Question

If A, B and C be three non-collinear points such that `vec ("AB") = hat i + 2 hat j - hat k and vec ("AC") = 2 hat i - 3 hat j`, then find the area of ΔABC.

Answer 1

The cross product is found using the determinant of a 3 × 3 matrix:

`vec ("AB") xx vec("AC") = |(hat i, hat j, hat k),(1, 2, -1),(2, -3, 0)|`

`vec ("AB") xx vec("AC") = hat i(0 - 3) - hat j(0 + 2) + hat k(-3 - 4)`

`vec ("AB") xx vec("AC") = 3 hat i - 2 hat j - 7 hat k`

Find the Magnitude of the Cross Product:

|`vec ("AB") xx vec ("AC")`| = `sqrt((-3)^2 + (-2)^2 + (-7)^2)`

= `sqrt (9 + 4 + 49)`

= `sqrt 62`

The area of triangle ABC is half the magnitude of the cross product of its two adjacent side vectors:

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

= `sqrt 62/2`

The area of ΔABC is `sqrt 62/2` square units.