Question

Find the angles of a triangle whose vertices are `A(0,-1,-2),B(3,1,4)` and `C` (5,7,1).

Answer 1

Here, `vec(AB) = vecB - vecA = 3hati+2hatj+6hatk`
`vec(BC) = 2hati+6hatj-3hatk`
`vec(AC) = 5hati+8hatj+3hatk`
`|vec(AB)| = sqrt(3^2+2^2+6^2) = 7`
`|vec(BC)| = sqrt(32^2+6^2+(-3)^2) = 7`
`|vec(AC)| = sqrt(5^2+8^2+3^2) = 7sqrt2`
Now, `vec(AB)*vec(BC) = 6+12-18 = 0`
It means, `AB` and `BC` are perpendicular.
`:./_ABC = 90^@`
Also, as `|vec(AB)| = |vec(BC)| `
`:. /_ACB = /_BAC`
As, `/_ABC = 90^@`
`:. /_ACB = /_BAC = 1/2*(180-90)^@ = 45^@.`