Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the direction cosines of the first line be $l_1=\frac{-\sqrt{3}}{4},m_1=\frac{1}{4},n_1=\frac{-\sqrt{3}}{2}$.

Let the direction cosines of the second line be $l_2=\frac{-\sqrt{3}}{4},m_2=\frac{1}{4},n_2=\frac{\sqrt{3}}{2}$.

The angle $\theta$ between the two lines is given by the formula:

$\cos \theta =l_1{l}_2+m_1{m}_2+n_1{n}_2$

Substituting the given values:

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

$\cos \theta =\frac{3}{16}+\frac{1}{16}-\frac{3}{4}$

$\cos \theta =\frac{4}{16}-\frac{3}{4}$

$\cos \theta =\frac{1}{4}-\frac{3}{4}$

$\cos \theta =-\frac{2}{4}=-\frac{1}{2}$

Since $\cos \theta =-\frac{1}{2}$, the angle is $\theta =120^{\circ }$.

Answer: $120^{\circ }$