Question

Consider the position vectors of A, B and C as `vec(OA) = 2hati - 2hatj + hatk, vec(OB) = hati + 2hatj - 2hatk` and `vec(OC) = 2hati - hatj + 4hatk`

1. Calculate `vec(AB)` and `vec(BC)`.   [1]
2. Find the projection of `vec(AB)` on `vec(BC)`.   [1]
3. Find the area of the triangle ABC whose sides are `vec(AB)` and `vec(BC)`.   [2]

Answer 1

`vec(OA) = 2hati - 2hatj + hatk`

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

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

i. `vec(AB) = vec(OB) - vec(OA)`

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

= `-hati + 4hatj - 3hatk`

`vec(BC) = vec(OC) - vec(OB)`

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

= `hati - 3hatj + 6hatk`

ii. Projection of `vec(AB)` on `vec(BC)`

= `(vec(AB) * vec(BC))/(|vec(BC)|`

= `((-hati + 4hatj - 3hatk) * (hati - 3hatj + 6hatk))/(|hati - 3hatj + 6hatk|)`

= `(-1 - 12 - 18)/sqrt(1 + 9 + 36)`

= `(-31)/sqrt(46)`

iii. Area of ΔABC whose sides are `vec(AB)` and `vec(BC)`

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

Area of ΔABC = `1/2 |(-hati + 4hatj - 3hatk) xx (hati - 3hatj + 6hatk)|`

= `1/2 |(hati, hatj, hatk),(-1, 4, -3),(1, -3, 6)|`

= `1/2 |hati(24 - 9) - hatj(-6 + 3) + hatk(3 - 4)|`

= `1/2 |15hati + 3hatj - hatk|`

Now, Area = `1/2 |15hati + 3hatj - hatk|`

`Δ = 1/2 sqrt(225 + 9 + 1)`

`Δ = 1/2 sqrt(235)` sq. units.