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.

Answer 1

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°