Answer
(a) $x=1$, $y=-t$ , $z=t$
(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 $z=0$ and solve for $x$ and $y$.
$x+y+z=1$
$x+y+0=1$
$x=1$ and $y=0$
Thus, the 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:
$cos \theta =\frac{n_1 \cdot n_2}{|n_1| |n_2|}$
$cos \theta =\dfrac{\lt 1,1,1\gt \cdot \lt 1,2,2\gt}{\sqrt {1^2+1^2+1^2}{\sqrt {1^2+2^2+2^2}}}=\frac{5}{3 \sqrt 3}$
$\theta = cos^{-1} \frac{5}{3 \sqrt 3} \approx 15.8 ^{\circ}$
