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 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 .

Find the vector product of the vectors 2 i ^ - j ^ + k ^ and 3 i ^ + j ^ - 2 k ^ .

Let a → = 2 i ^ - j ^ + k ^ and b → = 3 i ^ + j ^ - 2 k ^ a → × b → = i ^ j ^ k ^ 2 - 1 1 3 1 - 2 = - 2 - 1 - 1 1 i ^ - 2 - 2 - 3 1 j ^ + 2 1 - 3 - 1 k ^ = 2 - 1 i ^ - - 4 - 3 j ^ + 2 + 3 k ^ = i ^ + 7 j ^ + 5 k ^ ∴ The vector product of the vectors 2 i ^ - j ^ + k ^ and 3 i ^ + j ^ - 2 k ^ is i ^ + 7 j ^ + 5 k ^

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 angle between vectors 2 i ^ - j ^ and i ^ + 2 j ^ .

Let, u = 2 i ^ - j ^ and v = i ^ + 2 j ^ u · v = ( 2 i ^ - j ^ ) ( i ^ + 2 j ^ ) = 2 - 2 = 0 Since, u · v = | u | v ∣ cos θ ⇒ | u | | v | cos θ = 0 ⇒ cos θ = 0 ( ∵ u ≠ 0 , and | v | ≠ 0 ) ⇒ θ = π 2

If two sides of a triangle are represented by vectors i ^ + 2 j ^ + 2 k ^ and 3 i ^ - 2 j ^ + k ^ ,then find the area of the triangle.

Given two vectors 3 i ^ - 2 j ^ + 1 k ^ and i ^ + 2 j ^ + 2 k ^ Let A → = 3 i ^ - 2 j ^ + 1 k ^ and B → = i ^ + 2 j ^ + 2 k ^ . So, A → × B → = i ^ j ^ k ^ 1 2 2 3 - 2 1 = 6 i ^ + 5 j ^ - 8 k ^ We know that area of a triangle with two sides given is 1 2 × A → × B → Therefore, area is given by 6 i ^ + 5 j ^ - 8 k ^ 2 = 6 2 + 5 2 + ( - 8 ) 2 2 = 125 2 . Therefore, the area of the given triangle is 125 2 .

If a → × b → = c → × d → and a → × c → = b → × d → ,then prove that a → - d → is parallel to b → - c → .

Given, a → × b → = c → × d → … I a → × c → = b → × d → … II To show whether a → - d → is ∥ to b → - c → We are suppose to Eqn. I + Eqn. II a → × b → + b → × d → = c → × d → + a → × c → ⇒ a → × b → - d → × b → = - d → × c → + a → × c → ⇒ a → - d → × b → = - d → + a → × c → ⇒ a → - d → × b → = a → - d → × c → ⇒ a → - d → × b → - c → = 0 ⇒ a → - d → b → - c → sin θ = 0 So, θ = 0 ∴ a → - d → | | b → - c →

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.

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)

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

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.

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 ^

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 .

The four vertices of a tetrahedron are respectively O ( 0 , 0 , 0 ) , A ( 1 , 2 , 1 ) , B ( 2 , 1 , 3 ) and C ( 1 , 1 , 2 ) .Find the volume of the tetrahedron.

Given O ( 0 , 0 , 0 ) , A ( 1 , 2 , 1 ) , B ( 2 , 1 , 3 ) and C ( 1 , 1 , 2 ) Let a = i + 2 j + k b = 2 i + j + 3 k c = i + j + 2 k volume = ( 1 6 ) [ abc ] 1 2 1 2 1 3 1 1 2 = ( 1 6 ) [ 1 ( 2 - 3 ) - 2 ( 4 - 3 ) + 1 ( 2 - 1 ) ] = ( 1 6 ) [ - 1 - 2 + 1 ] = - 1 3 = 1 3 unit 3

E M B I B E

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

EASY

12th Rajasthan Board

IMPORTANT

Earn 100

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

Important Questions on Vector Algebra

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}

.

EASY

12th Rajasthan Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Rajasthan Board-2018>Q 3

Find the vector product of the vectors

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

and

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

.

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 6

Find the angle between vectors

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

and

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

.

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}

.

EASY

12th Rajasthan Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Rajasthan Board-2019>Q 8

If two sides of a triangle are represented by vectors

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

and

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

,then find the area of the triangle.

MEDIUM

12th Rajasthan Board

IMPORTANT

EMBIBE CHAPTER WISE PREVIOUS YEAR PAPERS FOR MATHEMATICS>Chapter 10 - Vector Algebra>Rajasthan Board-2019>Q 9

If $\vec{a} \times \vec{b} = \vec{c} \times \vec{d}$ and $\vec{a} \times \vec{c} = \vec{b} \times \vec{d}$ ,then prove that $\vec{a} - \vec{d}$ is parallel to $\vec{b} - \vec{c}$ .