Show that the straight lines whose direction cosines are given by 2l + 2m – n = 0 and mn + nl + lm = 0 are at right angles.


Given,


2l + 2m – n = 0 [1]


n = 2(l + m) [2]


and


mn + nl + lm = 0


2m(l + m) + 2(l + m)l + lm = 0


2lm + 2m2 + 2l2 + 2lm + lm = 0


2m2 + 5lm + 2l2 = 0


2m2 + 4lm + lm + 2l2 = 0


(2m + l)(m + 2l) = 0


So, we have two cases,


l = -2m


-4m + 2m – n = 0 [From 1]


n = 2m


Hence, direction ratios of one line is proportional to -2m, m, -2m or direction ratios are (2, 1, -2)


Another case is,


m = -2l


2l + 2(-2l) – n = 0


2l – 4l = n


n = -2l


Hence, direction ratios of another lines is proportional to l, -2l, -2l or direction ratios are (1, -2, -2)


Therefore, direction vectors of two lines are and


Also,


Angle between two lines, and is given by



Now,





cos θ = 0


θ = 90°


Hence, angle between given two lines is 90°


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