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Thread: Finding angle of triangle in R3

  1. #1
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    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:

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

    Are you suppose to use $\displaystyle \mathbf{b}\cdot \mathbf{c}$ since Angles A lies between side B and C?
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  2. #2
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    Quote Originally Posted by SyNtHeSiS View Post
    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:

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

    Are you suppose to use $\displaystyle \mathbf{b}\cdot \mathbf{c}$ 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).
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  3. #3
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    Hello, SyNtHeSiS!

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

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

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

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


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


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

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


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