Solveeit Logo
Login

Question

Question: Find the angle between two vectors \(a\,\,and\,\,b\,\,if|a + b| = a - b|\)....

Find the angle between two vectors a  and  b  if∣a+b∣=a−b∣a\,\,and\,\,b\,\,if|a + b| = a - b|.

Explanation

Solution

A vector is an object that has both a magnitude and a direction and we use the below formula to find the angle between the vectors.
∣a+b∣=a2+b2−2a.bcos⁡θ|a + b| = \sqrt {{a^2} + {b^2} - 2a.b\cos \theta } and ∣a−b∣+a2+b2−2a.bcos⁡θ|a - b| + \sqrt {{a^2} + {b^2} - 2a.b\cos \theta }

Complete step by step solution:
Let the angle between two vectors AAand BBbeθ\theta .
So, ∣a+b∣=∣a−b∣|a + b| = |a - b|
Now, by using the formula
∣a+b∣=(a)2+(b)2+2(a)(b)cos⁡θ|a + b| = \sqrt {{{(a)}^2} + {{(b)}^2} + 2(a)(b)\cos \theta }
∣a−b∣=∣a∣2+∣b∣2−2(a)(b)cos⁡θ|a - b| = \sqrt {|a{|^2} + |b{|^2} - 2(a)(b)\cos \theta }
Now, ∣a+b∣=∣a−b∣|a + b| = |a - b|
(a)2+(b)2+2a.bcos⁡θ=(a)2+(b)2−2(a)(b)cos⁡θ\sqrt {{{(a)}^2} + {{(b)}^2} + 2a.b\cos \theta } = \sqrt {{{(a)}^2} + {{(b)}^2} - 2(a)(b)\cos \theta }
Squaring both sides, we have
((a)2+(b)2+2ab.cos⁡θ)2=((a)2+(b)2=2(a)(b)cos⁡θ)2{\left( {\sqrt {{{(a)}^2} + {{(b)}^2} + 2ab.\cos \theta } } \right)^2} = {\left( {\sqrt {{{(a)}^2} + {{(b)}^2} = 2(a)(b)\cos \theta } } \right)^2}
(a)2+(b)2+2a.bcos⁡θ=a2+b2−2a+bcos⁡θ{(a)^2} + {(b)^2} + 2a.b\cos \theta = {a^2} + {b^2} - 2a + b\cos \theta
2abcos⁡θ+2abcos⁡θ=02ab\cos \theta + 2ab\cos \theta = 0
4a.bcos⁡θ=04a.b\cos \theta = 0
cos⁡θ=04ab\cos \theta = \dfrac{0 }{{4ab}}
cos⁡θ=0\cos \theta = 0
As we know that the value cos⁡90o=0\cos {90^o} = 0
So, cos⁡θ=90o\cos \theta = {90^o}
θ=90o\theta = {90^o}

Note: Dot product or scalar product is an algebraic operation that takes two equal-length sequences of numbers, returns a single number.