JEE MAIN - Mathematics (2021 - 25th February Morning Shift - No. 10)
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 :
$${{3 \over 8}}$$
$${{3 \over 4}}$$
$${{1 \over 2}}$$
$${{5 \over 8}}$$
Explanation
$${l^2} + {m^2} + {n^2} = 1$$
$$ \therefore $$ $$2{n^2} = 1 $$ ($$ \because $$ l2 + m2 $$-$$ n2 = 0)
$$\Rightarrow n = \pm {1 \over {\sqrt 2 }}$$
$$ \therefore $$ $${l^2} + {m^2} = {1 \over 2}$$ & $$l + m = {1 \over {\sqrt 2 }}$$
$$ \Rightarrow {1 \over 2} - 2lm = {1 \over 2}$$
$$ \Rightarrow lm = 0$$ or $$m = 0$$
$$ \therefore $$ $$l = 0,m = {1 \over {\sqrt 2 }}$$ or $$l = {1 \over {\sqrt 2 }}$$
$$ < 0,{1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }} > $$ or $$ < {1 \over {\sqrt 2 }},0,{1 \over {\sqrt 2 }} > $$
$$ \therefore $$ $$\cos \alpha = 0 + 0 + {1 \over 2} = {1 \over 2}$$
$$ \therefore $$ $${\sin ^4}\alpha + {\cos ^4}\alpha = 1 - {1 \over 2}{\sin ^2}(2\alpha ) = 1 - {1 \over 2}.{3 \over 4} = {5 \over 8}$$
$$ \therefore $$ $$2{n^2} = 1 $$ ($$ \because $$ l2 + m2 $$-$$ n2 = 0)
$$\Rightarrow n = \pm {1 \over {\sqrt 2 }}$$
$$ \therefore $$ $${l^2} + {m^2} = {1 \over 2}$$ & $$l + m = {1 \over {\sqrt 2 }}$$
$$ \Rightarrow {1 \over 2} - 2lm = {1 \over 2}$$
$$ \Rightarrow lm = 0$$ or $$m = 0$$
$$ \therefore $$ $$l = 0,m = {1 \over {\sqrt 2 }}$$ or $$l = {1 \over {\sqrt 2 }}$$
$$ < 0,{1 \over {\sqrt 2 }},{1 \over {\sqrt 2 }} > $$ or $$ < {1 \over {\sqrt 2 }},0,{1 \over {\sqrt 2 }} > $$
$$ \therefore $$ $$\cos \alpha = 0 + 0 + {1 \over 2} = {1 \over 2}$$
$$ \therefore $$ $${\sin ^4}\alpha + {\cos ^4}\alpha = 1 - {1 \over 2}{\sin ^2}(2\alpha ) = 1 - {1 \over 2}.{3 \over 4} = {5 \over 8}$$
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