Question

If $(\vec{a} \times \vec{b})^2+(\vec{a}.\vec{b})^2=144$ and $|\vec{a}|$ then $|\vec{b}|$=

• A. 16
• B. 8
• C. 3
• D. 12

Answer 1

Answer: (C)

3

Answer 2

The correct answer is C:3
Given, $\left(a \times b\right)^{2} + \left(a.b\right)^{2} = 144$
$|a|=4, |b|=?$
Let us consider the angle between $\vec{a}$ and $\vec{b}$ is $\theta$ then;
$|\vec{a}|^2|\vec{b}|^2sin^2\theta+|\vec{a}|^2|\vec{b}|^2.cos^2\theta=144$
⇒$|\vec{a}|^2|\vec{b}|^2=144 (sin^2\theta+cos^2\theta=1)$
⇒$4|\vec{b}|^2=144$
⇒$|\vec{b}|^2=\frac{144}{4}$
⇒$|\vec{b}|=\frac{12}{4}$
⇒$|b|=3$