Question

Find the angles of the triangle formed by vectors `vecP=5hati-3hatj+hatk`, `vecQ=-2hati+hatj+5hatk` and `vecR=9hati+5hatj+0hatk`?

Answer 1

The triangle formed is a right isoceles triangle and `m/_A=90^@`, `m/_B=m/_C=45^@`

Explanation:

Let the three vectors `vecP=5hati-3hatj+hatk`, `vecQ=-2hati+hatj+5hatk` and `vecR=9hati+5hatj+0hatk`, form a triangle with vertices `a`, `B` and `C` respectively.

Hence `|AB|=|vecQ-vecP|=|(-2-5)hati+(1-(-3))hatj+(5-1)hatk|`

= `|-7hati+4hatj+4hatk|=sqrt((-7)^2+4^2+4^2)=sqrt81=9`

`|BC|=|vecR-vecQ|=|(9-(-2))hati+(5-1)hatj+(0-5)hatk|`

= `|11hati+4hatj-5hatk|=sqrt(11^2+4^2+(-5)^2)=sqrt162=9sqrt2`

and `|CA|=|vecR-vecP|=|(9-5)hati+(5-(-3))hatj+(0-1)hatk|`

= `|4hati+8hatj-1hatk|=sqrt(4^2+8^2+(-1)^2)=sqrt81=9`

It is apparent that `AB^2+CA^2=BC^2`, where `AB=CA`

Hence the triangle formed is a right isoceles triangle and `m/_A=90^@`, `m/_B=m/_C=45^@`