The direction cosines of two lines are langle | vedclass

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(B) Let the direction cosines of the two lines be $(l_1, m_1, n_1) = (\frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4})$ and $(l_2, m_2, n_2) = (-\

Vedclass · Accepted Answer

$60$

Vedclass · Answer

(B) Let the direction cosines of the two lines be $(l_1, m_1, n_1) = (\frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4})$ and $(l_2, m_2, n_2) = (-\frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4})$.The cosine of the angle $	heta$ between the two lines is given by the formula $\cos 	heta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$.Substituting the values:$\cos 	heta = |(\frac{\sqrt{3}}{2})(-\frac{\sqrt{3}}{2}) + (\frac{1}{4})(\frac{1}{4}) + (\frac{\sqrt{3}}{4})(\frac{\sqrt{3}}{4})|$$\cos 	heta = |-\frac{3}{4} + \frac{1}{16} + \frac{3}{16}|$$\cos 	heta = |-\frac{12}{16} + \frac{4}{16}| = |-\frac{8}{16}| = |-\frac{1}{2}| = \frac{1}{2}$.Since $\cos 	heta = \frac{1}{2}$,we have $	heta = 60^{\circ}$.Thus,the angle between the lines is $60^{\circ}$.

The direction cosines of two lines are $\langle\frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\rangle$ and $\langle-\frac{\sqrt{3}}{2}, \frac{1}{4}, \frac{\sqrt{3}}{4}\rangle$. Then the angle between the lines is equal to (in $^{\circ}$)

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