Question

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

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

Answer 1

Answer: (A)

$(\vec{a} \times \vec{b})^{2}+(\vec{a} \cdot \vec{b})^{2}=676$
$(|\vec{a}| \cdot|\vec{b}| \sin \theta \hat{n})^{2}+(|\vec{a}| \cdot|\vec{b}| \cos \theta)^{2}=676$
$\Rightarrow a^{2} b^{2} \sin ^{2} \theta +a^{2} b^{2} \cos ^{2} \theta=676\left[(\hat{n})^{2}=1\right]$
$a^{2} b^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)=676$
$\Rightarrow a^{2}=\frac{676}{b^{2}}=\frac{676}{4}$
$|\vec{a}|=\sqrt{\frac{676}{4}}$
$\Rightarrow |\vec{a}|=\frac{26}{2}$
$\Rightarrow |\vec{a}|=13$