Answer
(a) $x=1+5t$, $y=1+11t$ , $z=7t$
(b) $cos^{-1} \frac{1}{14} \approx 85.9 ^{\circ}$
Work Step by Step
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: $cos \theta =\frac{n_1 \cdot n_2}{|n_1| |n_2|}$
$cos \theta =\dfrac{\lt 3,-2,1\gt \cdot\lt 2,1,-3\gt}{\sqrt {14}{\sqrt {14}}}=\frac{1}{14}$
$\theta = cos^{-1} \frac{1}{14} \approx 85.9 ^{\circ}$
