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. $\frac{1}{\left|\vec{\vec{b}}\right|}$
• 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 $\Delta P Q R$, we have $\tan 60^{\circ}=\frac{| a |}{| b |}$
$\Rightarrow \sqrt{3}=\frac{| a |}{| b |}$
$\Rightarrow | a |=\sqrt{3}| b |$