Question #94346

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Question #94346

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Answer 1 · 2015-07-01T18:46:43

$ˆ P Q R = {cos}^{- 1} (\frac{27}{\sqrt{1235}})$ $hat(PQR)=cos^(-1)(27/sqrt1235)$

Explanation:

Be two vectors $- - \to A B$ $vec(AB)$ and $- - \to A C$ $vec(AC)$ :

$- - \to A B \cdot - - \to A C = (A B) (A C) cos (ˆ B A C)$ $vec(AB) * vec(AC) = (AB)(AC)cos(hat(BAC))$
$= (x_{A B} x_{A C}) + (y_{A B} y_{A C}) + (z_{A B} z_{A C})$ $=(x_(AB)x_(AC))+(y_(AB)y_(AC))+(z_(AB)z_(AC))$

We have:

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

therefore

$- - \to Q P = (x_{P} - x_{Q}; y_{P} - y_{Q}; z_{P} - z_{Q}) = (3; - 1; - 3)$ $vec(QP)=(x_P-x_Q;y_P-y_Q;z_P-z_Q)=(3;-1;-3)$
$- - \to Q R = (x_{R} - x_{Q}; y_{R} - y_{Q}; z_{R} - z_{Q}) = (5; - 6; - 2)$ $vec(QR)=(x_R-x_Q;y_R-y_Q;z_R-z_Q)=(5;-6;-2)$

and

$(Q P) = \sqrt{{(x_{Q P})}_{Q P}^{2} + {(y_{Q P})}_{Q P}^{2} + {(z_{Q P})}_{Q P}^{2}} = \sqrt{9 + 1 + 9} = \sqrt{19}$ $(QP)=sqrt((x_(QP))^2+(y_(QP))^2+(z_(QP))^2)=sqrt(9+1+9)=sqrt(19)$

$(Q R) = \sqrt{{(x_{Q R})}_{Q R}^{2} + {(y_{Q R})}_{Q R}^{2} + {(z_{Q R})}_{Q R}^{2}} = \sqrt{25 + 36 + 4} = \sqrt{65}$ $(QR)=sqrt((x_(QR))^2+(y_(QR))^2+(z_(QR))^2)=sqrt(25+36+4)=sqrt(65)$

Therefore:

$- - \to Q P \cdot - - \to Q R = \sqrt{19} \sqrt{65} cos (ˆ P Q R)$ $vec(QP)*vec(QR)=sqrt19sqrt65cos(hat(PQR))$
$= (3 \cdot 5 + (- 1) (- 6) + (- 3) (- 2))$ $=(3*5+(-1)(-6)+(-3)(-2))$

$\to cos (ˆ P Q R) = \frac{15 + 6 + 6}{\sqrt{19} \sqrt{65}} = \frac{27}{\sqrt{1235}}$ $rarr cos(hat(PQR))=(15+6+6)/(sqrt19sqrt65)=27/sqrt1235$

$\to ˆ P Q R = {cos}^{- 1} (\frac{27}{\sqrt{1235}})$ $rarr hat(PQR)=cos^(-1)(27/sqrt1235)$

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Question #94346

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