Question

The sum of two unit vector is a vector having magnitude sqrt(3),then magnitude of their difference will be, Options:
a)sqrt(3) b)sqrt(3)/2
c)1 d)2

Answer 1

◘ Given ◘

The sum of two unit vectors is a vector having magnitude √3.

◘ To Find ◘

The magnitude of their difference.

◘ Solution ◘

According to the question,

$\sf{\overrightarrow{a}+\overrightarrow{b}=\sqrt{3}}$

We know,

$\sf{\sqrt{|\overrightarrow{a}|^2+|\overrightarrow{b}|^2+2|\overrightarrow{a}||\overrightarrow{b}|cos \theta}=\sqrt{3} }$

$\sf{\implies \sqrt{(1)^2+(1)^2+2\times1\times1\times cos\theta} =\sqrt{3} }\\\\\sf{\implies \sqrt{2+2cos\theta}=\sqrt{3} }\\\\\sf{\implies 2(1+cos\theta)=3}\\\\\sf{\implies 1+cos\theta=\dfrac{3}{2} }\\\\\sf{\implies cos\theta=\dfrac{3}{2}-1 }\\\\\sf{\implies cos\theta=\dfrac{1}{2} }\\\\\sf{\implies cos\theta=cos\ 60^\circ}\\\\\sf{\implies \theta=60^ \circ}$

Now,

$\sf{\overrightarrow{a}-\overrightarrow{b}=\sqrt{|\overrightarrow{a}|^2+|\overrightarrow{b}|^2-2|\overrightarrow{a}||\overrightarrow{b}|cos \theta} }\\\\\sf{\implies \overrightarrow{a}-\overrightarrow{b}=\sqrt{1+1 -2 \times cos\ 60^\circ} }\\\\\sf{\implies \overrightarrow{a}-\overrightarrow{b}=\sqrt{2-2\times\dfrac{1}{2} } }\\\\\sf{\implies \overrightarrow{a}-\overrightarrow{b}= \sqrt{2-1} }\\\\\sf{\implies \overrightarrow{a}-\overrightarrow{b}= \sqrt{1} }$

$\boxed{\boxed{\sf{\implies \overrightarrow{a}-\overrightarrow{b}=1}}}$

Therefore, the answer is c) 1.