Question

A line passes through the point (2, – 1, 3) and is perpendicular to the lines `vecr = (hati + hatj - hatk) + λ(2hati - 2hatj + hatk)` and `vecr = (2hati - hatj - 3hatk) + μ(hati + 2hatj + 2hatk)` obtain its equation.

Answer 1

Given lines

`vecr = (hati + hatj - hatk) + λ(2hati - 2hatj + hatk)`  ...(1)

and `vecr = (2hati - hatj - 3hatk) + μ(hati + 2hatj + 2hatk)`  ...(2)

Here `vecb_1 = 2hati - 2hatj + hatk`

and `vecb_2 = hati + 2hatj + 2hatk`

Now vector perpendicular to `vecb_1` and `vecb_2` is

`vecb = vecb_1 xx vecb_2 = |(hati, hatj, hatk),(2, -2, 1),(1, 2, 2)|`

`vecb = (-4 - 2)hati - (4 - 1)hatj + (4 + 2)hatk`

`vecb = -6hati - 3hatj + 6hatk`

Then equation of line passing through

`veca = (2, -1, 3)`

= `(2hati - hatj + 3hatk)`

and perpendicular to lines (1) and (2)

i.e., parallel to `vecb = vecb_1 xx vecb_2` is

`vecr = veca + tvecb`

`vecr = (2hati - hatj + 3hatk) + t(-6hati - 3hatj + 6hatk)`