Question

If $(\vec{a} \times \vec{b})^2+(\vec{a}.\vec{b})^2=144$ and $|\vec{a}|$ = 4 then $ |\vec{b}|$=

• A. 16
• B. 8
• C. 3
• D. 12

Answer 1

Answer: (C)

Given, $\left(a \times b\right)^{2} + \left(a· b\right)^{2} = 144$
$\Rightarrow \left(a^{2}b^{2}· 1·sin ^{2} \theta\right) + a^{2}b^{2} cos^{2} \theta = 144$
$\Rightarrow a^{2}b^{2} \left(sin^{2} \theta + cos^{2} \theta\right) = 144$
$\Rightarrow a^{2}b^{2} = 144$
$\Rightarrow 16b^{2} = 144$ $\left(\because \left|a\right| = 4\right)$
$\Rightarrow b^{2} = 9$
$\Rightarrow b=3$
or $\quad\left|b\right| = 3$