Question

The summation of two unit vectors is a third unit vector, then the modulus of the difference of these unit vectors is

• A. $\sqrt{3}$
• B. $1-\sqrt{3}$
• C. $1+\sqrt{3}$
• D. $-\sqrt{3}$

Answer 1

Answer: (A)

Since, $\vec{a}+\vec{b}=\vec{c}$
$\Rightarrow {\left(\vec{a}+\vec{b}\right)}^2={\vec{c}}^2$
$\Rightarrow {\left|\vec{a}\right|}^2+{\left|\vec{b}\right|}^2+2\left|\vec{a}\right|\left|\vec{b}\right|cos\theta ={\left|\vec{c}\right|}^2$
$\Rightarrow 2\left(1+cos\theta \right)=1$
$\Rightarrow cos\theta =-\frac{1}{2}$
Now, ${\left|\vec{a}-\vec{b}\right|}^2={\left|\vec{a}\right|}^2+{\left|\vec{b}\right|}^2-2\left|\vec{a}\right|\left|\vec{b}\right|cos\theta$
$=1+1-2\cdot \left(\frac{-1}{2}\right)$
$\Rightarrow \left|\vec{a}-\vec{b}\right|=\sqrt{3}$