Question

Find the area of the triangle whose vertices are A(3, –1, 2), B(1, –1, –3) and C(4, –3, 1)

Using vectors, find the area of the triangle whose vertices are: A(3, –1, 2), B(1, –1, –3) and C(4, –3, 1)

Answer 1

The given vertices of the triangle ABC are A(3, –1, 2), B(1, –1, –3) and C(4, –3, 1)

`vec"OA" = 3hat"i" - hat"j" + 2hat"k"`

`vec"OB" = hat"i" - hat"j" - 3hat"k"`

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

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

= `1/2 |vec"BA" xx vec"BC"|`

= `1/2 |vec"CA" xx vec"CB"|` 

`vec"AB" = vec"OB" - vec"OA"`

= `(hat"i" - hat"j" - 3hat"k") - (3hat"i" - hat"j" + 2hat"k")`

= `hat"i" - hat"j" - 3hat"k" - 3hat"i" + hat"j" - 2hat"k"`

`vec"AB" = -2hat"i" - 5hat"k"`

`vec"AC" = vec"OC" - vec"OA"`

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

= `4hat"i" - 3hat"j" + hat"k" - 3hat"i" + hat"j" - 2hat"k"`

`vec"AC" = hat"i" - 2hat"j" - hat"k"`

`vec"AB" xx vec"AC" = |(hat"i", hat"j", hat"k"),(-2, 0, -5),(1, -2, -1)|`

= `hat"i"(0 - 10) - hat"j"(2 + 5) + hat"k"(4 - 0)`

= `-10hat"i" - 7hat"j" + 4hat"k"`

`|vec"AB" xx vec"AC"| = |-10hat"i" - 7hat"j" + 4hat"k"|`

= `sqrt((-10)^2 + (-7)^2 + 4^2`

= `sqrt(100 + 49 + 16)`

= `sqrt(165)`

Area of the triangle ABC = `1/2 |vec"AB" xx vec"AC"|`

= `1/2 xx sqrt(165)`