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

May 30, 2017

The triangle formed is a right isoceles triangle and $m \angle A = {90}^{\circ}$, $m \angle B = m \angle C = {45}^{\circ}$

#### Explanation:

Let the three vectors $\vec{P} = 5 \hat{i} - 3 \hat{j} + \hat{k}$, $\vec{Q} = - 2 \hat{i} + \hat{j} + 5 \hat{k}$ and $\vec{R} = 9 \hat{i} + 5 \hat{j} + 0 \hat{k}$, form a triangle with vertices $a$, $B$ and $C$ respectively.

Hence $| A B | = | \vec{Q} - \vec{P} | = | \left(- 2 - 5\right) \hat{i} + \left(1 - \left(- 3\right)\right) \hat{j} + \left(5 - 1\right) \hat{k} |$

= $| - 7 \hat{i} + 4 \hat{j} + 4 \hat{k} | = \sqrt{{\left(- 7\right)}^{2} + {4}^{2} + {4}^{2}} = \sqrt{81} = 9$

$| B C | = | \vec{R} - \vec{Q} | = | \left(9 - \left(- 2\right)\right) \hat{i} + \left(5 - 1\right) \hat{j} + \left(0 - 5\right) \hat{k} |$

= $| 11 \hat{i} + 4 \hat{j} - 5 \hat{k} | = \sqrt{{11}^{2} + {4}^{2} + {\left(- 5\right)}^{2}} = \sqrt{162} = 9 \sqrt{2}$

and $| C A | = | \vec{R} - \vec{P} | = | \left(9 - 5\right) \hat{i} + \left(5 - \left(- 3\right)\right) \hat{j} + \left(0 - 1\right) \hat{k} |$

= $| 4 \hat{i} + 8 \hat{j} - 1 \hat{k} | = \sqrt{{4}^{2} + {8}^{2} + {\left(- 1\right)}^{2}} = \sqrt{81} = 9$

It is apparent that $A {B}^{2} + C {A}^{2} = B {C}^{2}$, where $A B = C A$

Hence the triangle formed is a right isoceles triangle and $m \angle A = {90}^{\circ}$, $m \angle B = m \angle C = {45}^{\circ}$