First slide

Vectors

Question

If $| \vec{A} \times \vec{B} | = \sqrt{3} (\vec{A} \cdot \vec{B})$ then the value of $| \vec{A} \times \vec{B} |$ is

Moderate

A

(A²+B² +AB)^1/2
B

${(A^{2} + B^{2} + \frac{AB}{\sqrt{3}})}^{1 / 2}$
C

A+B
D

${(A^{2} + B^{2} + \sqrt{3} AB)}^{1 / 2}$

Solution

Given that, $| \vec{A} \times \vec{B} | = \sqrt{3} \vec{A} \cdot \vec{B}$
$\begin{array}{l} ABsin θ = \sqrt{3} ABcos θ \\ \tan θ = \sqrt{3} = \tan 60^{\circ} \\ θ = 60^{\circ} \end{array}$
Now,
$\begin{array}{l} | A \times B | = \sqrt{[A^{2} + B^{2} + 2 ABcos 60^{\circ}]} \\ = \sqrt{[A^{2} + B^{2} + 2 AB \times (1 / 2)]} \\ = {[A^{2} + B^{2} + AB]}^{1 / 2} \end{array}$

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