Let α be the angle between the lines whose direction cosines satisfy the equations l+m−n=0 and .l2+m2−n2=0 Then the value of sin4α+cos4αis

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Let $α$ be the angle between the lines whose direction cosines satisfy the equations $l + m - n = 0$ and . $l^{2} + m^{2} - n^{2} = 0$ Then the value of $\sin^{4} α + \cos^{4} α$ is

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$\frac{1}{2}$

$\frac{5}{8}$

$\frac{3}{4}$

$\frac{3}{8}$

answer is B.

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Detailed Solution

The conditions are $l + m - n = 0, l^{2} + m^{2} - n^{2} = 0$

Suppose that $n = 1$

hence, $l + m = 1, l^{2} + m^{2} = 1$

observing the above two equations $l m = 0$

when $l = 0$ then $m = 1, n = 1$

when $m = 0$ then $l = 1, n = 1$

Hence, the direction ratios of two lines are $(1, 0, 1), (0, 1, 1)$

Since $α$ be the angle between the lines $\cos α = \frac{1}{2} \Rightarrow α = 60^{0}$

$\sin^{4} α + \cos^{4} α = \sin^{4} 60 + \cos^{4} 60 = {(\frac{\sqrt{3}}{2})}^{4} + {(\frac{1}{2})}^{4} = \frac{9 + 1}{16} = \frac{10}{16} = \frac{5}{8}$