Prove that a → × b → 2 = a → 2 b → 2 - a → · b → 2

To Prove: a → × b → 2 = a → 2 b → 2 - a → · b → 2 From the definition of cross product of two vectors, we know that a → × b → = a → b → sin θ , where θ is the angle between two vectors. ⇒ a → × b → 2 = a → 2 b → 2 sin 2 θ [On squaring both sides of the equation] ⇒ a → × b → 2 = a → 2 b → 2 1 - cos 2 θ ∵ sin 2 θ = 1 - cos 2 θ ⇒ a → × b → 2 = a → 2 b → 2 - a → 2 b → 2 cos 2 θ ⇒ a → × b → 2 = a → 2 b → 2 - a → · b → 2 ∵ a → · b → = a → b → cos θ Hence, Proved.

If a → = 2 i - 3 j - 5 k and b → = - 7 i + 6 j + 8 k , find a → × b →

Given that: a → = 2 i - 3 j - 5 k and b → = - 7 i + 6 j + 8 k , We have to find a → × b → Vector a = 2 i - 3 j - 5 k and vector b = - 7 i + 6 j + 8 k a → × b → = i j k 2 - 3 - 5 - 7 6 8 = i ( - 24 + 30 ) - j ( 16 - 35 ) + k ( 12 - 21 ) = 6 i + 19 j - 9 k

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

The formula to find the angle between two vectors 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 Substituting the values in the equation, cos θ = - 1 3 × 3 = - 1 3 ⇒ θ = cos - 1 - 1 3

Find a → - b → , if two vector a → and b → are such that a → = 2 , b → = 3 and a → · b → = 4 .

Given that, We have to find a → - b → , if two vectors a → and b → are such that a → = 2 , b → = 3 and a → . b → = 4 . Consider a → - b → 2 = a → - b → 2 = a → - b → . a → - b → . = a → . a → - a → . b → - b → . a → + b → . b → = a → 2 - a → . b → - a → . b → + b → 2 ∵ a → . b → = b → . a → = a → 2 - 2 a → . b → + b → 2 ⇒ a → - b → 2 = 2 2 - 2 × 4 + 3 2 = 5 ⇒ a → - b → = 5 We have considered positive root because magnitude is always positive. Therefore, the value of a → - b → is 5 .

E M B I B E

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

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

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

Attempt the practice questions on Chapter 10: Vector Algebra, Exercise 1: Bihar 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.