# Consider the line which passes through the point P(-1, 1, -4), and which is parallel to the line x=1+7t, y=2+3t, z=3+3t How do you find the point of intersection of this new line with each of the coordinate planes?

Jan 28, 2017

$\left(\begin{matrix}0 \\ \frac{10}{7} \\ - \frac{25}{7}\end{matrix}\right)$, $\left(\begin{matrix}- \frac{10}{3} \\ 0 \\ - 5\end{matrix}\right)$ and $\left(\begin{matrix}\frac{4}{3} \\ 5 \\ 0\end{matrix}\right)$

#### Explanation:

The vector equation of the line given by:

 x=1+7t; y=2+3t; z=3+3t

is:

$\vec{r} = \left(\begin{matrix}1 \\ 2 \\ 3\end{matrix}\right) + t \left(\begin{matrix}7 \\ 3 \\ 3\end{matrix}\right)$

So the line is in the direction of the vector $\left(\begin{matrix}7 \\ 3 \\ 3\end{matrix}\right)$

The new line, (whose equation we seek) passes through
$P \left(- 1 , 1 , - 4\right)$, and has the same direction vector, so it has the equation:

$\vec{s} = \left(\begin{matrix}- 1 \\ 1 \\ - 4\end{matrix}\right) + l a m \mathrm{da} \left(\begin{matrix}7 \\ 3 \\ 3\end{matrix}\right)$

So now we need to find the coordinates where this new line, given by $\vec{s}$ meets the coordinate planes,

For the $x$-axis, $- 1 + 7 l a m \mathrm{da} = 0 \implies l a m \mathrm{da} = \frac{1}{7}$, giving

$\vec{s} = \left(\begin{matrix}- 1 \\ 1 \\ - 4\end{matrix}\right) + \frac{1}{7} \left(\begin{matrix}7 \\ 3 \\ 3\end{matrix}\right) = \left(\begin{matrix}0 \\ \frac{10}{7} \\ - \frac{25}{7}\end{matrix}\right)$

For the $y$-axis, $1 + 3 l a m \mathrm{da} = 0 \implies l a m \mathrm{da} = - \frac{1}{3}$, giving

$\vec{s} = \left(\begin{matrix}- 1 \\ 1 \\ - 4\end{matrix}\right) - \frac{1}{3} \left(\begin{matrix}7 \\ 3 \\ 3\end{matrix}\right) = \left(\begin{matrix}- \frac{10}{3} \\ 0 \\ - 5\end{matrix}\right)$

For the $z$-axis, $- 4 + 3 l a m \mathrm{da} = 0 \implies l a m \mathrm{da} = \frac{4}{3}$, giving

$\vec{s} = \left(\begin{matrix}- 1 \\ 1 \\ - 4\end{matrix}\right) + \frac{4}{3} \left(\begin{matrix}7 \\ 3 \\ 3\end{matrix}\right) = \left(\begin{matrix}\frac{4}{3} \\ 5 \\ 0\end{matrix}\right)$

Hence our coordinates are:

$\left(\begin{matrix}0 \\ \frac{10}{7} \\ - \frac{25}{7}\end{matrix}\right)$, $\left(\begin{matrix}- \frac{10}{3} \\ 0 \\ - 5\end{matrix}\right)$ and $\left(\begin{matrix}\frac{4}{3} \\ 5 \\ 0\end{matrix}\right)$