Question

Using vectors, find the area of the triangle ABC with vertices A(1, 2, 3), B(2, – 1, 4) and C(4, 5, – 1).

Answer 1

Given that A(1, 2, 3), B(2, –1, 4) and C(4, 5, –1)

`vec"AB" = (2 - 1)hat"i" + (-1 - 2)hat"j" + (4 - 3)hat"k"`

`vec"AB" = hat"i" - 3hat"j" + hat"k"`

`vec"AC" = (4 - 1)hat"i" + (5 - 2)hat"j" + (-1 - 3)hat"k"`

= `3hat"i" + 3hat"j" - 4hat"k"`

Area of ΔABC = `1/2 |vec"AB" xx vec"AC"|`

= `1/2|(hat"i", vec"j", vec"k"),(1, -3, 1),(3, 3, -4)|`

= `1/2 [hat"i"(12 - 3) - hat"j"(-4 - 3) + hat"k"(3 + 9)]`

= `1/2|9hat"i" + 7hat"j" + 12hat"k"|`

= `1/12 sqrt((9)^2 + (7)^2 + (12)^2)`

= `1/2 sqrt(81 + 49 + 144)`

= `1/2 sqrt(274)`

Hence, the required area is `1/2 sqrt(274)`.

Answer 2

Given that, `vec(OA) = hati + 2hatj + 3hatk`

`vec(OB) = 2hati - hatj + 4hatk`

And `vec(OC) = 4hati + 5hatj - hatk`

Now, `vec(AB) = vec(OB) - vec(OA)`

= `2hati - hatj + 4hatk - hati - 2hatj - 3hatk`

= `hati - 3hatj + hatk`

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

= `4hati + 5hatj - hatk - hati - 2hatj - 3hatk`

= `3hati + 3hatj - 4hatk`

We know the area of the given triangle

= `1/2 |vec(AB) xx vec(AC)|`

Now, `vec(AB) xx vec(AC) = |(hati, hatj, hatk),(1, -3, 1),(3, 3, -4)|`

= `hati(12 - 3) + hatj(3 + 4) + hatk(3 + 9)`

= `9hati + 7hatj + 12hatk`

Therefore, `|vec(AB) xx vec(AC)| = sqrt((9)^2 + (7)^2 + (12)^2`

= `sqrt(81 + 49 + 144)`

= `sqrt(274)`

Hence, Required area = `1/2 sqrt(274)` units².