Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 12 - Vectors and the Geometry of Space - 12.5 Equations of Lines and Planes - 12.5 Exercises - Page 872: 57


(a) $x=1$, $y=-t$ , $z=t$ (b) (b) $ cos^{-1} \frac{5}{3 \sqrt 3} \approx 15.8 ^{\circ}$

Work Step by Step

(a) $x+y+z=1$ and $x+2y+2z=1$ In order to find a point of intersection we will set $z=0$, solve for $x$ and $y$. $x+y+z=1$ $x+y+0=1$ $x=1$ and $y=0$ Thus, point of intersection is: $(1,0,0)$ $n_1=\lt 1,1,1\gt$ and $n_2=\lt 1,2,2\gt$ $n_1 \times n_2=\lt 1,1,1\gt \times \lt 1,2,2\gt= \lt 0,-1,1\gt$ $x=1+(0)t$ $y=0+(-1)t$ $z=0+(1)t$ Hence, $x=1$, $y=-t$ , $z=t$ (b) $n_1=\lt 1,1,1\gt$ and $n_2=\lt 1,2,2\gt$ To find the cosine of the angle, use formula: $cos \theta =\frac{n_1 \cdot n_2}{|n_1| |n_2|}$ $cos \theta =\dfrac{\lt 1,1,1\gt \cdot \lt 1,2,2\gt}{\sqrt {1^2+1^2+1^2}{\sqrt {1^2+2^2+2^2}}}=\frac{5}{3 \sqrt 3}$ $\theta = cos^{-1} \frac{5}{3 \sqrt 3} \approx 15.8 ^{\circ}$
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