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=\frac{y+4}4=\frac{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(\frac43,\frac{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)`