Question

Find the vector equation of a plane which is at a distance of 7 units from the origin and which is normal to the vector `3hati + 5hatj - 6hatk`

Options

• 7

• `sqrt(70)`

• `7sqrt(70)`

• `70sqrt(7)`

Answer 1

`7sqrt(70)`

Explanation:

Let `vecn = 3hati + 5hatj - 6hatk`

`|vecn| = sqrt(3^2 + 5^2 (-6)^2) = sqrt(9 + 25 + 36) = sqrt(70)`

`hatn = vecn/|vecn| = 3/sqrt(70) hati + 5/sqrt(70)hatj - 6/sqrt(70) hatk`

The required equation of the plane is `vecr^'  hatn = d`

i.e., `vecr^'  (3/sqrt(70)hati + 5/sqrt(70) hatj - 3/sqrt(70)hatk)` = 7

or `vecr^'  (3hati + 5hatj - 6hatk) = 7sqrt(70)`