JEE MAIN - Mathematics (2021 - 27th August Evening Shift - No. 1)
The angle between the straight lines, whose direction cosines are given by the equations 2l + 2m $$-$$ n = 0 and mn + nl + lm = 0, is :
$${\pi \over 2}$$
$$\pi - {\cos ^{ - 1}}\left( {{4 \over 9}} \right)$$
$${\cos ^{ - 1}}\left( {{8 \over 9}} \right)$$
$${\pi \over 3}$$
Explanation
n = 2 (l + m)
lm + n(l + m) = 0
lm + 2(l + m)2 = 0
2l2 + 2m2 + 5ml = 0
$$2{\left( {{l \over m}} \right)^2} + 2 + 5\left( {{l \over m}} \right) = 0$$
2t2 + 5t + 2 = 0
(t + 2)(2t + 1) = 0
$$ \Rightarrow t = - 2; - {1 \over 2}$$
(i) $${l \over m} = - 2$$
$${n \over m} = - 2$$
($$-$$2m, m, $$-$$2m)
($$-$$2, 1, $$-$$2)
(ii) $${l \over m} = - {1 \over 2}$$
n = $$-$$2l
(l, $$-$$2l, $$-$$2l)
(1, $$-$$2, $$-$$2)
$$\cos \theta = {{ - 2 - 2 + 4} \over {\sqrt 9 \sqrt 9 }} = 0 \Rightarrow 0 = {\pi \over 2}$$
lm + n(l + m) = 0
lm + 2(l + m)2 = 0
2l2 + 2m2 + 5ml = 0
$$2{\left( {{l \over m}} \right)^2} + 2 + 5\left( {{l \over m}} \right) = 0$$
2t2 + 5t + 2 = 0
(t + 2)(2t + 1) = 0
$$ \Rightarrow t = - 2; - {1 \over 2}$$
(i) $${l \over m} = - 2$$
$${n \over m} = - 2$$
($$-$$2m, m, $$-$$2m)
($$-$$2, 1, $$-$$2)
(ii) $${l \over m} = - {1 \over 2}$$
n = $$-$$2l
(l, $$-$$2l, $$-$$2l)
(1, $$-$$2, $$-$$2)
$$\cos \theta = {{ - 2 - 2 + 4} \over {\sqrt 9 \sqrt 9 }} = 0 \Rightarrow 0 = {\pi \over 2}$$
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