If $\left| {A + B} \right| = \left| {A - B} \right|$, find the angle between the vectors A and B and show that the two vectors are perpendicular to each other. How to answer this question?

Hint: If two vectors are placed tail-to-tail then we measure the smallest angle between the two vectors. Since the angle $\theta $ lies in a triangle, we can compute it using a bit of trigonometry, namely, the law of cosines.

In this question, $\left| {A + B} \right| = \left| {A - B} \right|$ is given and we want to find the angle between the vector A and B. First, square both sides. After squaring both sides, multiply the brackets. Then apply the dot-product formula to find the angle. The formula of the dot product of two vectors is: $\overrightarrow A .\overrightarrow B = \left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos \theta $, where $\theta $ is the angle between the vectors.

Complete step-by-step answer:
In this question, the given expression is:
$ \Rightarrow \left| {A + B} \right| = \left| {A - B} \right|$
Let us square on both sides.
$ \Rightarrow {\left| {A + B} \right|^2} = {\left| {A - B} \right|^2}$
As we already know that ${\left| A \right|^2} = A.A$.
Therefore, we can write the above expression as
$ \Rightarrow \left( {A + B} \right)\left( {A + B} \right) = \left( {A - B} \right)\left( {A - B} \right)$
Let us apply the multiplication to remove the brackets. For that we will apply the distributive property.
$ \Rightarrow A.A + A.B + B.A + B.B = A.A - A.B - B.A + B.B$
Let us subtract $ - A.A + A.B + B.A - B.B$ on both sides.
$ \Rightarrow A.A + A.B + B.A + B.B - A.A + A.B + B.A - B.B = A.A - A.B - B.A + B.B - A.A + A.B + B.A - B.B$
The right-hand side will be 0. That is equal to,
$ \Rightarrow 4A.B = 0$
Let us divide both sides by 4.
$ \Rightarrow \dfrac{{4A.B}}{4} = \dfrac{0}{4}$
The division of 4 and 4 is 1 on the left-hand side and the division of 0 and 4 is equal to 0 on the right-hand side. That is equal to,
$ \Rightarrow A.B = 0$
Let us apply the dot product formula.
$ \Rightarrow \left| A \right|\left| B \right|\cos \theta = 0$
Since A and B are non-zero vectors, $\cos \theta $ must be 0.
$ \Rightarrow \cos \theta = 0$
So,
$ \Rightarrow \theta = \dfrac{\pi }{2}$
It implies that A and B are perpendicular as in the interval $\left[ {0,\pi } \right]$ at the angle $\dfrac{\pi }{2}$ .

Note:
The dot product is defined as the magnitude of the component of one of the vectors with respect to the second vector. If the dot product of any two vectors is zero, then they are perpendicular to each other.