# Math Help - Finding angle of triangle in R3

1. ## Finding angle of triangle in R3

Triangle ABC has vertices at A(2,1,2), B(3,0,2) and C(-1,2,1) respectively. Find Angle BAC.

I presume the question is asking to work out Angle A, but I am a bit confused. I know that you are suppose to use the dot product and formula:

$cos\theta = \frac {\mathbf{a}\cdot \mathbf{b}}{|\boldsymbol{a}||\mathbf{b}|}$

Are you suppose to use $\mathbf{b}\cdot \mathbf{c}$ since Angles A lies between side B and C?

2. Originally Posted by SyNtHeSiS
Triangle ABC has vertices at A(2,1,2), B(3,0,2) and C(-1,2,1) respectively. Find Angle BAC.

I presume the question is asking to work out Angle A, but I am a bit confused. I know that you are suppose to use the dot product and formula:

$cos\theta = \frac {\mathbf{a}\cdot \mathbf{b}}{|\boldsymbol{a}||\mathbf{b}|}$

Are you suppose to use $\mathbf{b}\cdot \mathbf{c}$ since Angles A lies between side B and C?
There are many ways of answering this question. A vector approach requires that you:

1. Get vector AB.

2. Get vector CA.

3. Get the angle between vector AB and vector CA.

4. Angle BAC = 180 degrees - (angle between vector AB and vector CA).

3. Hello, SyNtHeSiS!

$\text{Triangle }ABC\text{ has vertices: }\,A(2,1,2),\;B(3,0,2),\;C(-1,2,1)$
$\text{Find }\angle BAC.$

$\text{I presume the question is asking to work out Angle A.}$
$\text{I know that you are suppose to use the dot product formula:}$

. . $\cos\theta \:=\: \dfrac{\vec a \cdot \vec b}{ |\vec a|\,|\vec b|}$

$\text{Do we use }\vec b\cdot\vec c\,\text{ since }\angle A\text{ lies between sides }\vec b\text{ and }\vec c\;?$ . Yes!

We have: . $\begin{Bmatrix}\vec a &=& \overrightarrow{BC} &=& \langle \text{-}4,2,\text{-}1\rangle \\ \vec b &=& \overrightarrow{AC} &=& \langle \text{-}3,1,\text{-}1\rangle \\ \vec c &=& \overrightarrow{AB} &=& \langle 1,\text{-}1,0\rangle \end{Bmatrix}$

. . $\displaystyle \cos A \;=\;\frac{\vec b\cdot\vec c}{|\vec b|\,|\vec c|} \;=\;\frac{\langle \text{-}3,1,\text{-}1\rangle\cdot\langle1,\text{-}1,0\rangle} {\sqrt{3^2+1^2+1^2}\sqrt{1^2+1^2+0^2}}$

. . . . . . . $\displaystyle =\;\frac{\text{-}3-1+0}{\sqrt{11}\,\sqrt{2}} \;=\;-\frac{4}{\sqrt{22}} \;=\;-0.852802865$

Therefore: . $A \;=\;148.5178459^o \;\approx\;148.5^o$