First slide

Vectors

Question

If the magnitude of sum of two vectors is equal to the magnitude of difference of the two vectors, the angle between these vectors is

Easy

A

$45^{0}$
B

$180^{0}$
C

$0^{0}$
D

$90^{0}$

Solution

$Let the two vectors be \vec{A} and \vec{B} .$
$Then, magnitude of sum of \vec{A} and \vec{B},$
$| \vec{A} + \vec{B} | = \sqrt{A^{2} + B^{2} + 2 A B \cos θ}$
$and magnitude of difference of \vec{A} and \vec{B},$
$| \vec{A} - \vec{B} | = \sqrt{A^{2} + B^{2} - 2 A B \cos θ}, | \vec{A} + \vec{B} | = | \vec{A} - \vec{B} |$
$\begin{array}{l} or \sqrt{A^{2} + B^{2} + 2 A B \cos θ} = \sqrt{A^{2} + B^{2} - 2 A B \cos θ} \\ \Rightarrow 4 A B \cos θ = 0 \\ ∵ 4 A B \neq 0, ∴ \cos θ = 0 or θ = 90 ° \end{array}$

Get Instant Solutions

When in doubt download our app. Now available Google Play Store- Doubts App

Recieve an sms with download link

Doubts App

OR

SCAN ME

Doubts App

Doubts App