If a line L makes angles pi / 3 and pi / 4 wi | vedclass

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(D) Let the direction cosines of line $L$ be $(l, m, n)$.Given that $L$ makes angles $\pi / 3$ and $\pi / 4$ with $Y$-axis and $Z$-axis respectively.T

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$\cos^{-1} \left(\frac{\sqrt{2}+1}{\sqrt{6}}\right)$

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(D) Let the direction cosines of line $L$ be $(l, m, n)$.Given that $L$ makes angles $\pi / 3$ and $\pi / 4$ with $Y$-axis and $Z$-axis respectively.Therefore,$m = \cos(\pi / 3) = 1 / 2$ and $n = \cos(\pi / 4) = 1 / \sqrt{2}$.We know that $l^2 + m^2 + n^2 = 1$.Substituting the values,$l^2 + (1 / 2)^2 + (1 / \sqrt{2})^2 = 1 \Rightarrow l^2 + 1 / 4 + 1 / 2 = 1 \Rightarrow l^2 = 1 - 3 / 4 = 1 / 4$.Thus,$l = 1 / 2$ (taking the positive value).The direction ratios of the second line are $(1, 1, 1)$. Its direction cosines are $(1 / \sqrt{3}, 1 / \sqrt{3}, 1 / \sqrt{3})$.The angle $	heta$ between the two lines is given by $\cos 	heta = |l_1 l_2 + m_1 m_2 + n_1 n_2|$.$\cos 	heta = |(1 / 2)(1 / \sqrt{3}) + (1 / 2)(1 / \sqrt{3}) + (1 / \sqrt{2})(1 / \sqrt{3})| = |1 / (2 \sqrt{3}) + 1 / (2 \sqrt{3}) + 1 / \sqrt{6}|$.$\cos 	heta = |1 / \sqrt{3} + 1 / \sqrt{6}| = |\sqrt{2} / \sqrt{6} + 1 / \sqrt{6}| = (\sqrt{2} + 1) / \sqrt{6}$.Therefore,$	heta = \cos^{-1} \left( \frac{\sqrt{2} + 1}{\sqrt{6}} ight)$.

If a line $L$ makes angles $\pi / 3$ and $\pi / 4$ with $Y$-axis and $Z$-axis respectively,then the angle between $L$ and another line having direction ratios $(1, 1, 1)$ is

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