Question

If $a\cdot b=0$ and $a+b$ makes an angle of $60^{\circ }$ with $b$, then $|a|$ is equal to

• A. $0$
• B. $\frac{1}{\sqrt{3}}\left|\vec{b}\right|$
• C. $\frac{1}{\left|\overrightarrow{\stackrel{⃗}{b}}\right|}$
• D. $\left|\vec{b}\right|$
• E. $\sqrt{3}\left|\vec{b}\right|$

Answer 1

Answer: (E)

We have, $a\cdot b=0\Rightarrow a\perp b$
so, vectors $a,b$ and $a+b$ form a right angled triangle.

In $\Delta PQR$, we have $\tan 60^{\circ }=\frac{|a|}{|b|}$
$\Rightarrow \sqrt{3}=\frac{|a|}{|b|}$
$\Rightarrow |a|=\sqrt{3}|b|$