Answer
(a) $\frac{x-1}{-1}=\frac{y+5}{2}=\frac{z-6}{-3}$
(b) $(-1,-1,0)$, $(-3/2,0,-3/2)$, $(0,-3,3)$.
Work Step by Step
(a) Equation of the line is given by $r=r_0+tv$
$r_0=(1,-5,6)$ and $v= \lt 1,2,-3 \gt$
The symmetric equations are defined by:
$\frac{x-x_0}{a}=\frac{y-y_0}{b}=\frac{z-z_0}{c}$
Hence, the symmetric equations are:
$\frac{x-1}{-1}=\frac{y+5}{2}=\frac{z-6}{-3}$
(b) From part (a), we have
$\frac{x-1}{-1}=\frac{y+5}{2}=\frac{z-6}{-3}$
For the intersection with the $xy-plane$, set $z=0$ and solve for $x$ and $y$.
Therefore, the point of intersection with the $xy-plane$ is $(-1,-1,0)$.
For the intersection with the $xz-plane$, set $y=0$ and solve for $x$ and $z$.
Therefore, the point of intersection with the $xz-plane$ is $(-3/2,0,-3/2)$
For the intersection with the $yz-plane$, set $x=0$ and solve for $y$ and $z$.
Therefore, the point of intersection with the $yz-plane$ is $(0,-3,3)$.
Hence, $(-1,-1,0)$, $(-3/2,0,-3/2)$, $(0,-3,3)$.
