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

Hint: 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| = \sqrt {{a^2} + {b^2} - 2a.b\cos \theta } $and \[|a - b| + \sqrt {{a^2} + {b^2} - 2a.b\cos \theta } \]


Complete step by step solution:
Let the angle between two vectors $A$and $B$be$\theta $.
So, $|a + b| = |a - b|$
Now, by using the formula
\[|a + b| = \sqrt {{{(a)}^2} + {{(b)}^2} + 2(a)(b)\cos \theta } \]
$|a - b| = \sqrt {|a{|^2} + |b{|^2} - 2(a)(b)\cos \theta } $
Now, $|a + b| = |a - b|$
$\sqrt {{{(a)}^2} + {{(b)}^2} + 2a.b\cos \theta } = \sqrt {{{(a)}^2} + {{(b)}^2} - 2(a)(b)\cos \theta } $
Squaring both sides, we have
${\left( {\sqrt {{{(a)}^2} + {{(b)}^2} + 2ab.\cos \theta } } \right)^2} = {\left( {\sqrt {{{(a)}^2} + {{(b)}^2} = 2(a)(b)\cos \theta } } \right)^2}$
${(a)^2} + {(b)^2} + 2a.b\cos \theta = {a^2} + {b^2} - 2a + b\cos \theta $
$2ab\cos \theta + 2ab\cos \theta = 0$
$4a.b\cos \theta = 0$
$\cos \theta = \dfrac{0 }{{4ab}}$
$\cos \theta = 0$
As we know that the value $\cos {90^o} = 0$
So, $\cos \theta = {90^o}$
$\theta = {90^o}$


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