Search any question & find its solution
Question:
Answered & Verified by Expert
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
Options:
Solution:
1414 Upvotes
Verified Answer
The correct answer is:
\( 30^{\circ} \)
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
$3|\vec{a}|^{2}+|\vec{b}| \vec{r}-2 \sqrt{3}|\vec{a}| \cdot|\vec{b}|=(1)^{2}$
$3 \cdot 1+1-2 \sqrt{3} \cdot 1 \cdot 1 \cdot \cos \theta=1$
$\Rightarrow 2 \sqrt{3} \cos \theta=3$
$\Rightarrow \cos \theta=\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2}$
$\Rightarrow \theta=30^{\circ}$
$|\vec{a}|=|\vec{b}|=1$
also, $|\sqrt{3} \vec{a}-\vec{b}|=1 \rightarrow(1)$
which is also unit vector.
Squaring both the sides of Eq. (1), we get
$3|\vec{a}|^{2}+|\vec{b}| \vec{r}-2 \sqrt{3}|\vec{a}| \cdot|\vec{b}|=(1)^{2}$
$3 \cdot 1+1-2 \sqrt{3} \cdot 1 \cdot 1 \cdot \cos \theta=1$
$\Rightarrow 2 \sqrt{3} \cos \theta=3$
$\Rightarrow \cos \theta=\frac{3}{2 \sqrt{3}}=\frac{\sqrt{3}}{2}$
$\Rightarrow \theta=30^{\circ}$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.