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)

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