Question

If $\overrightarrow{a}$ and $\overrightarrow{b}$ are adjacent side of parallelogram , then $|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}-\overrightarrow{b}|$ is necessary and sufficient condition for a parallelogram to be a

• A. rhombus
• B. trapezium
• C. square
• D. rectangle

Answer 1

Answer: (A)

rhombus

$|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}-\overrightarrow{b}|$

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

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

$\because \left|\overrightarrow{a}\right|\ne 0\ne \left|\overrightarrow{b}\right|$

$\therefore$ $\cos \theta =0$

$\Rightarrow \theta =\pi /2$

Hence the parallelogram with sides $\overrightarrow{a}\&\overrightarrow{b}$ obeying

$|\overrightarrow{a}+\overrightarrow{b}|=|\overrightarrow{a}-\overrightarrow{b}|$

is a rhombus

Option A.