If a → = 5 i ^ - j ^ + 7 k ^ and b → = i ^ - j ^ - λ k ^ , find the value of λ for which a → + b → and a → - b → are perpendicular to each other.

Given : a → = 5 i ^ - j ^ + 7 k ^ and b → = i ^ - j ^ - λ k ^ ⇒ a → + b → = 5 i ^ - j ^ + 7 k ^ + i ^ - j ^ - λ k ^ ⇒ a → + b → = 6 ı ^ - 2 ȷ ^ + ( 7 - λ ) k ^ Similarly a → - b → = 5 i ^ - j ^ + 7 k ^ - i ^ - j ^ - λ k ^ ⇒ a → - b → = 4 i ^ + 0 j ^ + ( 7 + λ ) k ^ We know that if two vectors are perpendicular then their dot product is zero. Given that a → + b → and a → - b → are perpendicular to each other. ( a → + b → ) . ( a → - b → ) = 0 ⇔ [ 6 ı ^ - 2 ȷ ^ + ( 7 - λ ) k ^ ] . [ 4 i ^ + 0 j ^ + ( 7 + λ ) k ^ ] = 0 ⇒ 6 × ( 4 ) + ( - 2 ) × 0 + ( 7 - λ ) ( 7 + λ ) = 0 ⇒ 24 + 7 2 - λ 2 = 0 ⇒ 73 - λ 2 = 0 ⇒ λ 2 = 73 ⇒ λ = ± 73 Hence required value of λ is ± 73 .

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

Given a → = 3 i ^ + j ^ - 2 k ^ and b → = 2 i ^ + 3 j ^ - k ^ The unit vector perpendicular to both the vectors a → and b → is ± a → × b → a → × b → . Cross product of vectors a → and b → is a → × b → = i ^ j ^ k ^ 3 1 - 2 2 3 - 1 ⇒ a → × b → = i ^ - 1 + 6 - j ^ - 3 + 4 + k ^ 9 - 2 ⇒ a → × b → = 5 i ^ - j ^ + 7 k ^ The unit vector perpendicular to both the vectors a → and b → is ± a → × b → a → × b → = ± 5 i ^ - j ^ + 7 k ^ 5 2 + - 1 2 + 7 2 = ± 5 i ^ - j ^ + 7 k ^ 25 + 1 + 49 = ± 5 i ^ - j ^ + 7 k ^ 75 = ± 5 i ^ - j ^ + 7 k ^ 5 3 Hence, unit vector perpendicular to a → = 3 i ^ + j ^ - 2 k ^ and b → = 2 i ^ + 3 j ^ - k ^ is ± 5 i ^ - j ^ + 7 k ^ 5 3 .

If a → , b → , c → are unit vectors such that a → + b → + c → = 0 , find the value of a → . b → + b → . c → + c → . a → .

Given, a → , b → and c → are unit vectors. So, a → = b → = c → = 1 Also given that a → + b → + c → = 0 → So, a → + b → + c → = 0 → = 0 Now, a → + b → + c → 2 = a → + b → + c → ⋅ a → + b → + c → = a → · a → + a → · b → + a → · c → + b → · a → + b → · b → + b → · c → + c → · a → + c → · b → + c → · c → [Using distributivity of scalar product] Now, using the commutative property of dot product i.e. a → · b → = b → · a → ⇒ a → + b → + c → 2 = a → · a → + a → · b → + c → · a → + a → · b → + b → · b → + b → · c → + c → · a → + b → · c → + c → · c → ⇒ 0 = a → · a → + b → · b → + c → · c → + 2 a → · b → + b → · c → + c → · a → ⇒ 0 = a → 2 + b → 2 + c → 2 + 2 a → · b → + b → · c → + c → · a → [ Using a → · a → = a → 2 ] ⇒ 0 = 1 + 1 + 1 + 2 a → · b → + b → · c → + c → · a → ⇒ 0 = 3 + 2 a → · b → + b → · c → + c → · a → ⇒ a → ⋅ b → + b → ⋅ c → + c → ⋅ a → = - 3 2

Find the unit vector in the direction of the vector a → = j ^ + i ^ + 2 k ^ .

The unit vector a ^ in the direction of vector a → = i ^ + j ^ + 2 k ^ is given by a ^ = a → | a | | a → | = 1 2 + 1 2 + 2 2 = 1 + 1 + 4 = 6 ∴ a ^ = a → | a → | = i ^ + j ^ + 2 k ^ 6 = 1 6 i ^ + 1 6 j ^ + 2 6 k ^

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

If two vectors a → and b → are given in component form as a 1 i ^ + a 2 j ^ + a 3 k ^ and b 1 i ^ + b 2 j ^ + b 3 k ^ , then their scalar product is given as a → · b → = a 1 b 1 + a 2 b 2 + a 3 b 3 . Given, a → = i ^ + 4 j ^ + 2 k ^ , b → = 3 i ^ - 2 j ^ + 7 k ^ and c → = 2 i ^ - j ^ + 4 k ^ Let d → = d 1 i ^ + d 2 j ^ + d 3 k ^ Two vectors are perpendicular implies that dot product of the vectors is zero. d → · a → = 0 ⇒ d 1 + 4 d 2 + 2 d 3 = 0 .............(i) d → · b → = 0 ⇒ 3 d 1 - 2 d 2 + 7 d 3 = 0 ...........(ii) c → · d → = 15 ⇒ 2 d 1 - d 2 + 4 d 3 = 15 ........(iii) From equation (i) and (ii) we have d 1 28 + 4 = d 2 6 - 7 = d 3 - 2 - 12 = k ⇒ d 1 32 = d 2 - 1 = d 3 - 14 = k ⇒ d 1 = 32 k , d 2 = - k , and d 3 = - 14 k Substituting the values of d 1 , d 2 , and d 3 in equation (iii) 2 × 32 k - - k + 4 - 14 k = 15 ⇒ 64 k + k - 56 k = 15 ⇒ 9 k = 15 ⇒ k = 5 3 ∴ d 1 = 32 × 5 3 = 160 3 , d 2 = - 5 3 a n d d 3 = - 14 × 5 3 = - 70 3 Hence, d → = 160 3 i ^ - 5 3 j ^ - 70 3 k ^ = 1 3 ( 160 i ^ - 5 j ^ - 70 k ^ )

E M B I B E

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

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

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

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

EASY

12th Meghalaya Board

IMPORTANT

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

Prove that the points $A (2, 0, - 3), B (3, 2, - 1)$ and $C (1, - 2, - 5)$ are collinear.

MEDIUM

12th Meghalaya Board

IMPORTANT

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

If $\vec{a} = 5 \hat{i} - \hat{j} + 7 \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} - λ \hat{k}$ , find the value of $λ$ for which $\vec{a} + \vec{b}$ and $\vec{a} - \vec{b}$ are perpendicular to each other.

MEDIUM

12th Meghalaya Board

IMPORTANT

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

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

MEDIUM

12th Meghalaya Board

IMPORTANT

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

If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = 0$ , find the value of $\vec{a} . \vec{b} + \vec{b} . \vec{c} + \vec{c} . \vec{a}$ .