# Question 94346

Jul 1, 2015

$\hat{P Q R} = {\cos}^{- 1} \left(\frac{27}{\sqrt{1235}}\right)$

#### Explanation:

Be two vectors $\vec{A B}$ and $\vec{A C}$:

$\vec{A B} \cdot \vec{A C} = \left(A B\right) \left(A C\right) \cos \left(\hat{B A C}\right)$
$= \left({x}_{A B} {x}_{A C}\right) + \left({y}_{A B} {y}_{A C}\right) + \left({z}_{A B} {z}_{A C}\right)$

We have:

P=(1;1;1)
Q=(-2;2;4)
R=(3;-4;2)

therefore

vec(QP)=(x_P-x_Q;y_P-y_Q;z_P-z_Q)=(3;-1;-3)
vec(QR)=(x_R-x_Q;y_R-y_Q;z_R-z_Q)=(5;-6;-2)#

and

$\left(Q P\right) = \sqrt{{\left({x}_{Q P}\right)}^{2} + {\left({y}_{Q P}\right)}^{2} + {\left({z}_{Q P}\right)}^{2}} = \sqrt{9 + 1 + 9} = \sqrt{19}$

$\left(Q R\right) = \sqrt{{\left({x}_{Q R}\right)}^{2} + {\left({y}_{Q R}\right)}^{2} + {\left({z}_{Q R}\right)}^{2}} = \sqrt{25 + 36 + 4} = \sqrt{65}$

Therefore:

$\vec{Q P} \cdot \vec{Q R} = \sqrt{19} \sqrt{65} \cos \left(\hat{P Q R}\right)$
$= \left(3 \cdot 5 + \left(- 1\right) \left(- 6\right) + \left(- 3\right) \left(- 2\right)\right)$

$\rightarrow \cos \left(\hat{P Q R}\right) = \frac{15 + 6 + 6}{\sqrt{19} \sqrt{65}} = \frac{27}{\sqrt{1235}}$

$\rightarrow \hat{P Q R} = {\cos}^{- 1} \left(\frac{27}{\sqrt{1235}}\right)$