## Calculus 8th Edition

(a) $x=1+5t$, $y=1+11t$ , $z=7t$ (b) $cos^{-1} \frac{1}{14} \approx 85.9 ^{\circ}$
a) $3x-2y+z=1$ and $2x+y-3z=3$ In order to find a point of intersection we will set $z=0$, solve for $x$ and $y$. ' $3x-2y+z=1$ $3x-2y+0=1$ $x=1$ and $y=1$ Thus, point of intersection is: $(1,1,0)$ $n_1=\lt 3,-2,1\gt$ and $n_2=\lt 2,1,-3\gt$ $n_1 \times n_2=\lt 3,-2,1\gt \times \lt 2,1,-3\gt= \lt 5,11,7\gt$ $x=1+(5)t$ $y=1+(11)t$ $z=0+(7)t$ Hence, $x=1+5t$, $y=1+11t$ , $z=7t$ (b) $n_1=\lt 3,-2,1\gt$ and $n_2=\lt 2,1,-3\gt$ To find the cosine of the angle, use formula: $cos \theta =\frac{n_1 \cdot n_2}{|n_1| |n_2|}$ $cos \theta =\dfrac{\lt 3,-2,1\gt \cdot\lt 2,1,-3\gt}{\sqrt {14}{\sqrt {14}}}=\frac{1}{14}$ $\theta = cos^{-1} \frac{1}{14} \approx 85.9 ^{\circ}$