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(D) The angle $	heta$ between two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ is given by $\cos 	heta = |l_1 l_2 + m_1 m_2

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(D) The angle $	heta$ between two lines with direction cosines $(l_1, m_1, n_1)$ and $(l_2, m_2, n_2)$ is given by $\cos 	heta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$.Given direction cosines are:Line $1: (\frac{-\sqrt{3}}{4}, \frac{1}{4}, \frac{-\sqrt{3}}{2})$Line $2: (\frac{-\sqrt{3}}{4}, \frac{1}{4}, \frac{\sqrt{3}}{2})$Calculating the dot product:$\cos 	heta = |(\frac{-\sqrt{3}}{4})(\frac{-\sqrt{3}}{4}) + (\frac{1}{4})(\frac{1}{4}) + (\frac{-\sqrt{3}}{2})(\frac{\sqrt{3}}{2})|$$\cos 	heta = |\frac{3}{16} + \frac{1}{16} - \frac{3}{4}|$$\cos 	heta = |\frac{4}{16} - \frac{12}{16}| = |\frac{-8}{16}| = |-\frac{1}{2}| = \frac{1}{2}$.Since $\cos 	heta = \frac{1}{2}$,the angle $	heta = 60^{\circ}$.Thus,the correct option is $D$.

The angle $\theta$ between the lines whose direction cosines are $l_1 = \frac{-\sqrt{3}}{4}, m_1 = \frac{1}{4}, n_1 = \frac{-\sqrt{3}}{2}$ and $l_2 = \frac{-\sqrt{3}}{4}, m_2 = \frac{1}{4}, n_2 = \frac{\sqrt{3}}{2}$ is: (in $^{\circ}$)

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