Question

A unit vector perpendicular to the plane ABC, where A, B and C are respectively the points (3, –1, 2), (1, –1, –3) and (4, –3, 1), is

Options

• `- 1/sqrt(29) (2hati + 5hatk)`

• `1/sqrt(6) (hati - 2hatj - hatk)`

• `1/sqrt(26) (4hati - 3hatj + hatk)`

• `- 1/sqrt(165) (10hati + 7hatj - 4hatk)`

Answer 1

`- 1/sqrt(165) (10hati + 7hatj - 4hatk)`

Explanation:

Let O be the origin. 

Then `vec(OA) = 3hati - hatj + 2hatk, vec(OB) = hati - hatj - 3hatk` and `vec(OC) = 4hati - 3hatj + hatk`

∴ `vec(AB) = vec(OB) - vec(OA) = (hati - hatj - 3hatk) - (3hati - hatj + 2hatk) = 2hati + 0hatj - 5hatk`

`vec(AC) = vec(OC) - vec(OA) = (4hati - 3hatj + hatk) - (3hati - hatj + 2hatk) = hati - 2hatj - hatk`

Now, `vec(AB) xx vec(AC) = |(hati, hatj, hatk),(-2, 0, -5),(1, -2, -1)|` 

= `(0 - 10)hati - (2 + 5)hatj + (4 - 0)hatk`

= `- 10hati - 7hatj + 4hatk`

⇒ `|vec(AB) xx vec(AC) = sqrt((-10)^2 + (-7)^2 + (4)^2|`

= `sqrt(100 + 49 + 16)`

= `sqrt(165)`

A unit vector perpendicular to the plane of ΔABC is perpendicular to both `vec(AB)` and `vec(AC)`. Hence, a unit vector perpendicular, to the plane of 

ΔABC = `(vec(AB) xx vec(AC))/(|vec(AB) xx vec(AC)|)`

= `(-10hati - 7hatj + 4hatk)/sqrt(165)`

= `- 1/sqrt(165) (10hati + 7hatj - 4hatk)`.