Question

Find the distance of the point (1, –2, 0) from the point of the line `vecr = 4hati + 2hatj + 7hatk + λ(3hati + 4hatj + 2hatk)` and the point `vecr.(hati - hatj + hatk)` = 10.

Answer 1

Given line is `vecr = 4hati + 2hatj + 7hatk + λ(3hati + 4hatj + 2hatk)`  ...(i)

and plane is `vecr.(hati - hatj + hatk)` = 10  ...(ii)

From (i) and (ii), we get

`4hati + 2hatj + 7hatk + λ(3hati + 4hatj + 2hatk).(hati - hatj + hatk)` = 10

⇒ `(4 + 3λ)hati + (2 + 4λ)hatj + (7 + 2hatk)(hati - hatj + hatk)` = 10

⇒ (4 + 3λ).1 + (2 + 4λ)(–1) + (7 + 2λ).1 = 10

⇒ 4 + 3λ – 2 – 4λ + 7 + 2λ = 10

⇒ λ + 9 = 10

⇒ λ = 1

Now, from (i), we get

`vecr = (4hati + 2hatj + 7hatk) + 1(3hati + 4hatj + 2hatk)`

or, `vecr = 7hati + 6hatj + 9hatk`

or, `xhati + yhatj + zhatk = 7hati + 6hatj + 9hatk`

∴ Required point is (7, 6, 9)

Now, distance between (7, 6, 9) and (1, –2, 9) is

d = `sqrt((7 - 1)^2 + (6 + 2)^2 + (9 - 9)^2)`

= `sqrt(6^2 + 8^2 + 0^2)`

= `sqrt(36 + 64)`

= `sqrt(100)`

= 10 units