If a → = 6 i ^ + 8 j ^ and b → = 3 j ^ + 4 k ^ , find the vector component of a → along b → .

The vector component of a → along b → can be obtained using the formula as a → · b → b → 2 b → . = 6 i ^ + 8 j ^ · 3 j ^ + 4 k ^ 3 2 + 4 2 2 3 j ^ + 4 k ^ = 6 × 0 + 8 × 3 + 0 × 4 9 + 16 2 3 j ^ + 4 k ^ = 24 25 2 3 j ^ + 4 k ^ = 24 25 3 j ^ + 4 k ^

Find a unit vector perpendicular to each of the vector a → + b → and a → - b → , where a → = 3 i ^ + 2 j ^ + 2 k ^ and b → = i ^ + 2 j ^ - 2 k ^ .

Let a → and b → be two vectors given in component form as a 1 i ^ + a 2 j ^ + a 3 k ^ and b 1 i ^ + b 2 j ^ + b 3 k ^ , respectively. Then their cross product may be given by a → × b → = i ^ j ^ k ^ a 1 a 2 a 3 b 1 b 2 b 3 In this question, a → = 3 i ^ + 2 j ^ + 2 k ^ and b → = i ^ + 2 j ^ - 2 k ^ According to given problem a → + b → = 4 i ^ + 4 j ^ , a → - b → = 2 i ^ + 4 k ^ ( a → + b → ) × ( a → - b → ) = i ^ j ^ k ^ 4 4 0 2 0 4 = i ^ ( 16 - 0 ) - j ^ ( 16 - 0 ) + k ^ ( 0 - 8 ) = 16 i ^ - 16 j ^ - 8 k ^ | ( a → + b → ) × ( a → - b → ) | = 16 2 + ( - 16 ) 2 + ( - 8 ) 2 = 2 2 × 8 2 + 2 2 × 8 2 + 8 2 = 8 2 2 + 2 2 + 1 = 8 9 = 8 × 3 = 24 For a given vector, the vector a ^ = a → a → gives the unit vector in the direction of a → . So, the unit vector in the direction of ( a → + b → ) × ( a → - b → ) = ± ( a → + b → ) × ( a → - b → ) | ( a → + b → ) × ( a → - b → ) | = ± 16 i ^ - 16 j ^ - 8 k ^ 24 = ± 2 i ^ - 2 j ^ - k ^ 3 = ± 2 3 i ^ ∓ 2 3 j ^ ∓ 1 3 k ^ Hence, the unit vector perpendicular to each of the vector a → + b → and a → - b → is given by ± 2 3 i ^ ∓ 2 3 j ^ ∓ 1 3 k ^ .

The projections of a line on the axes are 3 , 4 and 2 6 . Find the length of the line.

Given that, projections of line on axes are 3 , 4 and 2 6 Let d l = 3 , d m = 4 and d n = 2 6 , where d is the length of the line and l , m , n are direction cosines of the line. d 2 l 2 + d 2 m 2 + d 2 n 2 = 9 + 16 + 24 ⇒ d 2 l 2 + m 2 + n 2 = 49 ⇒ d 2 = 49 ⇒ d = 7 ∴ Length of line = 7 units

Find a unit vector perpendicular to each of the vector a → + b → and a → - b → , where a → = 3 i ^ + 2 j ^ + 2 k ^ and b → = i ^ + 2 j ^ - 2 k ^ .

Let a → and b → be two vectors given in component form as a 1 i ^ + a 2 j ^ + a 3 k ^ and b 1 i ^ + b 2 j ^ + b 3 k ^ , respectively. Then their cross product may be given by a → × b → = i ^ j ^ k ^ a 1 a 2 a 3 b 1 b 2 b 3 In this question, a → = 3 i ^ + 2 j ^ + 2 k ^ and b → = i ^ + 2 j ^ - 2 k ^ According to given problem a → + b → = 4 i ^ + 4 j ^ , a → - b → = 2 i ^ + 4 k ^ ( a → + b → ) × ( a → - b → ) = i ^ j ^ k ^ 4 4 0 2 0 4 = i ^ ( 16 - 0 ) - j ^ ( 16 - 0 ) + k ^ ( 0 - 8 ) = 16 i ^ - 16 j ^ - 8 k ^ | ( a → + b → ) × ( a → - b → ) | = 16 2 + ( - 16 ) 2 + ( - 8 ) 2 = 2 2 × 8 2 + 2 2 × 8 2 + 8 2 = 8 2 2 + 2 2 + 1 = 8 9 = 8 × 3 = 24 For a given vector, the vector a ^ = a → a → gives the unit vector in the direction of a → . So, the unit vector in the direction of ( a → + b → ) × ( a → - b → ) = ± ( a → + b → ) × ( a → - b → ) | ( a → + b → ) × ( a → - b → ) | = ± 16 i ^ - 16 j ^ - 8 k ^ 24 = ± 2 i ^ - 2 j ^ - k ^ 3 = ± 2 3 i ^ ∓ 2 3 j ^ ∓ 1 3 k ^ Hence, the unit vector perpendicular to each of the vector a → + b → and a → - b → is given by ± 2 3 i ^ ∓ 2 3 j ^ ∓ 1 3 k ^ .

Let a → = i ^ + 4 j ^ + 2 k ^ , b → = 3 i ^ - 2 j ^ + 7 k ^ and c → = 2 i ^ - j ^ + 4 k ^ . Find a vector d → which is perpendicular to both a → and b → and c → · d → = 15 .

Given a → = i ^ + 4 j ^ + 2 k ^ , b → = 3 i ^ - 2 j ^ + 7 k ^ and c → = 2 i ^ - j ^ + 4 k ^ since d is perpendicular to both a and b , it is parallel to a → × b → . Suppose d = λ ( a → × b → ) for some scalar λ d = λ i ^ j ^ k ^ 1 4 2 3 - 2 7 = λ [ ( 28 + 4 ) i ^ - ( 7 - 6 ) j ^ + ( - 2 - 12 ) k ^ ] = λ [ 32 i ^ - j ^ - 14 k ^ ] c → · d → = 15 (Given) ( 2 i ^ - j ^ + 4 k ^ ) · λ ( 32 i ^ - j ^ - 14 k ^ ) = 15 λ ( 64 + 1 - 56 ) = 15 λ = 5 3 ∴ d → = 5 3 ( 32 i ^ - j ^ - 14 k ^ ) ⇒ d → = 1 3 ( 160 i ^ - 5 j ^ - 70 k ^ )

The two adjacent sides of a parallelogram are 2 i ^ - 4 j ^ + 5 k ^ and i ^ - 2 j ^ - 3 k ^ .Find the unit vector parallel to its diagonal. Also, find its area.

Diagonal of a parallelogram is a → + b → a → + b → = ( 2 + 1 ) i ^ + ( - 4 - 2 ) j ^ + ( 5 - 3 ) k ^ = 3 i ^ - 6 j ^ + 2 k ^ So, the unit vector parallel to diagonal is a → + b → | a → + b → | = 3 i ^ - 6 j ^ + 2 k ^ 3 2 + ( - 6 ) 2 + 2 2 = 3 i ^ - 6 j ^ + 2 k ^ 9 + 36 + 4 = 3 i ^ - 6 j ^ + 2 k ^ 7 = 3 7 i ^ - 6 7 j ^ + 2 7 k ^ Area of a parallelogram = a → × b → a → × b → = i ^ j ^ k ^ 2 - 4 5 1 - 2 - 3 = i ^ ( 12 + 10 ) - j ^ ( - 6 - 5 ) + k ^ ( - 4 + 4 ) = 22 ı ^ + 11 ȷ ^ = 11 ( 2 i ^ + j ^ ) ∴ | a → × b → | = 11 2 2 + 1 2 = 11 5 So, area of parallelogram is 11 5 sq units

For any two vectors a → and b → ,prove that | a → + b → | ≤ | a → | + | b → |

From dot product, | a → + b → | 2 = ( a → + b → ) · ( a → + b → ) = a → · a → + a → · b → + b → · a → + b → . b → ∵ a → · b → = b → · a → = | a → | 2 + 2 a → · b → + | b → | 2 = | a → | 2 + 2 | a → | | b → | | cos θ | + | b → | 2 We know that 0 ≤ cos θ ≤ 1 Considering cos θ ≤ 1 ⇒ 2 | a → | | b → | cos θ ≤ 2 | a → | | b → | ⇒ | a → | 2 + 2 | a → | | b → | | cos θ | + | b → | 2 ≤ | a → | 2 + 2 | a → | | b → | + | b → | 2 ⇒ a → + b → 2 ≤ a → 2 + 2 a → b → + b → 2 ⇒ a → + b → 2 ≤ a → + b → 2 ∴ | a → + b → | ≤ | a → | + | b → | Hence proved.

E M B I B E

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Assam Board-2020

Embibe Experts Solutions for Chapter: Vector Algebra, Exercise 1: Assam Board-2019

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Vector Algebra, Exercise 1: Assam Board-2019

Attempt the free practice questions on Chapter 10: Vector Algebra, Exercise 1: Assam Board-2019 with hints and solutions to strengthen your understanding. EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Vector Algebra, Exercise 1: Assam Board-2019 with Hints & Solutions

MEDIUM

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Assam Board-2019>Q 3

If $\vec{a} = 6 \hat{i} + 8 \hat{j}$ and $\vec{b} = 3 \hat{j} + 4 \hat{k}$ , find the vector component of $\vec{a}$ along $\vec{b}$ .

MEDIUM

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Assam Board-2019>Q 4

Find a unit vector perpendicular to each of the vector $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b},$ where $\vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $\vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ .

EASY

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Assam Board-2019>Q 5

Prove that $(\vec{a} - \vec{b}) \times (\vec{a} + \vec{b}) = 2 (\vec{a} \times \vec{b}) .$ Hence find the area of the parallelogram whose diagonals are the vectors $3 \hat{i} + \hat{j} - 2 \hat{k}$ and $\hat{i} - 3 \hat{j} + 4 \hat{k}$ .

EASY

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Assam Board-2020>Q 6

The projections of a line on the axes are $3, 4$ and $2 \sqrt{6}$ . Find the length of the line.

MEDIUM

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Assam Board-2020>Q 7

Find a unit vector perpendicular to each of the vector $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b},$ where $\vec{a} = 3 \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $\vec{b} = \hat{i} + 2 \hat{j} - 2 \hat{k}$ .

HARD

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Assam Board-2020>Q 8

Let $\vec{a} = \hat{i} + 4 \hat{j} + 2 \hat{k}, \vec{b} = 3 \hat{i} - 2 \hat{j} + 7 \hat{k}$ and $\vec{c} = 2 \hat{i} - \hat{j} + 4 \hat{k} .$ Find a vector $\vec{d}$ which is perpendicular to both $\vec{a}$ and $\vec{b}$ and $\vec{c} \cdot \vec{d} = 15$ .

MEDIUM

12th Assam Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Assam Board-2020>Q 9

The two adjacent sides of a parallelogram are $2 \hat{i} - 4 \hat{j} + 5 \hat{k}$ and $\hat{i} - 2 \hat{j} - 3 \hat{k}$ .Find the unit vector parallel to its diagonal. Also, find its area.