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Question
Show that the straight lines whose direction cosines are given by 2l + 2m – n = 0 and mn + nl + lm = 0 are at right angles.
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Solution
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
`b_1 = -2hati + hatj - 2hatk` and `b_2 = hati - 2hatj - 2hatk`
Also, angle between two lines,
`vecr = veca_1 + lambdavecb_1` and `veca_2 + muvecb_2` is given by
`costheta = |(vecb_1 * vecb_2)/(|vecb_1||vecb_2|)|`
Now, `vecb_1 * vecb_2 = 2(1) + 1(-2) + (-2)(-2)`
= 2 – 2 + 4
= 0
⇒ cos θ = 0
⇒ θ = 90°
Hence, angle between given two lines is 90°
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