Question

a. Find symmetric equations for the line that passes through the point (4, −4, 8) and is parallel to the vector −1, 4, −3b. Find the points in which the required line in part (a) intersects the coordinate planes.i. point of intersection with xy-planeii. point of intersection with yz-planeiii. point of intersection with xz-plane

Answer 1

The line has equation

$\vec r(t)=(4,-4,8)+(-1,4,-3)t$

where $t$ is any real number. $(-1,4,-3)t$ is the line containing all scalar multiples of the vector (-1, 4, -3); we add (4, -4, 8) to shift the line so that it passes through this point while remaining parallel to the the line.

To get the symmetric form, we have

$\begin{cases}x(t)=4-t\\y(t)=-4+4t\\z(t)=8-3t\end{cases}$

Solving for $t$ in each equation gives the symmetric form,

$t=\boxed{4-x=\dfrac{y+4}4=\dfrac{8-z}3}$

The line has intercepts in the coordinate planes wherever either the $x$, $y$, or $z$ coordinate is 0.

$xy$-plane:

$z=0\implies\begin{cases}4-x=\frac83\\\frac{y+4}4=\frac83\end{cases}\implies\left(\dfrac43,\dfrac{20}3,0\right)$

$yz$-plane:

$x=0\implies\begin{cases}\frac{y+4}4=4\\\frac{8-z}3=4\end{cases}\implies\left(0,12,-4\right)$

$xz$-plane:

$y=0\implies\begin{cases}4-x=1\\\frac{8-z}3=1\end{cases}\implies\left(3,0,5\right)$