Question

Find the area of the triangle with vertices A(1, l, 2), (2, 3, 5) and (1, 5, 5).

Options

• `sqrt(61)`

• `sqrt(61)`

• `1/2sqrt(61)`

• 61

Answer 1

`1/2sqrt(61)`

Explanation:

The position vectors of vertices of ΔABC are (1, l, 2) (1, 3, 5) and C(1, 5, 5)

∴ `vec(AB) = vec(OB) - vec(OA)`

= `(2hati + 3hatj + 5hatk) + (hati + hatj + 2hatk)`

= `hati + 2hatj + 3hatk`

`vec(AC) = vec(OC) - vec(OA)`

= `(hati + 5hatj + 5hatk) - (hati + hatj + 2hatk)`

= `0hati + 4hatj + 3hatk`

= `4hatj + 3hatk`

`vec(AB) xx vec(AC) = |(hati, hatj, hatk),(1, 2, 3),(0, 4, 3)| = (6 - 12)hati + (3)hatj + 4hatk`

`|vec(AB) xx vec(AC)| = |-6hati - 3hatj + 4hatk|`

= `sqrt((-6)^2 + (-3)^2 + 4^2)`

= `sqrt(36 + 9 + 16)`

= `sqrt(61)`

Area of the ΔABC = `1/2|vec(AB) xx vec(AC)| = 1/2 sqrt(61)`