Question

If $\left(\vec{a}\times\vec{b}\right)^{2}+\left(\vec{a}.\vec{b}\right)^{2}=676$ and $\left|\vec{b}\right|=2,$ then $\left|\vec{a}\right|$ is equal to

• A. $13$
• B. $26$
• C. $39$
• D. None of these

Answer 1

Answer: (A)

$13$

Answer 2

Since, $\left(\vec{a}\times\vec{b}\right)^{2}+\left(\vec{a}.\vec{b}\right)^{2}=676$
$\Rightarrow \left(\left|\vec{a}\right|.\left|\vec{b}\right|sin\,\theta\hat{n}\right)^{2}+\left(\left|\vec{a}\right|.\left|\vec{b}\right|cos\,\theta \hat{n}\right)^{2}=676$
$\Rightarrow \left|\vec{a}\right|^{2}.\left|\vec{b}\right|^{2}\left(sin^{2}\,\theta+cos^{2}\,\theta\right)=676$
$\Rightarrow \left|\vec{a}\right|^{2}\left(2\right)^{2}=676 \Rightarrow \left|\vec{a}\right|^{2}=169$
$\Rightarrow \left|\vec{a}\right|=13$