Question

The distance of the point `2hati + hatj - hatk` from the plane `vecr.(hati - 2hatj + 4hatk)` = 9 will be ______.

Options

• 13

• `13/sqrt(21)`

• 21

• `21/sqrt(13)`

Answer 1

The distance of the point `2hati + hatj - hatk` from the plane `vecr.(hati - 2hatj + 4hatk)` = 9 will be `underlinebb(13/sqrt(21))`.

Explanation:

Plane `vecr.(hati - 2hatj + 4hatk)` = 9  ...(1)

and Point `veca = 2hati + hatj - hatk`

Here `vecn = hati - 2hatj + 4hatk`

`|vecn| = sqrt(1^2 + (-2)^2 + 4^2)`

= `sqrt(1 + 4 + 16)`

= `sqrt(21)`

Then distance of point `veca` from plane (1)

= `(|veca.vecn - 9|)/|vecn|`

= `(|(2)(1) + (1)(-2) + (-1)(4) - 9|)/sqrt(21)`

= `(|2 - 2 - 4 - 9|)/sqrt(21)`

= `13/sqrt(21)` units.