Using vectors, find the area of the `triangleABC`, whose vertices are `A(1, 2, 3),B(2, -1, 4) and C(4, 5, -1)`.

Explanation

We have
Position vector of `A= ( hat(i) + 2 hat(j) + 3 hat(k)),`
position vector of `B = ( 2 hat(i) - hat(j) + 4 hat(k)) and `
position vector of `C = ( 4 hat(i) + 5 hat(j)- 4 hat(k)).`
`:. vec(AB) = " ( position vector of B) - ( position vector of A) "`
`= (2 hat(i) - hat(j) + 4 hat(k))- ( hat(i) + 2 hat(j) + 3 hat(k))= ( hat(i) - 3 hat(j) + hat(k))`.
`vec(AC) = " (position vector of C) - ( position vector of A) "`
` = ( 4 hat(i) + 5 hat(j) - hat(k)) - ( hat(i) +2 hat(j) + 3 hat(k))= ( 3 hat(i) + 3 hat(j) - 4 hat(k))`.
`:." area of " triangle ABC = | 1/2 ( vec(AB) xx vec(AC))|.`
Now, `vec(AB) xx vec(AC)= |(hat(i) , hat(j), hat(k)),(1,-3,1),(3,3,-4)|`
`=(12-3) hat(i) - (-4-3)hat(j) + ( 3+9) hat(k)`
` =(9 hat(i) + 7 hat(j) + 12 hat(k)).`
`:." area of "triangle ABC = 1/2|vec(AB) xx vec(AC)|`
`=1/2|(9 hat(i) + 7 hat(j) + 12 hat(k))|= 1/2.sqrt((9)^(2)+7^(2) + (12)^(2))`
`=1/2 .sqrt((81+ 49+144))=1/2 sqrt(274) " sq units ".`
Hence, the area of `" the area of " triangle ABC is 1/2sqrt(274)" sq units "`