Question

The angle between the lines whose cosines are $(\frac{-\sqrt{3}}{4},\frac{1}{4},\frac{-\sqrt{3}}{2})$ and $(\frac{-\sqrt{3}}{4},\frac{1}{4},\frac{-\sqrt{3}}{2})$, is given by ___________.

• A. ${30}^o$
• B. $0^o$
• C. ${45}^o$
• D. ${60}^o$

Answer 1

Answer: (B)

$0^o$

Given cosines are $\left(\frac{-\sqrt{3}}{4},\frac{1}{4},\frac{-\sqrt{3}}{2}\right)$ and $\left(\frac{-\sqrt{3}}{4},\frac{1}{4},\frac{-\sqrt{3}}{2}\right)$,

If, $os\theta =\left|l_1{l}_2+m_1{m}_m+n_1{n}_2\right|$.

then,

$=\left|\left(\frac{-\sqrt{3}}{4}\right)\left(\frac{-\sqrt{3}}{4}\right)+\left(\frac{1}{4}\right)\left(\frac{1}{4}\right)+\left(\frac{-\sqrt{3}}{2}\right)\left(\frac{-\sqrt{3}}{2}\right)\right|$

$=\frac{3}{16}+\frac{1}{16}+\frac{3}{4}$

=$\frac{16}{16}$=1

$\therefore \cos \theta =1$

$\theta =0$

So, the correct option is B