Question

The $d$.cs of two lines are $\frac{\sqrt{3}}{4},\frac{1}{4},\frac{\sqrt{3}}{2}$ and $\frac{\sqrt{3}}{4},\frac{1}{4},k$ lf the angle between the lines is ${120}^{\circ }$, then $k=$

• A. $-\sqrt{3}$
• B. $\frac{2}{\sqrt{3}}$
• C. $\frac{-1}{2\sqrt{3}}$
• D. $\frac{-\sqrt{3}}{2}$

Answer 1

The correct option is D $\frac{-\sqrt{3}}{2}$

As we know that
$\cos \theta$ $=$ dot product of direction cosine
Therefore,
$\cos {120}^o=$ $\frac{\sqrt{3}}{4}$.$\frac{\sqrt{3}}{4}$+$\frac{1}{4}$.$\frac{1}{4}$+$\frac{\sqrt{3}}{2}.k$
$\cos {120}^o=\frac{3}{16}+\frac{1}{16}+\frac{\sqrt{3}}{2}k$
$\Rightarrow \frac{-1}{2}$ $=$ $\frac{1}{4}$+$\frac{\sqrt{3}}{2}k$

$\Rightarrow \frac{-1}{2}$-$\frac{1}{4}$ $=$ $\frac{\sqrt{3}}{2}k$

$\Rightarrow \frac{-3}{4}$ $=$ $\frac{\sqrt{3}}{2}k$

$\Rightarrow k=-\frac{\sqrt{3}}{2}$