MCQ
Choose the correct answer:$\text{Let}\ \vec{\text{a}}\ \text{and}\ \vec{\text{b}}$ be two unit vectors and $\theta$ is the angle between them. Then $\vec{\text{a}}+\vec{\text{b}}$ is a unit vector if,
  • A
    $\theta=\frac{\pi}{4}$
  • B
    $\theta=\frac{\pi}{3}$
  • C
    $\theta=\frac{\pi}{2}$
  • $\theta=\frac{2\pi}{3}$

Answer

Correct option: D.
$\theta=\frac{2\pi}{3}$
$\text{Let}\ \vec{\text{a}}\ \text{and}\ \vec{\text{b}}$ be two unit vectors and $\theta$ be the angle between them.
$\text{Then},\ \big|\vec{\text{a}}\big|=\Big|\vec{\text{b}}\Big|=1.$
$\text{Now},\ \vec{\text{a}}+\vec{\text{b}}$ is a unit vector if $\Big|\vec{\text{a}}+\vec{\text{b}}\Big|=1.$
$\Big|\vec{\text{a}}+\vec{\text{b}}\Big|=1$
$\Rightarrow\Big(\vec{\text{a}}+\vec{\text{b}}\Big)^2=1$
$\Rightarrow\Big(\vec{\text{a}}+\vec{\text{b}}\Big)\cdot\Big(\vec{\text{a}}+\vec{\text{b}}\Big)=1$
$\Rightarrow\vec{\text{a}}.\vec{\text{a}}+\vec{\text{a}}.\vec{\text{b}}+\vec{\text{b}}.\vec{\text{a}}+\vec{\text{b}}.\vec{\text{b}}=1$
$\Rightarrow\Big|\vec{\text{a}}\Big|^2+2\vec{\text{a}}.\vec{\text{b}}+\Big|\vec{\text{b}}\Big|^2=1$
$\Rightarrow1^2+2\Big|\vec{\text{a}}\Big|\Big|\vec{\text{b}}\Big|\cos\theta+1^2=1$
$\Rightarrow1+2.1.1\cos\theta+1=1$
$\Rightarrow\cos\theta=-\frac{1}{2}$
$\Rightarrow\theta=-\frac{2\pi}{3}$
Hence, $\vec{\text{a}}+\vec{\text{b}}$ is a unit vector if $\theta=\frac{2\pi}{3}.$
The correct answer is D.

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