Let $\alpha$ 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 sin4$\alpha$ + cos4$\alpha$ is :
Solution
${l^2} + {m^2} + {n^2} = 1$<br><br>$\therefore$ $2{n^2} = 1$ ($\because$ l<sup>2</sup> + m<sup>2</sup> $-$ n<sup>2</sup> = 0)
<br><br>$\Rightarrow n = \pm {1 \over {\sqrt 2 }}$<br><br>$\therefore$ ${l^2} + {m^2} = {1 \over 2}$ & $l + m = {1 \over {\sqrt 2 }}$<br><br>$\Rightarrow {1 \over 2} - 2lm = {1 \over 2}$<br><br>$\Rightarrow lm = 0$ or $m = 0$<br><br>$\therefore$ $l = 0,m = {1 \over {\sqrt 2 }}$ or $l = {1 \over {\sqrt 2 }}$<br><br>$< 0,{1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }} >$ or $< {1 \over {\sqrt 2 }},0,{1 \over {\sqrt 2 }} >$<br><br>$\therefore$ $\cos \alpha = 0 + 0 + {1 \over 2} = {1 \over 2}$<br><br>$\therefore$ $${\sin ^4}\alpha + {\cos ^4}\alpha = 1 - {1 \over 2}{\sin ^2}(2\alpha ) = 1 - {1 \over 2}.{3 \over 4} = {5 \over 8}$$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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