Question

Find the angle between vector P and vector Q if $\left|\overrightarrow{p}\right|=\left|\overrightarrow{q}\right|\&\left|\overrightarrow{p}-\overrightarrow{q}\right|=\left|\overrightarrow{p}+\overrightarrow{q}\right|$

• A. $45°$
• B. $90°$
• C. $135°$
• D. $150°$

Answer 1

The correct option is B

$90°$

Explanation for correct option:

Finding the angle:

Step-1: Apply formula $\mathbf{|}\overrightarrow{\mathbf{a}}\mathbf{\pm }\overrightarrow{\mathbf{b}}{\mathbf{|}}^{\mathbf{2}}\mathbf{=}\sqrt{{\mathbf{\left|}\overrightarrow{\mathbf{a}}\mathbf{\right|}}^{\mathbf{2}}\mathbf{+}{\mathbf{\left|}\overrightarrow{\mathbf{b}}\mathbf{\right|}}^{\mathbf{2}}\mathbf{\pm }\mathbf{2}\mathbf{\left|}\overrightarrow{\mathbf{a}}\mathbf{\right|}\mathbf{\left|}\overrightarrow{\mathbf{b}}\mathbf{\right|}\mathbf{c}\mathbf{o}\mathbf{s}\mathbf{\left(}\mathbf{\theta }\mathbf{\right)}}$

Given, $\left|\overrightarrow{p}\right|=\left|\overrightarrow{q}\right|\&\left|\overrightarrow{p}-\overrightarrow{q}\right|=\left|\overrightarrow{p}+\overrightarrow{q}\right|$

$\Rightarrow$$\sqrt{{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \left(\theta \right)}=\sqrt{{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2-2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \left(\theta \right)}$

Step-2: Squaring both sides.

$\Rightarrow$${\left(\sqrt{{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \left(\theta \right)}\right)}^2={\left(\sqrt{{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2-2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \left(\theta \right)}\right)}^2$

$\Rightarrow$ ${\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2+2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \left(\theta \right)={\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2-2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \left(\theta \right)$

$\Rightarrow$ $4\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|\cos \left(\theta \right)=0$

$\Rightarrow$ $\theta =90°$

Hence, correct answer is option $\left(\mathit{B}\right)$.