Given $P = A + B$ and $Q = A - B$ . If the magnitudes of vectors P and Q are equal, what is the angle between vectors A and B?

Hint: The magnitude of the vector is just the length of the vector. It is expressed in the form of mode. Here, we will find the magnitude of both the vectors individually and will equate them to find the required angle.

Complete step by step answer: Given that : The magnitudes of the vectors $P = A + B$ and $Q = A - B$ are equal.
$\overrightarrow P = A + B$
Take magnitude of the above equation –
$\left| {\overrightarrow P } \right| = \left| {A + B} \right|$
Do the simplification of the mode in the above equation –
\[{\left| {\overrightarrow P } \right|^2} = {\left| {A + B} \right|^2}\]
Apply the algebraic formula to expand the square of sum of the two terms.
$\left| {{P^2}} \right| = {A^2} + {B^2} + 2AB\cos \theta {\text{ }}.....{\text{(1)}}$
Similarly find the magnitude of the second given vector
$Q = A - B$
Take mode on both the sides of the equation –
$\left| {\overrightarrow Q } \right| = \left| {A - B} \right|$
Apply the algebraic formula to expand the square of the difference of the two terms.
$ \Rightarrow \left| {{Q^2}} \right| = {A^2} + {B^2} - 2AB\cos \theta {\text{ }}....{\text{(2)}}$
Given that the magnitude of both the vectors are equal.
Therefore, by using equations $(1){\text{ and (2)}}$
$ \Rightarrow {A^2} + {B^2} + 2AB\cos \theta = {A^2} + {B^2} - 2AB\cos \theta $
Like terms from both the sides of the equations cancel each other.
$ \Rightarrow 2AB\cos \theta = - 2AB\cos \theta $
Take common factors$(2AB)$from both the sides of the equation and remove.
\[
   \Rightarrow \cos \theta = - \cos \theta \\
   \Rightarrow 2\cos \theta = 0 \\
 \]
Take $2$ on the division of the left hand side of the equation –
$ \Rightarrow \cos \theta = \dfrac{0}{2}$
Any number upon zero is always zero.
$ \Rightarrow \cos \theta = 0$
Make the required angle the subject –
$
   \Rightarrow \theta = {\cos ^{ - 1}}0 \\
   \Rightarrow \theta = \dfrac{\pi }{2} \\
 $
Therefore, the required answer - the angle between vectors A and B is $90^\circ $

Note: Remember that the vector quantities are quantities where both its direction and magnitude are important. Here we are asked to find the angle and it is always better to remember the trigonometric properties and its values for the angles such as $0^\circ ,30^\circ ,45^\circ ,60^\circ ,90^\circ $ for easy substitutions. Also remember basic algebraic formulas to expand the square of the sum of two terms and difference of two terms.