Calculate the area of the triangle formed by the tips of vectors `vecP=2hati+hatj-3hatk, vecQ=3hati-hatj+2hatk` and
`vecR=4hati-3hatj+hatk`

Explanation

Let the triangle formed be PQR
`vec(QP)=vecP-vecQ=(2hati+hatj-3hatk)-(3hati-hatj+2hatk)`
`= -hati+2hatj-5hatk`
`vec(QR)=vecR-vecQ`
`=(4hati-3hatj+hatk)-(3hati-hatj+2hatk)`
`=hati-2hatj-hatk`
Now, `vec(QP)xxvec(QR)=|(hati,hatj,hatk),(-1,2,-5),(1,-2,-1)|`
`=[2xx(-1)-(-5)xx(-2)]hati+[1xx(-5)-(-1)xx(-1)]hatj+[(-1)xx(-2)-2xx1]hatk`
`=(-2-10)hati+(-5-1)hatj+(2-2)hatk`
`=-12hati-6hatj+0hatk`
`=-12hati-6hatj+0hatk`
`|vec(QP)xxvec(QR)|=sqrt((-12)^(2)+(-6)^(2)+0^(2))`
`=sqrt(144+36)=sqrt(180)=13.42`
`therefore` Area of triangle PQR`=(1)/(2)|vec(QP)xxvec(QR)|`
`=(1)/(2)xx13.42=6.71` square unit