Question

Find the distance of the point (–1, –5, –10) from the point of intersection of the line `vecr = 2hati - hatj - 2hatk + lambda(3hati + 4hatj + 2hatk)` and the plane `vecr * (hati - hatj + hatk)` = 5

Options

• 10

• 12

• 13

• 14

Answer 1

13

Explanation:

The line of the plane are

`vecr = 2hati - hatj - 2hatk + lambda(3hati + 4hatj + 2hatk)`  ......(i)

And `vecr * (hati - hatj + hatk)` = 5  ......(ii)

Solving the equations (i) and (ii)

`[2hati - hatj - 2hatk + lambda(3hati + 4hatj + 2hatk)] * (hati - hatj + hatk)` = 5

or `(2hati - hatj + 2hatk) * (hati - hatj + hatk) + lambda(23 - 4hatj + 2hatk)`

`(hati - hatj + hatk)` = 5

or `(2 + 1 + 2) + lambda(3 - 4 + 2)` = 5

or `5 + (1)` = 5 ⇒ `lambda` = 0

∴ The other point is `(-1, -5, -10) = hati - 5hatj  - 10hatk`

Distance between these points = `sqrt([2 (-1)^2 + (-1 + 5)^2 + [2 - (- 10)]]^2`

= `sqrt(3^2 + 4^2 + 12^2)`

= `sqrt(9 + 16 + 144)`

= 13