so it is a unit vector.

Key Concepts:
- Unit vector
- Trigonometric identities

Memory Aids:
- $\sin^2 \theta + \cos^2 \theta = 1$

Quick Tips:
- Remember the definition of a unit vector: its magnitude must be 1.

98 Easy

98 of 257

Application +4 | -1

If the diagonals of a parallelogram are

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

and

\hat i - 3\hat j + 4\hat k

, then the lengths of its sides are

View answer and explanation

Correct option: B

\sqrt 6 ,\;\sqrt {14}

Option B (√6, √14) is correct because the lengths of the sides are calculated using the parallelogram law of vector addition and subtraction.

Solution approach

Correct Answer + Core Reason: √6, √14. The lengths of the sides are found by using the diagonals to solve for the sides as vectors, then finding their magnitudes.

Step-by-Step Solution:

Step 1: Let $\vec{a}$ and $\vec{b}$ be the sides of the parallelogram. The diagonals are $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ .

Step 2: Given diagonals are $3\hat{i} + \hat{j} - 2\hat{k}$ and $\hat{i} - 3\hat{j} + 4\hat{k}$ .\nSo, $\vec{a} + \vec{b} = 3\hat{i} + \hat{j} - 2\hat{k}$ and $\vec{a} - \vec{b} = \hat{i} - 3\hat{j} + 4\hat{k}$ .

Step 3: Adding the two equations, we get $2\vec{a} = 4\hat{i} - 2\hat{j} + 2\hat{k} \Rightarrow \vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ .\nSubtracting the two equations, we get $2\vec{b} = 2\hat{i} + 4\hat{j} - 6\hat{k} \Rightarrow \vec{b} = \hat{i} + 2\hat{j} - 3\hat{k}$ .

Step 4: Find the magnitudes of $\vec{a}$ and $\vec{b}$ :
$|\vec{a}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$
$|\vec{b}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$

Why other options are wrong:
- Option A: Incorrect vector addition/subtraction or magnitude calculation.
- Option C: Incorrect vector addition/subtraction or magnitude calculation.
- Option D: Incorrect vector addition/subtraction or magnitude calculation.

Key Concepts:
- Vector addition and subtraction.
- Magnitude of a vector.
- Parallelogram law.

Memory Aids:
- Remember how to find the magnitude of a vector: $|\vec{v}| = \sqrt{x^2 + y^2 + z^2}$

Quick Tips:
- Use the parallelogram law to relate sides and diagonals.

99 Easy

99 of 257

Analyzing +4 | -1

Let the position vectors of the points

A, B, C

and

D

be

5 \hat{i}+5 \hat{j} +2 \lambda \hat{k}, \hat{i}+2 \hat{j}+3 \hat{k},-2 \hat{i}+\lambda \hat{j}+4 \hat{k}

and

-\hat{i}+5 \hat{j}+6 \hat{k}

. Let the set

S=\{\lambda \in \mathbb{R}

: the points

A

,

B, C

and

D

are coplanar}. Then

\sum\limits_{{\lambda \in S}}(\lambda+2)^{2}

is equal to

View answer and explanation

Correct option: D

41

The correct answer is D (41) because the sum of

(\lambda+2)^2

for the values of

\lambda

that make the points coplanar is 41.

Solution approach

Correct Answer + Core Reason: 41. The values of $\lambda$ that make the points coplanar are found by setting the scalar triple product of the vectors formed by the points to zero, and then summing $(\lambda+2)^2$ for these values.

Step-by-Step Solution:

Step 1: Define the points A, B, C, and D as vectors: $A = 5 \hat{i}+5 \hat{j} +2 \lambda \hat{k}$ , $B = \hat{i}+2 \hat{j}+3 \hat{k}$ , $C = -2 \hat{i}+\lambda \hat{j}+4 \hat{k}$ , and $D = -\hat{i}+5 \hat{j}+6 \hat{k}$ .

Step 2: Find the vectors $\overrightarrow{AB}$ , $\overrightarrow{AC}$ , and $\overrightarrow{AD}$ .
$\overrightarrow{AB} = B - A = (1-5)\hat{i} + (2-5)\hat{j} + (3-2\lambda)\hat{k} = -4\hat{i} - 3\hat{j} + (3-2\lambda)\hat{k}$
$\overrightarrow{AC} = C - A = (-2-5)\hat{i} + (\lambda-5)\hat{j} + (4-2\lambda)\hat{k} = -7\hat{i} + (\lambda-5)\hat{j} + (4-2\lambda)\hat{k}$
$\overrightarrow{AD} = D - A = (-1-5)\hat{i} + (5-5)\hat{j} + (6-2\lambda)\hat{k} = -6\hat{i} + 0\hat{j} + (6-2\lambda)\hat{k}$

Step 3: For the points to be coplanar, the scalar triple product of these vectors must be zero: $[\overrightarrow{AB} \\, \overrightarrow{AC} \\, \overrightarrow{AD}]$

100 Easy

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Analyzing +4 | -1

Let

A, B, C

be three points whose position vectors respectively are

\vec{a}=\hat{i}+4 \hat{j}+3 \hat{k}

\vec{b}=2 \hat{i}+\alpha \hat{j}+4 \hat{k}, \alpha \in \mathbb{R}

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

If

\alpha

is the smallest positive integer for which

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

are non collinear, then the length of the median, in

\triangle A B C

, through

A

is:

View answer and explanation

Correct option: A

\frac{\sqrt{82}}{2}

Option A (

\frac{\sqrt{82}}{2}

) is correct because it is the calculated length of the median AM after determining the smallest positive integer value of

\alpha

for which the points are non-collinear.

Solution approach

Correct Answer + Core Reason: $\frac{\sqrt{82}}{2}$ . This is the length of the median AM after finding the smallest positive integer value of $\alpha$ for which the points are non-collinear.

Step-by-Step Solution:

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

Step 2: Find the vectors $\overrightarrow{A B}$ and $\overrightarrow{A C}$ :
$\overrightarrow{A B} = \vec{b} - \vec{a} = (2-1)\hat{i} + (\alpha-4)\hat{j} + (4-3)\hat{k} = \hat{i} + (\alpha-4)\hat{j} + \hat{k}$
$\overrightarrow{A C} = \vec{c} - \vec{a} = (3-1)\hat{i} + (-2-4)\hat{j} + (5-3)\hat{k} = 2\hat{i} - 6\hat{j} + 2\hat{k}$

Step 3: For A, B, and C to be collinear, $\overrightarrow{A B}$ must be parallel to $\overrightarrow{A C}$ . This means their components are proportional:
$\frac{1}{2} = \frac{\alpha-4}{-6} = \frac{1}{2}$

Step 4: Solve for $\alpha$ :
$\frac{1}{2} = \frac{\alpha-4}{-6} \Rightarrow -6 = 2(\alpha-4) \Rightarrow -6 = 2\alpha - 8 \Rightarrow 2\alpha = 2 \Rightarrow \alpha = 1$

Step 5: Since we need the smallest positive integer for which A, B, C are non-collinear, $\alpha = 2$ .

Step 6: Find the midpoint M of BC:
$\vec{m} = \frac{\vec{b} + \vec{c}}{2} = \frac{(2\hat{i}+2 \hat{j}+4 \hat{k}) + (3\hat{i}-2 \hat{j}+5 \hat{k})}{2} = \frac{5}{2}\hat{i} + 0\hat{j} + \frac{9}{2}\hat{k}$

Step 7: Find the vector $\overrightarrow{A M}$ :
$\overrightarrow{A M} = \vec{m} - \vec{a} = (\frac{5}{2}-1)\hat{i} + (0-4)\hat{j} + (\frac{9}{2}-3)\hat{k} = \frac{3}{2}\hat{i} - 4\hat{j} + \frac{3}{2}\hat{k}$

Step 8: Find the length of the median AM:
$AM = |\overrightarrow{A M}| = \sqrt{(\frac{3}{2})^2 + (-4)^2 + (\frac{3}{2})^2} = \sqrt{\frac{9}{4} + 16 + \frac{9}{4}} = \sqrt{\frac{18}{4} + 16} = \sqrt{\frac{9}{2} + 16} = \sqrt{\frac{9+32}{2}} = \sqrt{\frac{41}{2}} = \sqrt{\frac{82}{4}} = \frac{\sqrt{82}}{2}$

Why other options are wrong:
- Option B: $\frac{\sqrt{62}}{2}$ - Incorrect calculation of the median length.
- Option C: $\frac{\sqrt{69}}{2}$ - Incorrect calculation of the median length.
- Option D: $\frac{\sqrt{66}}{2}$ - Incorrect calculation of the median length.

Key Concepts:
- Position Vectors
- Collinearity of Points
- Median of a Triangle
- Vector Operations

Memory Aids:
- Remember the formula for the midpoint of a line segment.
- Remember the condition for collinearity: Vectors are parallel.

Quick Tips:
- Draw a diagram to visualize the problem.
- Use vector operations to find the required vectors and lengths.

101 Easy

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Understanding +4 | -1

In the triangle

A B C, \overrightarrow{A B}=\mathbf{a}, \overrightarrow{A C}=\mathbf{c}, \overrightarrow{B C}=\mathbf{b}

, then

View answer and explanation

Correct option: B

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

Option B (

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

) is correct because it represents the correct vector addition around the triangle, where

\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = 0

.

Solution approach

Correct Answer + Core Reason: $\mathbf{a} + \mathbf{b} - \mathbf{c} = \mathbf{0}$ . This represents the correct vector addition around the triangle.

Step-by-Step Solution:

Step 1: Write the vector equation for the triangle ABC: $\overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \mathbf{0}$ .

Step 2: Express $\overrightarrow{CA}$ in terms of $\overrightarrow{AC}$ : $\overrightarrow{CA} = -\overrightarrow{AC}$ .

Step 3: Substitute the given vectors: $\mathbf{a} + \mathbf{b} + (-\mathbf{c}) = \mathbf{0}$ .

Step 4: Simplify the equation: $\mathbf{a} + \mathbf{b} - \mathbf{c} = \mathbf{0}$ .

Why other options are wrong:
- Option A: This is incorrect because it implies $\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}$ , which is not true for the given triangle.
- Option C: This is incorrect because it implies $\mathbf{a} - \mathbf{b} + \mathbf{c} = \mathbf{0}$ , which is not true for the given triangle.
- Option D: This is incorrect because it implies $-\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}$ , which is not true for the given triangle.

Key Concepts:
- Vector addition
- Triangle law of vector addition

Memory Aids:
- Remember that the sum of vectors around a closed triangle is zero.

Quick Tips:
- Pay attention to the direction of the vectors when applying the triangle law.

102 Easy

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Application +4 | -1

A line segment has length 63 and direction ratios are 3, -2 and 6. The components of line vector are

View answer and explanation

Correct option: B

27, -18, 54

Option B (27, -18, 54) is correct because the components are proportional to the direction ratios and the length of the segment is 63.

Solution approach

Correct Answer + Core Reason: 27, -18, 54. The components are calculated using the given length and direction ratios.

Step-by-Step Solution:

Step 1: Let the components of the line segment on the axes be x, y, and z.

Step 2: We have ${x^2} + {y^2} + {z^2} = {63^2}$ .

Step 3: The direction ratios are 3, -2, and 6, so $\frac{x}{3} = \frac{y}{-2} = \frac{z}{6} = k$ .

Step 4: Thus, $x = 3k$ , $y = -2k$ , and $z = 6k$ .

Step 5: Substitute these into the equation from Step 2: ${(3k)^2} + {(-2k)^2} + {(6k)^2} = {63^2}$ , which simplifies to $9{k^2} + 4{k^2} + 36{k^2} = {63^2}$ , or $49{k^2} = {63^2}$ .

Step 6: Solve for k: $k^2 = \frac{{63^2}}{{49}} = \frac{{(9 \cdot 7)^2}}{{7^2}} = 81$ , so $k = \pm 9$ .

Step 7: Calculate the components for both values of k:\nFor $k = 9$ : $x = 3(9) = 27$ , $y = -2(9) = -18$ , $z = 6(9) = 54$ .\nFor $k = -9$ : $x = 3(-9) = -27$ , $y = -2(-9) = 18$ , $z = 6(-9) = -54$ .

Step 8: Therefore, the components are (27, -18, 54) or (-27, 18, -54).

Why other options are wrong:
- Option A (-27, 18, 54): This is incorrect because the z-component should be -54 if x and y are -27 and 18 respectively.
- Option C (27, -18, -54): This is incorrect because the z-component should be 54 if x and y are 27 and -18 respectively.
- Option D (-27, -18, -54): This is incorrect because the x and y components should be 27 and 18 respectively.

Key Concepts:
- Direction ratios
- Components of a vector

Memory Aids:
- Remember that the components are proportional to the direction ratios.

Quick Tips:
- Always check the magnitude of the resulting vector to ensure it matches the given length.

103 Easy

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Analyzing +4 | -1

Let P be the plane passing through the intersection of the planes

\vec r \cdot (\hat i + 3\hat j - \hat k) = 5

and

\vec r \cdot (2\hat i - \hat j + \hat k) = 3,

and the point

(2,1, - 2).

Let the position vectors of the points X and Y be

\hat i - 2\hat j + 4\hat k

and

5\hat i - \hat j + 2\hat k

respectively. Then the points

View answer and explanation

Correct option: C

X and Y are on the opposite sides of P

Option C (X and Y are on the opposite sides of P) is correct because by finding the equation of the plane and substituting the coordinates of X and Y, we can determine their positions relative to the plane.

Solution approach

Correct Answer + Core Reason: X and Y are on the opposite sides of P. This is determined by finding the equation of the plane and checking the sign of the expression when the coordinates of X and Y are substituted.

Step-by-Step Solution:

Step 1: The equation of the plane passing through the intersection of the planes $\vec r \cdot (\hat i + 3\hat j - \hat k) = 5$ and $\vec r \cdot (2\hat i - \hat j + \hat k) = 3$ is given by:
$(\vec r \cdot (\hat i + 3\hat j - \hat k) - 5) + \lambda(\vec r \cdot (2\hat i - \hat j + \hat k) - 3) = 0$ \nIn Cartesian coordinates, this is:
$(x + 3y - z - 5) + \lambda(2x - y + z - 3) = 0$

Step 2: The plane passes through the point (2, 1, -2). Substitute these coordinates into the equation:
$(2 + 3(1) - (-2) - 5) + \lambda(2(2) - 1 + (-2) - 3) = 0$
$(2 + 3 + 2 - 5) + \lambda(4 - 1 - 2 - 3) = 0$
$2 + \lambda(-2) = 0$
$\lambda = 1$

Step 3: Substitute $\lambda = 1$ back into the equation of the plane:
$(x + 3y - z - 5) + (2x - y + z - 3) = 0$
$3x + 2y - 8 = 0$

Step 4: Now, let's check the positions of points X and Y with respect to the plane. The position vector of X is $\hat i - 2\hat j + 4\hat k$ , so X = (1, -2, 4). The position vector of Y is $5\hat i - \hat j + 2\hat k$ , so Y = (5, -1, 2).

Step 5: Substitute the coordinates of X into the equation of the plane:
$3(1) + 2(-2) - 8 = 3 - 4 - 8 = -9 < 0$

Step 6: Substitute the coordinates of Y into the equation of the plane:
$3(5) + 2(-1) - 8 = 15 - 2 - 8 = 5 > 0$

Step 7: Since the values have opposite signs, X and Y are on opposite sides of the plane.

Why other options are wrong:
- Option A (X and X + Y are on the same side of P): X + Y = (6, -3, 6). 3(6) + 2(-3) - 8 = 18 - 6 - 8 = 4 > 0. Since X is negative and X+Y is positive, they are on opposite sides.
- Option B (Y and Y – X are on the opposite sides of P): Y - X = (4, 1, -2). 3(4) + 2(1) - 8 = 12 + 2 - 8 = 6 > 0. Since Y is positive and Y-X is positive, they are on the same side.
- Option D (X + Y and X – Y are on the same side of P): X - Y = (-4, -1, 2). 3(-4) + 2(-1) - 8 = -12 - 2 - 8 = -22 < 0. Since X+Y is positive and X-Y is negative, they are on opposite sides.

Key Concepts:
- Equation of a plane
- Intersection of planes
- Position of a point with respect to a plane

Memory Aids:
- The equation of a plane passing through the intersection of two planes can be written as a linear combination of the two plane equations.

Quick Tips:
- To determine the position of a point with respect to a plane, substitute the coordinates of the point into the equation of the plane. The sign of the result indicates which side of the plane the point is on.

104 Easy

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Application +4 | -1

If

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

be any three non-coplanar vectors, then

[a+b ~~~b+c~~~~ c+a]=

View answer and explanation

Correct option: B

2[a b c]

Option B (2[a b c]) is correct because expanding the scalar triple product results in 2[a b c].

Solution approach

Correct Answer + Core Reason: $2[a b c]$ . Expanding the scalar triple product results in $2[a b c]$ .

Step-by-Step Solution:

Step 1: Expand the scalar triple product:
$[a+b \ b+c \ c+a] = (a+b) \cdot ((b+c) \times (c+a))$ .

Step 2: Expand the cross product:
$(b+c) \times (c+a) = b \times c + b \times a + c \times c + c \times a = b \times c + b \times a + 0 + c \times a = b \times c + b \times a + c \times a$ .

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

Step 4: Simplify using properties of scalar triple product:
$a \cdot (b \times c) = [a b c]$
$a \cdot (b \times a) = 0$
$a \cdot (c \times a) = 0$
$b \cdot (b \times c) = 0$
$b \cdot (b \times a) = 0$
$b \cdot (c \times a) = [b c a] = [a b c]$

Step 5: Combine the terms:
$[a b c] + 0 + 0 + 0 + 0 + [a b c] = 2[a b c]$ .

Step 6: Final Answer: $2[a b c]$

Why other options are wrong:
- **Option A ( $|a b c|$ ):** This is incorrect because the correct expansion yields $2[a b c]$ .
- **Option C ( $[\mathbf{a b c}]^2$ ):** This is incorrect because the correct expansion yields $2[a b c]$ .
- **Option D ( $2[a b c]^2$ ):** This is incorrect because the correct expansion yields $2[a b c]$ .

Key Concepts:
- Scalar triple product
- Properties of cross product
- Properties of dot product

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

Quick Tips:
- Remember the properties of the scalar triple product to simplify calculations.

105 Easy

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Understanding +4 | -1

Which of the following is/are true?

I. In a zero vector, initial and terminal points coincide.

II. Zero vector is denoted as

\mathrm{O}

.

III. Zero vector has zero magnitude.

View answer and explanation

Correct option: D

All are true

All are true because a zero vector has coinciding initial and terminal points, is denoted as O, and has zero magnitude.

Solution approach

Correct Answer + Core Reason: All are true because all three statements accurately describe the properties of a zero vector.

Step-by-Step Solution:

Step 1: Analyze statement I: In a zero vector, initial and terminal points coincide. This is true by definition.

Step 2: Analyze statement II: Zero vector is denoted as $\mathrm{O}$ . This is the standard notation for a zero vector.

Step 3: Analyze statement III: Zero vector has zero magnitude. This is also true by definition.

Step 4: Conclude that all three statements are true.

Why other options are wrong:
- Option A: I and III are also true.
- Option B: II is also true.
- Option C: I is also true.

Key Concepts:
- Zero vector
- Magnitude of a vector

Memory Aids:
- Think of a zero vector as a point.

Quick Tips:
- Remember the properties of a zero vector.

106 Easy

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Understanding +4 | -1

The value of

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

where

\left| {\vec a} \right| = 1,\left| {\vec b} \right| = 5,\left| {\vec c} \right| = 3,

is

View answer and explanation

Correct option: A

0

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

Solution approach

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

Step-by-Step Solution:

Step 1: Recognize the scalar triple product: $\left[ {\vec a - \vec b,\vec b - \vec c,\vec c - \vec a} \right]$ .

Step 2: Expand the scalar triple product:
$\left[ {\vec a - \vec b,\vec b - \vec c,\vec c - \vec a} \right] = (\vec a - \vec b) \cdot ((\vec b - \vec c) \times (\vec c - \vec a))$

Step 3: Simplify the cross product:
$(\vec b - \vec c) \times (\vec c - \vec a) = \vec b \times \vec c - \vec b \times \vec a - \vec c \times \vec c + \vec c \times \vec a = \vec b \times \vec c - \vec b \times \vec a + \vec c \times \vec a$ (since $\vec c \times \vec c = 0$ )

Step 4: Simplify the dot product:
$(\vec a - \vec b) \cdot (\vec b \times \vec c - \vec b \times \vec a + \vec c \times \vec a) = \vec a \cdot (\vec b \times \vec c) - \vec a \cdot (\vec b \times \vec a) + \vec a \cdot (\vec c \times \vec a) - \vec b \cdot (\vec b \times \vec c) + \vec b \cdot (\vec b \times \vec a) - \vec b \cdot (\vec c \times \vec a) = \vec a \cdot (\vec b \times \vec c) - 0 + 0 - 0 + 0 - \vec b \cdot (\vec c \times \vec a) = \left[ {\vec a\,\vec b\,\vec c} \right] - \left[ {\vec b\,\vec c\,\vec a} \right] = \left[ {\vec a\,\vec b\,\vec c} \right] - \left[ {\vec a\,\vec b\,\vec c} \right] = 0$

Step 5: Alternatively, note that $\vec a - \vec b + \vec b - \vec c + \vec c - \vec a = 0$ , which means the vectors $\vec a - \vec b$ , $\vec b - \vec c$ , and $\vec c - \vec a$ are coplanar. Therefore, their scalar triple product is zero.

Why other options are wrong:
- Option B: Incorrect. The scalar triple product is 0, not 1.
- Option C: Incorrect. The scalar triple product is 0, not 6.
- Option D: Incorrect. The correct answer is 0.

Key Concepts:
- Scalar triple product
- Coplanar vectors

Memory Aids:
- If three vectors sum to zero, they are coplanar.
- The scalar triple product of coplanar vectors is zero.

Quick Tips:
- If the sum of the vectors in a scalar triple product is zero, the result is zero.

107 Easy

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Analyzing +4 | -1

The vector

\mathbf{\vec a} = \mathrm{\alpha }\widehat \mathbf{i} + 2\widehat \mathbf{j} + \mathrm{\beta }\mathbf{\hat k}

lies in the plane of the vectors

\mathbf{\vec b} = \widehat \mathbf{i} + \widehat \mathbf{j}

and

\vec c = \widehat \mathbf{j} + \mathbf{\hat k}

and bisects the angle between

{\mathbf{\vec b}}

and

{\mathbf{\vec c}}

Then, which one of the following gives possible value of

\mathrm{\alpha }

and

\mathrm{\beta }

?

View answer and explanation

Correct option: A

\mathrm{\alpha } = 1,\mathrm{\;\beta } = 1

Option A (

\mathrm{\alpha } = 1,\mathrm{\;\beta } = 1

) is correct because the condition for coplanarity leads to the equation

\alpha + \beta = 2

, which is satisfied by

\alpha = 1

and

\beta = 1

.

Solution approach

Correct Answer + Core Reason: $\mathrm{\alpha } = 1,\mathrm{\;\beta } = 1$ . The condition for coplanarity leads to the equation $\alpha + \beta = 2$ , which is satisfied by $\alpha = 1$ and $\beta = 1$ .

Step-by-Step Solution:

Step 1: Given $\mathbf{\vec a} = \mathrm{\alpha }\widehat \mathbf{i} + 2\widehat \mathbf{j} + \mathrm{\beta }\mathbf{\hat k}$ , $\mathbf{\vec b} = \widehat \mathbf{i} + \widehat \mathbf{j}$ , and $\vec c = \widehat \mathbf{j} + \mathbf{\hat k}$ .

Step 2: Since $\mathbf{\vec a}$ , $\mathbf{\vec b}$ , and $\mathbf{\vec c}$ are coplanar, their scalar triple product is zero: $\left| {\begin{array}{cccccccccccccccccccc}\mathrm{\alpha }&2&\mathrm{\beta }}\\1&1&0\\0&1&1\end{array}} \right| = 0$ .

Step 3: Evaluate the determinant: $\mathrm{\alpha }(1 - 0) - 2(1 - 0) + \mathrm{\beta }(1 - 0) = 0 \Rightarrow \mathrm{\alpha } - 2 + \mathrm{\beta } = 0 \Rightarrow \mathrm{\alpha } + \mathrm{\beta } = 2$ .

Step 4: Check the given options:
- Option A: $\mathrm{\alpha } = 1$ , $\mathrm{\beta } = 1 \Rightarrow 1 + 1 = 2$ . This satisfies the equation.
- Option B: $\mathrm{\alpha } = 2$ , $\mathrm{\beta } = 2 \Rightarrow 2 + 2 = 4 \neq 2$ .
- Option C: $\mathrm{\alpha } = 1$ , $\mathrm{\beta } = 2 \Rightarrow 1 + 2 = 3 \neq 2$ .
- Option D: $\mathrm{\alpha } = 2$ , $\mathrm{\beta } = 1 \Rightarrow 2 + 1 = 3 \neq 2$ .

Step 5: Therefore, the only possible values are $\mathrm{\alpha } = 1$ and $\mathrm{\beta } = 1$ .

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

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

Memory Aids:
- The scalar triple product of coplanar vectors is zero.

Quick Tips:
- Remember the condition for coplanarity using the scalar triple product.

108 Easy

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Application +4 | -1

If vectors

3\hat i + 2\hat j - \hat k\,\,\,\,{\text{and}}\,\,\,6\hat i - 4x\hat j + y\hat k

are parallel, then the value of x and y will be

View answer and explanation

Correct option: A

-1, -2

Option A (-1, -2) is correct because it represents the values of x and y for which the given vectors are parallel, calculated using the condition for parallel vectors.

Solution approach

Correct Answer + Core Reason: -1, -2. The values of x and y are calculated using the condition for parallel vectors.

Step-by-Step Solution:

Step 1: If two vectors are parallel, their components are proportional. Therefore:
$\frac{3}{6} = \frac{2}{-4x} = \frac{-1}{y}$

Step 2: Solve for x and y.\nFrom $\frac{3}{6} = \frac{2}{-4x}$ , we get:
$\frac{1}{2} = \frac{2}{-4x}$
$-4x = 4$
$x = -1$ \nFrom $\frac{3}{6} = \frac{-1}{y}$ , we get:
$\frac{1}{2} = \frac{-1}{y}$
$y = -2$

Why other options are wrong:
- Option B: $1, -2$ : This is incorrect because x should be -1.
- Option C: $-1, 2$ : This is incorrect because y should be -2.
- Option D: $1, 2$ : This is incorrect because both x and y are incorrect.

Key Concepts:
- Parallel vectors

Memory Aids:
- Parallel vectors are scalar multiples of each other.

Quick Tips:
- Remember that if two vectors are parallel, their corresponding components are proportional.

109 Easy

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Application +4 | -1

In a trapezium, the vector

BC = \lambda AD.

We will then find that p = AC + BD is collinear with AD. If

p = \mu AD,

then

View answer and explanation

Correct option: A

\mu = \lambda + 1

Option A (

\mu = \lambda + 1

) is correct because by expressing AC + BD in terms of AD and BC, and using the given relationship BC = λAD, we can derive the relationship between μ and λ.

Solution approach

Correct Answer + Core Reason: $\mu = \lambda + 1$ . The relationship between $\mu$ and $\lambda$ is found by expressing AC + BD in terms of AD.

Step-by-Step Solution:

Step 1: Given that $BC = \lambda AD$ .

Step 2: We have $p = AC + BD = AC + BC + CD$ .

Step 3: Substitute $BC = \lambda AD$ : $p = AC + \lambda AD + CD$ .

Step 4: Rewrite $AC + CD$ as $AD$ : $p = \lambda AD + AD$ .

Step 5: Factor out AD: $p = (\lambda + 1)AD$ .

Step 6: Given that $p = \mu AD$ , we have $\mu AD = (\lambda + 1)AD$ .

Step 7: Therefore, $\mu = \lambda + 1$ .

Why other options are wrong:
- **Option B ( $\lambda = \mu + 1$ ):** This is incorrect because it implies $\mu = \lambda - 1$ , which contradicts the correct relationship.
- **Option C ( $\lambda + \mu = 1$ ):** This is incorrect because it implies $\mu = 1 - \lambda$ , which contradicts the correct relationship.
- **Option D ( $\mu = 2 + \lambda$ ):** This is incorrect because it implies $\mu = \lambda + 2$ , which contradicts the correct relationship.

Key Concepts:
- Vector addition
- Collinear vectors

Memory Aids:
- Remember that $AC + CD = AD$ .

Quick Tips:
- Always express vectors in terms of a common vector to simplify the equation.

110 Easy

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Application +4 | -1

The position vectors of the points

A,B,C

are

\left( {2\mathbf{\hat i} + \mathbf{\hat j} - \mathbf{\hat k}} \right),\mathrm{\;}\left( {3\mathbf{\hat i} - 2\mathbf{\hat j} + \mathbf{\hat k}} \right)

and

\left( {\mathbf{\hat i} + 4\mathbf{\hat j} - 3\mathbf{\hat k}} \right)

respectively. These points

View answer and explanation

Correct option: A

Form an isosceles triangle

Option A (Form an isosceles triangle) is correct because two sides of the triangle formed by the points have equal lengths.

Solution approach

Correct Answer + Core Reason: Form an isosceles triangle. Two sides of the triangle have equal lengths.

Step-by-Step Solution:

Step 1: Given the position vectors of points A, B, and C: $\overrightarrow {OA} = 2\hat i + \hat j - \hat k$ , $\overrightarrow {OB} = 3\hat i - 2\hat j + \hat k$ , and $\overrightarrow {OC} = \hat i + 4\hat j - 3\hat k$ .

Step 2: Find the vectors $\overrightarrow {AB}$ , $\overrightarrow {BC}$ , and $\overrightarrow {CA}$ .
\overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} = (3-2)\hat i + (-2-1)\hat j + (1-(-1))\hat k = \hat i - 3\hat j + 2\hat k $.\n\overrightarrow {BC} = \overrightarrow {OC} - \overrightarrow {OB} = (1-3)\hat i + (4-(-2))\hat j + (-3-1)\hat k = -2\hat i + 6\hat j - 4\hat k$ .
\overrightarrow {CA} = \overrightarrow {OA} - \overrightarrow {OC} = (2-1)\hat i + (1-4)\hat j + (-1-(-3))\hat k = \hat i - 3\hat j + 2\hat k $.\n\n**Step 3:** Calculate the magnitudes of the vectors.\n\left| {\overrightarrow {AB} } \right| = \sqrt {1^2 + (-3)^2 + 2^2} = \sqrt {1 + 9 + 4} = \sqrt {14}.\\\\ \left| {\overrightarrow {BC} } \right| = \sqrt {(-2)^2 + 6^2 + (-4)^2} = \sqrt {4 + 36 + 16} = \sqrt {56}.\\\\ \left| {\overrightarrow {CA} } \right| = \sqrt {1^2 + (-3)^2 + 2^2} = \sqrt {1 + 9 + 4} = \sqrt {14}.\\\\ \n**Step 4:** Compare the magnitudes. Since$ \left| {\overrightarrow {AB} } \right| = \left| {\overrightarrow {CA} } \right| = $\sqrt {14}$ $, the triangle is isosceles.\n\n**Why other options are wrong:**\n- **Option B (Form a right angled triangle):** This is incorrect because the sides do not satisfy the Pythagorean theorem.\n- **Option C (Are collinear):** This is incorrect because the points form a triangle, not a straight line.\n- **Option D (Form a scalene triangle):** This is incorrect because two sides are equal, meaning it's not scalene.\n\n**Key Concepts:**\n- Position vectors\n- Vector subtraction\n- Magnitude of a vector\n- Types of triangles\n\n**Memory Aids:**\n- Isosceles triangle: Two sides are equal.\n- Scalene triangle: All sides are different.\n- Collinear points: Points lie on the same line.\n\n**Quick Tips:**\n- To determine the type of triangle, calculate the lengths of all three sides.$

111 Easy

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Application +4 | -1

If three points

A, B

and

C

have position vectors

(1, x, 3),(3,4,7)

and

(y,-2,-5)

respectively and if they are collinear, then

(x, y)=

View answer and explanation

Correct option: A

(2,-3)

Option A ((2,-3)) is correct because if the points are collinear, the vectors formed by them are proportional, allowing us to solve for x and y.

Solution approach

Correct Answer + Core Reason: (2,-3). Collinear points imply proportional vectors.

Step-by-Step Solution:

Step 1: Find the vectors $\overrightarrow{AB}$ and $\overrightarrow{BC}$ :
$\overrightarrow{AB} = (3-1)\mathbf{i} + (4-x)\mathbf{j} + (7-3)\mathbf{k} = 2\mathbf{i} + (4-x)\mathbf{j} + 4\mathbf{k}$
$\overrightarrow{BC} = (y-3)\mathbf{i} + (-2-4)\mathbf{j} + (-5-7)\mathbf{k} = (y-3)\mathbf{i} - 6\mathbf{j} - 12\mathbf{k}$

Step 2: Since A, B, and C are collinear, $\overrightarrow{AB} = \lambda \overrightarrow{BC}$ for some scalar $\lambda$ :
$2\mathbf{i} + (4-x)\mathbf{j} + 4\mathbf{k} = \lambda [(y-3)\mathbf{i} - 6\mathbf{j} - 12\mathbf{k}]$

Step 3: Equate the components:
$2 = \lambda(y-3)$ ...(1)
$4-x = -6\lambda$ ...(2)
$4 = -12\lambda$ ...(3)

Step 4: Solve for $\lambda$ from equation (3):
$\lambda = -\frac{1}{3}$

Step 5: Substitute $\lambda$ into equation (1):
$2 = -\frac{1}{3}(y-3)$
$-6 = y-3$
$y = -3$

Step 6: Substitute $\lambda$ into equation (2):
$4-x = -6(-\frac{1}{3})$
$4-x = 2$
$x = 2$

Step 7: Therefore, $(x, y) = (2, -3)$

Why other options are wrong:
- Option B ((-2,3)): This is incorrect because substituting these values into the vector equations does not satisfy the collinearity condition.
- Option C ((2,3)): This is incorrect because substituting these values into the vector equations does not satisfy the collinearity condition.
- Option D ((-2,-3)): This is incorrect because substituting these values into the vector equations does not satisfy the collinearity condition.

Key Concepts:
- Collinear points
- Vector representation
- Scalar multiplication of vectors

Memory Aids:
- Collinear points lie on the same line.

Quick Tips:
- Express the points as vectors and use the property that collinear vectors are scalar multiples of each other.

112 Easy

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Understanding +4 | -1

The vectors a, b and a + b are

View answer and explanation

Correct option: B

coplanar

Option B (coplanar) is correct because the vector a+b lies in the plane defined by vectors a and b.

Solution approach

Correct Answer + Core Reason: coplanar. The vector a+b lies in the plane defined by vectors a and b.

Step-by-Step Solution:

Step 1: Consider the vectors a, b, and a+b.

Step 2: The vector a+b is the resultant of adding vectors a and b.

Step 3: The resultant vector a+b lies in the same plane as vectors a and b.

Step 4: Therefore, the vectors a, b, and a+b are coplanar.

Why other options are wrong:
- Option A (collinear): This is incorrect because a, b, and a+b are not necessarily along the same line.
- Option C (non – coplanar): This is incorrect because a+b is formed by a and b, thus they must be on the same plane.
- Option D (None of these): This is incorrect because the vectors are coplanar.

Key Concepts:
- Vector Addition
- Coplanar Vectors

Memory Aids:
- The sum of two vectors always lies in the plane defined by those two vectors.

Quick Tips:
- Visualize the vectors in a plane to understand their relationship.

113 Easy

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Application +4 | -1

If

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

and

\alpha \vec{a}+\beta \vec{b}+\gamma \vec{c}=-3(\hat{i}-\hat{k})

, then the number triple

(\alpha, \beta, \gamma)

is

View answer and explanation

Correct option: A

(2,-1,-1)

Option A ((2,-1,-1)) is correct because it satisfies the given vector equation when the components are equated and solved.

Solution approach

Correct Answer + Core Reason: (2,-1,-1). The correct triple satisfies the vector equation when components are equated and solved.

Step-by-Step Solution:

Step 1: Set up the equation: $\alpha(\hat{i}+2 \hat{j}+3 \hat{k})+\beta(2 \hat{i}+3 \hat{j}+\hat{k})+\gamma(3 \hat{i}+\hat{j}+2 \hat{k}) = -3(\hat{i}-\hat{k})$

Step 2: Equate the coefficients of $\hat{i}, \hat{j}, \hat{k}$ :
* $\alpha + 2\beta + 3\gamma = -3$ ...(i)
* $2\alpha + 3\beta + \gamma = 0$ ...(ii)
* $3\alpha + \beta + 2\gamma = 3$ ...(iii)

Step 3: Solve the system of equations. From (ii), $\gamma = -2\alpha - 3\beta$ . Substitute into (i) and (iii):
* $\alpha + 2\beta + 3(-2\alpha - 3\beta) = -3 => -5\alpha - 7\beta = -3$ ...(iv)
* $3\alpha + \beta + 2(-2\alpha - 3\beta) = 3 => -\alpha - 5\beta = 3$ ...(v)

Step 4: Solve (iv) and (v). From (v), $\alpha = -5\beta - 3$ . Substitute into (iv):
* $-5(-5\beta - 3) - 7\beta = -3 => 25\beta + 15 - 7\beta = -3 => 18\beta = -18 => \beta = -1$

Step 5: Find $\alpha$ and $\gamma$ :
* $\alpha = -5(-1) - 3 = 2$
* $\gamma = -2(2) - 3(-1) = -4 + 3 = -1$

Step 6: The solution is $(\alpha, \beta, \gamma) = (2, -1, -1)$

Why other options are wrong:
- Option B ((-2,1,1)): Substituting these values into the equations does not satisfy the original vector equation.
- Option C ((-2,-1,1)): Substituting these values into the equations does not satisfy the original vector equation.
- Option D ((2,1,-1)): Substituting these values into the equations does not satisfy the original vector equation.

Key Concepts:
- Vector addition
- Solving systems of linear equations
- Vector components

Memory Aids:
- Remember to equate coefficients of each unit vector.

Quick Tips:
- Always check your solution by substituting back into the original equations.
- When solving systems of equations, look for the easiest variable to isolate first.

114 Easy

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Analyzing +4 | -1

Five points given by

A,B,C,D,E

are in plane. Three forces

\overrightarrow {\mathbf{AC}} ,\;\overrightarrow {\;\mathbf{AD}}

and

\overrightarrow {\mathbf{AE}}

act a

A

and three forces

\overrightarrow {\mathbf{CB}} ,\overrightarrow {\mathbf{DB}} ,\overrightarrow {\mathbf{EB}}

act at

B

. Then, their resultant is

View answer and explanation

Correct option: B

3\overrightarrow {\mathbf{AB}}

Option B (

3\overrightarrow {\mathbf{AB}}

) is correct because the resultant of the given forces simplifies to

3\overrightarrow {\mathbf{AB}}

using vector addition.

Solution approach

Correct Answer + Core Reason: $3\overrightarrow {\mathbf{AB}}$ . The resultant of the given forces simplifies to $3\overrightarrow {\mathbf{AB}}$ using vector addition.

Step-by-Step Solution:

Step 1: Given forces $\overrightarrow {\mathbf{AC}} ,\;\overrightarrow {\;\mathbf{AD}} , \overrightarrow {\mathbf{AE}}$ act at A and $\overrightarrow {\mathbf{CB}} ,\overrightarrow {\mathbf{DB}} ,\overrightarrow {\mathbf{EB}}$ act at B.

Step 2: The resultant force is the sum of all forces: $\overrightarrow {\mathbf{AC}} + \overrightarrow {\mathbf{AD}} + \overrightarrow {\mathbf{AE}} + \overrightarrow {\mathbf{CB}} + \overrightarrow {\mathbf{DB}} + \overrightarrow {\mathbf{EB}}$ .

Step 3: Rearrange the terms: $(\overrightarrow {\mathbf{AC}} + \overrightarrow {\mathbf{CB}} ) + (\overrightarrow {\mathbf{AD}} + \overrightarrow {\mathbf{DB}} ) + (\overrightarrow {\mathbf{AE}} + \overrightarrow {\mathbf{EB}} )$ .

Step 4: Use the vector addition rule: $\overrightarrow {\mathbf{AC}} + \overrightarrow {\mathbf{CB}} = \overrightarrow {\mathbf{AB}}$ , $\overrightarrow {\mathbf{AD}} + \overrightarrow {\mathbf{DB}} = \overrightarrow {\mathbf{AB}}$ , $\overrightarrow {\mathbf{AE}} + \overrightarrow {\mathbf{EB}} = \overrightarrow {\mathbf{AB}}$ .

Step 5: Substitute back into the expression: $\overrightarrow {\mathbf{AB}} + \overrightarrow {\mathbf{AB}} + \overrightarrow {\mathbf{AB}} = 3\overrightarrow {\mathbf{AB}}$ .

Why other options are wrong:
- Option A: This option is incorrect because it only considers the force $\overrightarrow {\mathbf{AC}}$ and doesn't account for the other forces.
- Option C: This option is incorrect because it only considers the force $\overrightarrow {\mathbf{DB}}$ and doesn't account for the other forces.
- Option D: This option is incorrect because it only considers the force $\overrightarrow {\mathbf{BC}}$ and doesn't account for the other forces.

Key Concepts:
- Vector addition
- Resultant force

Memory Aids:
- Remember the triangle law of vector addition: $\overrightarrow {\mathbf{AB}} + \overrightarrow {\mathbf{BC}} = \overrightarrow {\mathbf{AC}}$ .

Quick Tips:
- When dealing with vector addition, rearrange terms to simplify the expression.

115 Easy

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Application +4 | -1

A

vector a has the components

2 p

and 1 w.r.t. a rectangular cartesian system. This system is rotated through a certain angle about the origin in the counter clockwise sense. If with respect to a new system, a has components

(p+1)

and 1 , then

View answer and explanation

Correct option: B

p=1

or

p=-\frac{1}{3}

Option B is correct because the magnitudes of the vector before and after rotation are equal, leading to the equation

4p^2 + 1 = (p+1)^2 + 1

, which solves to

p=1

or

p=-\frac{1}{3}

.

Solution approach

Correct Answer + Core Reason: $p=1$ or $p=-\frac{1}{3}$ . The magnitudes of the vector before and after rotation are equal.

Step-by-Step Solution:

Step 1: The vector $\mathbf{a}$ has components $2p$ and $1$ in the original system, so $\mathbf{a} = 2p \hat{\mathbf{i}} + \hat{\mathbf{j}}$ .

Step 2: After rotation, the vector has components $p+1$ and $1$ , so $\mathbf{b} = (p+1) \hat{\mathbf{i}} + \hat{\mathbf{j}}$ .

Step 3: Since rotation preserves the magnitude of the vector, $|\mathbf{a}| = |\mathbf{b}|$ , which means $|\mathbf{a}|^2 = |\mathbf{b}|^2$ .

Step 4: Calculate the magnitudes squared: $|\mathbf{a}|^2 = (2p)^2 + 1^2 = 4p^2 + 1$ and $|\mathbf{b}|^2 = (p+1)^2 + 1^2 = p^2 + 2p + 1 + 1 = p^2 + 2p + 2$ .

Step 5: Set the magnitudes squared equal: $4p^2 + 1 = p^2 + 2p + 2$ .

Step 6: Simplify the equation: $3p^2 - 2p - 1 = 0$ .

Step 7: Solve the quadratic equation for $p$ : $(3p + 1)(p - 1) = 0$ , so $p = 1$ or $p = -\frac{1}{3}$ .

Why other options are wrong:
- Option A: This option is incorrect because $p=0$ does not satisfy the equation $3p^2 - 2p - 1 = 0$ .
- Option C: This option is incorrect because $p=-1$ or $p=\frac{1}{3}$ do not satisfy the equation $3p^2 - 2p - 1 = 0$ .
- Option D: This option is incorrect because $p=1$ or $p=-1$ do not both satisfy the equation $3p^2 - 2p - 1 = 0$ .

Key Concepts:
- Vector components
- Rotation of coordinate systems
- Magnitude of a vector

Memory Aids:
- Remember that rotation preserves the magnitude of a vector.

Quick Tips:
- When dealing with rotations, equate the magnitudes of the vectors before and after rotation.

116 Easy

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Application +4 | -1

The position vectors of A and B are

2\hat i - 9\hat j - 4\hat k\,\,\,{\text{and}}\,\,6\hat i - 3\hat j + 8\hat k

respectively, then the magnitude of AB is

View answer and explanation

Correct option: D

14

Option D (14) is correct because the magnitude of the vector AB is calculated to be 14.

Solution approach

Correct Answer + Core Reason: 14. The magnitude of vector AB is calculated.

Step-by-Step Solution:

Step 1: Given position vectors $A = 2\widehat i - 9\widehat j - 4\widehat k$ and $B = 6\widehat i - 3\widehat j + 8\widehat k$ .

Step 2: Find the vector $AB = B - A = (6 - 2)\widehat i + (-3 - (-9))\widehat j + (8 - (-4))\widehat k = 4\widehat i + 6\widehat j + 12\widehat k$ .

Step 3: Calculate the magnitude of $AB$ : $|AB| = \sqrt{4^2 + 6^2 + 12^2} = \sqrt{16 + 36 + 144} = \sqrt{196} = 14$ .

Why other options are wrong:
- Option A (11): This is incorrect because it is not the correct magnitude.
- Option B (12): This is incorrect because it is not the correct magnitude.
- Option C (13): This is incorrect because it is not the correct magnitude.

Key Concepts:
- Position vector
- Vector subtraction
- Magnitude of a vector

Memory Aids:
- Remember that the magnitude of a vector $v = x\widehat i + y\widehat j + z\widehat k$ is $|v| = \sqrt{x^2 + y^2 + z^2}$ .

Quick Tips:
- Find the vector AB first, then calculate its magnitude.

117 Easy

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Analyzing +4 | -1

If

a = \hat i + \hat j + \hat k,b = 4\hat i + 3\hat j + 4\hat k\,\,\,\,{\text{and}}\,\,\,\,c = \hat i + \alpha \hat j + \beta \hat k

are linearly dependent vectors and

\left| c \right| = \sqrt 3

then

View answer and explanation

Correct option: D

\alpha = \pm 1,\beta = 1

Option D (

\alpha = \pm 1,\beta = 1

) is correct because, for the vectors to be linearly dependent and satisfy the magnitude condition, these are the only possible values for

\alpha

and

\beta

.

Solution approach

Correct Answer + Core Reason: $\alpha = \pm 1,\beta = 1$ . This satisfies the linear dependence and magnitude conditions.

Step-by-Step Solution:

Step 1: Since the vectors $a = \hat i + \hat j + \hat k, b = 4\hat i + 3\hat j + 4\hat k,$ and $c = \hat i + \alpha \hat j + \beta \hat k$ are linearly dependent, there exist scalars x, y, and z (not all zero) such that $xa + yb + zc = 0$ .

Step 2: This gives $x(\hat i + \hat j + \hat k) + y(4\hat i + 3\hat j + 4\hat k) + z(\hat i + \alpha \hat j + \beta \hat k) = 0$ , which implies $(x + 4y + z)\hat i + (x + 3y + \alpha z)\hat j + (x + 4y + \beta z)\hat k = 0$ .

Step 3: So, we have the system of equations: $x + 4y + z = 0, x + 3y + \alpha z = 0, x + 4y + \beta z = 0$ .

Step 4: For a non-trivial solution, the determinant of the coefficients must be zero: $\begin{vmatrix} 1 & 4 & 1 \\ 1 & 3 & \alpha \\ 1 & 4 & \beta \end{vmatrix} = 0$ . This simplifies to $\beta = 1$ .

Step 5: Given that $\left| c \right| = \sqrt 3$ , we have $1 + \alpha^2 + \beta^2 = 3$ , so $\alpha^2 + \beta^2 = 2$ . Since $\beta = 1$ , we get $\alpha^2 = 1$ , which means $\alpha = \pm 1$ .

Step 6: Therefore, $\alpha = \pm 1$ and $\beta = 1$ .

Why other options are wrong:
- Option A: $\alpha = 1, \beta = -1$ does not satisfy the condition $\alpha^2 + \beta^2 = 2$ .
- Option B: $\alpha = 1, \beta = \pm 1$ does not satisfy the determinant condition for linear dependence.
- Option C: $\alpha = -1, \beta = \pm 1$ does not satisfy the determinant condition for linear dependence.

Key Concepts:
- Linear dependence of vectors
- Determinant of a matrix
- Magnitude of a vector

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

Quick Tips:
- Use the determinant condition to check for linear dependence.
- Remember the formula for the magnitude of a vector.

118 Easy

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Application +4 | -1

If

O

is origin of

C

is the mid point of

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

and

B\left( { - 4,\;3} \right)

. Then, the value of

\overrightarrow {\mathbf{OC}}

is

View answer and explanation

Correct option: C

- \mathbf{\hat i} + \mathbf{\hat j}

Option C (

\mathbf{-\hat i + \hat j}

) is correct because the midpoint C of points A and B has coordinates (-1, 1), thus the position vector

\overrightarrow{\mathbf{OC}}

is

-\mathbf{\hat i + \hat j}

.

Solution approach

Correct Answer + Core Reason: $- \mathbf{\hat i} + \mathbf{\hat j}$ . The position vector $\overrightarrow{\mathbf{OC}}$ is $- \mathbf{\hat i} + \mathbf{\hat j}$ since C is the midpoint of A and B.

Step-by-Step Solution:

Step 1: Given A(2, -1) and B(-4, 3), and C is the midpoint of AB.

Step 2: Find the coordinates of C using the midpoint formula.
$C = (\frac{{x_1 + x_2}}{2}, \frac{{y_1 + y_2}}{2}) = (\frac{{2 + (-4)}}{2}, \frac{{-1 + 3}}{2}) = (\frac{{-2}}{2}, \frac{2}{2}) = (-1, 1)$

Step 3: Find the position vector $\overrightarrow{\mathbf{OC}}$ .
$\overrightarrow{\mathbf{OC}} = -\widehat{\mathbf{i}} + \widehat{\mathbf{j}}$

Why other options are wrong:
- Option A: Incorrect. The signs are incorrect.
- Option B: Incorrect. The signs are incorrect.
- Option D: Incorrect. The sign of the j component is incorrect.

Key Concepts:
- Midpoint formula
- Position vectors

Memory Aids:
- Midpoint formula: $(\frac{{x_1 + x_2}}{2}, \frac{{y_1 + y_2}}{2})$

Quick Tips:
- Remember the midpoint formula correctly.
- Write the position vector based on the coordinates.

119 Easy

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Application +4 | -1

Vectors

\vec{a}

and

\vec{b}

are non-zero and non-collinear. The value of

x

for which

\vec{c}=(x-2) \vec{a}+\vec{b}

and

\bar{d}=(2 x+1) \vec{a}-\vec{b}

are collinear, is

View answer and explanation

Correct option: B

\frac{1}{3}

Option B (

\frac{1}{3}

) is correct because when vectors

\vec{c}

and

\vec{d}

are collinear,

\vec{c} = \lambda \vec{d}

, which leads to

x = \frac{1}{3}

.

Solution approach

Correct Answer + Core Reason: $\frac{1}{3}$ . The value of x is found by setting $\vec{c} = \lambda \vec{d}$ and solving for x.

Step-by-Step Solution:

Step 1: Set up the collinearity condition: $\vec{c} = \lambda \vec{d}$ , where $\lambda$ is a scalar.

Step 2: Substitute the given expressions for $\vec{c}$ and $\vec{d}$ : $(x-2)\vec{a} + \vec{b} = \lambda[(2x+1)\vec{a} - \vec{b}]$ .

Step 3: Equate the coefficients of $\vec{a}$ and $\vec{b}$ : $x-2 = \lambda(2x+1)$ and $1 = -\lambda$ .

Step 4: Solve for $\lambda$ : From the second equation, $\lambda = -1$ .

Step 5: Substitute $\lambda = -1$ into the first equation: $x-2 = -1(2x+1)$ .

Step 6: Simplify and solve for x: $x-2 = -2x-1 \Rightarrow 3x = 1 \Rightarrow x = \frac{1}{3}$ .

Why other options are wrong:
- Option A (1): Substituting x=1 into the equations does not make the vectors collinear.
- **Option C ( $\frac{1}{2}$ ):** Substituting x= $\frac{1}{2}$ into the equations does not make the vectors collinear.
- Option D (2): Substituting x=2 into the equations does not make the vectors collinear.

Key Concepts:
- Collinear vectors
- Scalar multiplication of vectors
- Vector equality

Memory Aids:
- Collinear vectors are scalar multiples of each other.

Quick Tips:
- When vectors are collinear, their corresponding components are proportional.

120 Easy

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Application +4 | -1

Let

P(3,2,6)

be a point in space and

Q

be a point on the line

\mathbf{r}=(\hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}})+\mu(-3 \hat{\mathbf{i}}+\hat{\mathbf{j}}+5 \hat{\mathbf{k}})

. Then the value of

\mu

for which the vector

P Q

is parallel to the plane

x-4 y+3 z=1

is

View answer and explanation

Correct option: A

\frac{1}{4}

Option A (

\frac{{1}}{{4}}

) is correct because it represents the value of

\mu

for which the vector

PQ

is parallel to the given plane, calculated by setting the dot product of the normal vector of the plane and the vector

PQ

to zero.

Solution approach

Correct Answer + Core Reason: $\frac{{1}}{{4}}$ . The value of $\mu$ is found by ensuring the vector PQ is parallel to the plane.

Step-by-Step Solution:

Step 1: Find the coordinates of point $Q$ on the line.\nThe line is given by $\mathbf{r} = (\hat{\mathbf{i}} - \hat{\mathbf{j}} + 2\hat{\mathbf{k}}) + \mu(-3\hat{\mathbf{i}} + \hat{\mathbf{j}} + 5\hat{\mathbf{k}})$ .\nSo, the coordinates of point $Q$ are $(1 - 3\mu, -1 + \mu, 2 + 5\mu)$ .

Step 2: Find the vector $\mathbf{PQ}$ .
$P(3, 2, 6)$ and $Q(1 - 3\mu, -1 + \mu, 2 + 5\mu)$ .
$\mathbf{PQ} = (1 - 3\mu - 3)\hat{\mathbf{i}} + (-1 + \mu - 2)\hat{\mathbf{j}} + (2 + 5\mu - 6)\hat{\mathbf{k}} = (-2 - 3\mu)\hat{\mathbf{i}} + (-3 + \mu)\hat{\mathbf{j}} + (-4 + 5\mu)\hat{\mathbf{k}}$ .

Step 3: Use the fact that $\mathbf{PQ}$ is parallel to the plane.\nThe plane is given by $x - 4y + 3z = 1$ .\nThe normal vector to the plane is $\mathbf{n} = \hat{\mathbf{i}} - 4\hat{\mathbf{j}} + 3\hat{\mathbf{k}}$ .\nSince $\mathbf{PQ}$ is parallel to the plane, $\mathbf{PQ} \cdot \mathbf{n} = 0$ .
$(-2 - 3\mu)(1) + (-3 + \mu)(-4) + (-4 + 5\mu)(3) = 0$ .
$-2 - 3\mu + 12 - 4\mu - 12 + 15\mu = 0$ .
$8\mu - 2 = 0 \Rightarrow 8\mu = 2 \Rightarrow \mu = \frac{1}{4}$ .

Step 4: Final Answer\nThe value of $\mu$ is $\frac{1}{4}$ .

Why other options are wrong:
- **Option B ( $\frac{{-1}}{{4}}$ ):** This is incorrect because it would result in $\mathbf{PQ}$ not being parallel to the plane.
- **Option C ( $\frac{{1}}{{8}}$ ):** This is incorrect because it would result in $\mathbf{PQ}$ not being parallel to the plane.
- **Option D ( $\frac{{-1}}{{8}}$ ):** This is incorrect because it would result in $\mathbf{PQ}$ not being parallel to the plane.

Key Concepts:
- Vector equation of a line
- Normal vector to a plane
- Dot product of vectors
- Parallel vector and plane

Memory Aids:
- If a vector is parallel to a plane, its dot product with the normal vector to the plane is zero.

Quick Tips:
- Remember to find the vector $\mathbf{PQ}$ correctly by subtracting the coordinates of $P$ from $Q$ .
- Ensure that the dot product is calculated correctly.

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