Q111 / 180 · KCET 2017 · Mathematics

If \( \vec{a} \& \vec{b} \) are unit vectors, then the angle between \( \vec{a} \) and \( \vec{b} \) for \( \sqrt{3} \vec{a}-\vec{b} \) to be unit vector is

  • Correct answer
    A
    \( 30^{\circ} \)
  • B
    \( 45^{\circ} \)
  • C
    \( 60^{\circ} \)
  • D
    \( 90^{\circ} \)

Solution

Since $\vec{a}$ and $\vec{b}$ are unit vectors. Then, $|\vec{a}|=|\vec{b}|=1$ also, $|\sqrt{3} \vec{a}-\vec{b}|=1 \rightarrow(1)$ which is also unit vector. Squaring both the sides of Eq. (1), we get $3|\vec{a}|^{2}+|\vec{b}| \vec{r}-2 \sqrt{3}|\vec{a}| \cdot|\vec{b}|=(1)^{2}$ $3 \cdot 1+1-2 \sqrt{3} \cdot 1 \cdot 1 \cdot \cos \theta=1$ $\Rightarrow 2 \sqrt{3} \cos \theta=3$ $\Rightarrow \cos \theta=\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2}$ $\Rightarrow \theta=30^{\circ}$

Hence, the correct answer is option: 'A'.

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