Write the component of b → along a →

Component of a given vector b → along a → is given by the length of b → on a → . Let θ be the angle between both the vectors. So the length of b → on a → is given as: b → cos θ We know that, The formula to find the angle between two vectors a → and b → is given by cos θ = a → · b → a → b → Therefore, b → cos θ = b → × a → · b → a → b → = a → · b → a → .

Find the angle between the vectors a → = i ^ - j ^ + k ^ and b → = i ^ + j ^ - k ^ .

The formula to find the angle between two vectors a → and b → is given by cos θ = a → · b → a → b → Let a → = i ^ - j ^ + k ^ and b → = i ^ + j ^ - k ^ . Now, a → · b → = ( i ^ - j ^ + k ^ ) · ( i ^ + j ^ - k ^ ) = ( 1 × 1 ) + ( - 1 × 1 ) + ( 1 × - 1 ) = - 1 The magnitude of the vectors is a → = 1 2 + 1 2 + ( - 1 ) 2 = 1 + 1 + 1 = 3 and b → = 1 2 + ( - 1 ) 2 + 1 2 = 1 + 1 + 1 = 3 Now, cos θ = - 1 3 × 3 = - 1 3 ⇒ θ = cos - 1 - 1 3

Find the projection of a → on b → , if a → · b → = 8 and b → = 2 i ^ + 6 j ^ + 3 k ^

Given that, a → · b → = 8 and b → = 2 i ^ + 6 j ^ + 3 k ^ We know that, the projection of a → on b → = a → · b → | b → | = 8 2 i ^ + 6 j ^ + 3 k ^ = 8 4 + 36 + 9 = 8 7 units. Hence, the projection of a → on b → is 8 7 units.

Write the value of p for which a → = 3 i ^ + 2 j ^ + 9 k ^ and b → = i ^ + p j ^ + 3 k ^ are parallel vectors .

We have, a → = 3 i ^ + 2 j ^ + 9 k ^ and b → = i ^ + p j ^ + 3 k ^ Given that a → and b → are parallel . ⇒ a → = t b → , t ≠ 0 ⇒ 3 i ^ + 2 j ^ + 9 k ^ = t ( i ^ + p j ^ + 3 k ^ ) ⇒ 3 i ^ + 2 j ^ + 9 k ^ = t i ^ + pt j ^ + 3 t k ^ Comparing both sides, we get 3 = t , 2 = pt and 9 = 3 t ⇒ t = 3 and pt = 2 ∴ p = 2 3

If a → = 2 , and a → · b → = 3 find the projection of b → on a → .

We have, | a → | = 2 and a → · b → = 3 So, the projection of b → on a → is a → · b → | a → | = 3 2

Write the angle between two vectors a → and b → with magnitudes 3 and 2 respectively having a → · b → = 6

Let θ be the angle between a → and b → . Given, | a → | = 3 ; | b → | = 2 ; a → · b → = 6 The angle between a → and b → is given by cos θ = a → · b → a → b → ⇒ cosθ = 6 2 3 = 1 2 ⇒ θ = cos - 1 1 2 = π 4

If the vectors a → and b → are such that a → = 3 , b → = 2 3 and a → × b → is a unit vector, then write the angle between a → and b →

Given, a → = 3 and b → = 2 3 , and a → × b → is a unit vector. ∴ | a → × b → | = 1 We know that a → × b → = a → b → sin θ n ^ , where θ is the angle between a → and b → and n ^ is a unit vector perpendicular to both a → and b → , 0 ≤ θ ≤ π . ⇒ a → × b → = a → b → sin θ n ^ = 1 ⇒ a → b → sin θ n ^ = 1 ⇒ | a → | | b → | sin θ = 1 [ ∵ n ^ = 1 ] ⇒ 3 × 2 3 × sin θ = 1 [Substituting the given values] ⇒ sin θ = 1 2 ⇒ θ = π 6 Hence, the required angle is π 6 .

Find the magnitude of each of the two vectors a → and b → , having the same magnitude such that the angle between them is 60 ° and their scalar product is 9 2 .

Given that, a → and b → two vectors such that, | a → | = | b → | ; a → · b → = 9 2 If θ is the angle between two vectors a → and b → , then cos θ = a → · b → | a → | | b → | Therefore, cos 60 ° = 9 2 | a → | | a → | ⇒ 9 2 = | a → | 2 · 1 2 ⇒ | a → | 2 = 9 ⇒ | a → | = 3 [The magnitude of a vector can never be negative] ∴ | a → | = | b → | = 3 .

E M B I B E

MATHEMATICS CLASS XII VOLUME-2>Chapter 5 - Scalar or Dot Product>VERY SHORT ANSWER QUESTIONS (VSAQs)

VERY SHORT ANSWER QUESTIONS (VSAQs)

Author:R D Sharma

R D Sharma Mathematics Solutions for Exercise - VERY SHORT ANSWER QUESTIONS (VSAQs)

Simple step-by-step solutions to VERY SHORT ANSWER QUESTIONS (VSAQs) questions of Scalar or Dot Product from MATHEMATICS CLASS XII VOLUME-2. Also get 3D topic explainers, cheat sheets, and unlimited doubts solving on EMBIBE.

Questions from VERY SHORT ANSWER QUESTIONS (VSAQs) with Hints & Solutions

EASY

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 5 - Scalar or Dot Product>VERY SHORT ANSWER QUESTIONS (VSAQs)>Q 18

Write the component of $\vec{b}$ along $\vec{a}$

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 5 - Scalar or Dot Product>VERY SHORT ANSWER QUESTIONS (VSAQs)>Q 26

Find the angle between the vectors $\vec{a} = \hat{i} - \hat{j} + \hat{k} and \vec{b} = \hat{i} + \hat{j} - \hat{k}$ .

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 5 - Scalar or Dot Product>VERY SHORT ANSWER QUESTIONS (VSAQs)>Q 28

Find the projection of $\vec{a} on \vec{b}, if \vec{a} \cdot \vec{b} = 8 and \vec{b} = 2 \hat{i} + 6 \hat{j} + 3 \hat{k}$

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 5 - Scalar or Dot Product>VERY SHORT ANSWER QUESTIONS (VSAQs)>Q 29

Write the value of $p$ for which $\vec{a} = 3 \hat{i} + 2 \hat{j} + 9 \hat{k} and \vec{b} = \hat{i} + p \hat{j} + 3 \hat{k}$ are parallel vectors .

EASY

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 5 - Scalar or Dot Product>VERY SHORT ANSWER QUESTIONS (VSAQs)>Q 31

If $|\vec{a}| = 2, and \vec{a} \cdot \vec{b} = 3$ find the projection of $\vec{b} on \vec{a}$ .

EASY

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 5 - Scalar or Dot Product>VERY SHORT ANSWER QUESTIONS (VSAQs)>Q 32

Write the angle between two vectors $\vec{a}$ and $\vec{b}$ with magnitudes $\sqrt{3}$ and $2$ respectively having $\vec{a} \cdot \vec{b} = \sqrt{6}$

HARD

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 5 - Scalar or Dot Product>VERY SHORT ANSWER QUESTIONS (VSAQs)>Q 40

If the vectors $\vec{a}$ and $\vec{b}$ are such that $|\vec{a}| = 3, |\vec{b}| = \frac{2}{3}$ and $\vec{a} \times \vec{b}$ is a unit vector, then write the angle between $\vec{a}$ and $\vec{b}$

MEDIUM

12th CBSE

IMPORTANT

MATHEMATICS CLASS XII VOLUME-2>Chapter 5 - Scalar or Dot Product>VERY SHORT ANSWER QUESTIONS (VSAQs)>Q 43

Find the magnitude of each of the two vectors $\vec{a}$ and $\vec{b}$ , having the same magnitude such that the angle between them is $60 °$ and their scalar product is $\frac{9}{2}$ .

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