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, ${\left(a\times b\right)}^2+{\left(a\cdot b\right)}^2=144$
$\Rightarrow \left(a^2{b}^2\cdot 1\cdot si{n}^2\theta \right)+a^2{b}^{2}co{s}^2\theta =144$
$\Rightarrow a^2{b}^2\left(si{n}^2\theta +co{s}^2\theta \right)=144$
$\Rightarrow a^2{b}^2=144$
$\Rightarrow 16b^2=144$ $\left(\because \left|a\right|=4\right)$
$\Rightarrow b^2=9$
$\Rightarrow b=3$
or $\left|b\right|=3$