Find a vector of magnitude 5 units along the vector i ^ - 2 j ^ + 2 k ^ .

Let a = ı ^ - 2 ȷ ^ + 2 k ^ We know that, a ^ = a | a | | a | = 1 2 + ( - 2 ) 2 + 2 2 = 3 ∴ a ^ = 1 3 ( ı ^ - 2 ȷ ^ + 2 k ^ ) Given, the vector of magnitude 5 units along the vector a will be 5 a ^ = 5 × 1 3 ( ı ^ - 2 ȷ ^ + 2 k ^ ) = 5 3 ( ı ^ - 2 ȷ ^ + 2 k ^ )

Find the projection of the vector i ^ - j ^ on the vector i ^ + j ^ .

Let a → = i ^ - j ^ and b → = i ^ + j ^ . To find: the projection of the vector i ^ - j ^ on the vector i ^ + j ^ Now, projection of vector a → on b → = 1 b → a → ⋅ b → We know that, If P → = a 1 i ^ + a 2 j ^ + a 3 k ^ and Q → = b 1 i ^ + b 2 j ^ + b 3 k ^ are two vectors, then their dot product is denoted by P → · Q → and is given by P → · Q → = a 1 b 1 + a 2 b 2 + a 3 b 3 and the magnitude of Q → is given by Q → = b 1 2 + b 2 2 + b 3 2 . Therefore, a → · b → = 1 · 1 + - 1 · 1 = 1 - 1 = 0 and b → = 1 2 + 1 2 = 1 + 1 = 2 . So, the projection of a → on b → = 1 b → a → ⋅ b → = 0 2 = 0 .

For any vector a → , prove that a → × i ^ 2 + a → × j ^ 2 + a → × k ^ 2 = 2 a 2 .

Let, a → = a x i ^ + a y j ^ + a z k ^ Using the properties of cross product i × i = j × j = k × k = 0 , i × j = k , j × k = i , k × i = j and j × i = - k , i × k = - j , k × j = - i we have to find a → × i ^ , a → × j ^ and a → × k ^ . Now, a → × i ^ = a x i ^ + a y j ^ + a z k ^ × i ^ = a x i ^ × i ^ + a y j ^ × i ^ + a z k ^ × i ^ = a y - k ^ + a z j ^ a → × j ^ = a x i ^ + a y j ^ + a z k ^ × j ^ = a x i ^ × j ^ + a y j ^ × j ^ + a z k ^ × j ^ = a x k ^ + a z - i ^ a → × k ^ = a x i ^ + a y j ^ + a z k ^ × k ^ = a x i ^ × k ^ + a y j ^ × k ^ + a z k ^ × k ^ = a x - j ^ + a y i ^ We know that, a → = a x 2 + a y 2 + a z 2 Then, a → × i ^ 2 = - a y k ^ + a z j ^ 2 = - a y 2 + a z 2 2 = a y 2 + a z 2 Similarly a → × j ^ 2 = a x k ^ - a z i ^ 2 = a x 2 + a z 2 a → × k ^ 2 = - a x j ^ + a y i ^ 2 = a y 2 + a z 2 and a → 2 = a x i ^ + a y j ^ + a z k ^ 2 = a x 2 + a y 2 + a z 2 LHS = | a → × i ^ | 2 + | a → × j ^ | 2 + | a → × k ^ | 2 = a y 2 + a z 2 + a x 2 + a z 2 + a y 2 + a z 2 = 2 a x 2 + a y 2 + a z 2 = 2 a → 2 = RHS (Hence Proved)

For any vector a → , prove that a → = a → . i ^ i ^ + a → . j ^ j ^ + a → . k ^ k ^ .

To prove: a → = ( a → · i ^ ) i ^ + ( a → · j ^ ) j ^ + ( a → · k ^ ) k ^ for any vector a → Let, a → = a 1 i ^ + a 2 j ^ + a 3 k ^ a → · i ^ = a 1 i ^ + a 2 j ^ + a 3 k ^ i ^ = a 1 [by using the properties of dot product i . i = j . j = k . k = 1 , i . j = j . k = k . i = 0 ] Similarly a → · j ^ = a 1 i ^ + a 2 j ^ + a 3 k ^ j ^ = a 2 a → · k ^ = a 1 i ^ + a 2 j ^ + a 3 k ^ · k ^ = a 3 Now, put the value of a → · i ^ , a → · j ^ and a → · k ^ in ( a → · i ^ ) i ^ + ( a → · j ^ ) j ^ + ( a → · k ^ ) k ^ ∴ ( a → · i ^ ) i ^ + ( a → · j ^ ) j ^ + ( a → · k ^ ) k ^ = a 1 i ^ + a 2 j ^ + a 3 k ^ ⇒ ( a → · i ^ ) i ^ + ( a → · j ^ ) j ^ + ( a → · k ^ ) k ^ = a → Hence, proved.

Find the value of 2 i ^ - 3 j ^ + 4 k ^ × 3 i ^ + 4 j ^ - 4 k ^

( 2 i ^ - 3 ȷ ^ + 4 k ^ ) × ( 3 ı ^ + 4 ȷ ^ - 4 k ^ ) = i ^ j ^ k ^ 2 - 3 4 3 4 - 4 = ( 12 - 16 ) i ^ - ( - 8 - 12 ) ȷ ^ + ( 8 - - 9 ) k ^ = - 4 ı ^ + 20 ȷ ^ + 17 k ^ Hence, the required value - 4 i ^ + 20 ȷ ^ + 17 k ^

Prove that: a → + b → b → + c → c → + a → = 2 a → b → c →

LHS = a → + b → b → + c → c → + a → = a → + b → · b → + c → × c → + a → = a → + b → · b → × c → + b → × a → + c → × c → + c → × a → = a → + b → · b → × c → + b → × a → + c → × a → = a → · b → × c → + b → · b → × c → + a → · b → × a → + b → · b → × a → + a → · c → × a → + b → · c → × a → = a → · b → × c → + b → · c → × a → = a → b → c → + b → c → a → = a → b → c → + a → b → c → = 2 a → b → c → = RHS Hence, proved.

Prove that: ( a → × b ) → ( b → × c → ) ( c → × a ) → = a → b → c → 2

Given LHS = ( a → × b ) → ( b → × c → ) ( c → × a ) → We know that a → b → c → = a → · ( b → × c → ) ⇒ LHS = ( a → × b ) → ( b → × c → ) ( c → × a ) → We know that a → × ( b → × c → ) = ( a → · c → ) · b → - ( a → · b → ) c → ⇒ LHS = ( a → × b → ) ( ( ( b → × c → ) · a → ) c → - ( ( b → × c → ) · c → ) a → ) We know that b → × c → = a → then c → perpendicular to b → or c → perpendicular to a → ⇒ LHS = ( a → × b → ) ( ( ( b → × c → ) · a → ) c → - ( ( a → · c → ) a → ) If two vectors are perpendicular then their dot product is zero. ⇒ LHS = ( a → × b → ) ( ( ( b → × c → ) · a → ) c → ) ⇒ LHS = ( ( a → × b → ) · c → ) ( ( b → × c → ) · a → ) ⇒ LHS = [ a → b → c → ] [ b → c → a → ] ⇒ LHS = [ a → b → c → ] [ a → b → c → ] ⇒ LHS = [ a → b → c → ] 2 = RHS

E M B I B E

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

Embibe Experts Solutions for Chapter: Vector Algebra, Exercise 1: Rajasthan Board-2018

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Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Vector Algebra, Exercise 1: Rajasthan Board-2018

Attempt the free practice questions on Chapter 10: Vector Algebra, Exercise 1: Rajasthan Board-2018 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: Rajasthan Board-2018 with Hints & Solutions

EASY

12th Rajasthan Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Rajasthan Board-2018>Q 1

Find a vector of magnitude $5$ units along the vector $\hat{i} - 2 \hat{j} + 2 \hat{k}$ .

EASY

12th Rajasthan Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Rajasthan Board-2018>Q 2

Find the projection of the vector $\hat{i} - \hat{j}$ on the vector $\hat{i} + \hat{j}$ .

MEDIUM

12th Rajasthan Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Rajasthan Board-2018>Q 4

For any vector $\vec{a}$ , prove that ${|\vec{a} \times \hat{i}|}^{2} + {|\vec{a} \times \hat{j}|}^{2} + {|\vec{a} \times \hat{k}|}^{2} = 2 {|a|}^{2}$ .

MEDIUM

12th Rajasthan Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Rajasthan Board-2018>Q 5

For any vector $\vec{a}$ , prove that $\vec{a} = (\vec{a} . \hat{i}) \hat{i} + (\vec{a} . \hat{j}) \hat{j} + (\vec{a} . \hat{k}) \hat{k}$ .

EASY

12th Rajasthan Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Rajasthan Board-2019>Q 7

If $| \vec{a} | = 10, | \vec{b} | = 2$ and $\vec{a} \cdot \vec{b} = 12$ ,then find the value of $\sin θ$ ,where $θ$ is the angle between vectors $\vec{a}$ and $\vec{b}$ .