Cartesian equation

Quote:

Originally Posted by iExcavate

Hey I am studying Linear and Abstract Algebra and came across this:

Find the Cartesian equation of the line that passes through the point $P(2, 0, -5)$ and is perpendicular to both the vectors $u = (0, 1, 4)$ and $v(-2, -1, 3)$

I am aware that Cartesian equations are in the form: ${(x - x0)/a} = {(y - y0)/b} = {(z - z0)/c}$ , where a,b,c do not equal 0. However, I am not sure how to write the above as a Cartesian equation. If you could help, I thank you!

The vector form of lines is always easiest.

To find a vector that's perpendicular to both $\mathbf{u}$ and $\mathbf{v}$ is $\mathbf{u}\times\mathbf{v}$ .

$\mathbf{u}\times\mathbf{v} = \left|\begin{matrix}\phantom{-}\mathbf{i} & \phantom{-}\mathbf{j} & \phantom{-}\mathbf{k} \\ \phantom{-}0 & \phantom{-}1 & \phantom{-}4 \\ -2 & -1 & \phantom{-}3\end{matrix}\right|$

$= \mathbf{i}\left|\begin{matrix}\phantom{-}1 & 4 \\ -1 & 3\end{matrix}\right| - \mathbf{j}\left|\begin{matrix}\phantom{-}0 & 4 \\ -2 & 3\end{matrix}\right| + \mathbf{k}\left|\begin{matrix}\phantom{-}0 & \phantom{-}1 \\ -2 & -1\end{matrix}\right|$

$= \mathbf{i}[1(3) - 4(-1)] - \mathbf{j}[0(3) - 4(-2) ] + \mathbf{k}[0(-1) - 1(-2)]$

$= 7\mathbf{i} - 8\mathbf{j} + 2\mathbf{k}$

$= (7, -8, 2)$ .

Therefore the line that passes through $P(2, 0, -5)$ and perpendicular to both $\mathbf{u}$ and $\mathbf{v}$ is

$(2, 0, 5) + t(7, -8, 2)$

$= (2 + 7t, -8t, 5 + 2t)$ .

So $x = 2 + 7t, y = -8t, z = 5 + 2t$

Therefore:

$t = \frac{x - 2}{7} = -\frac{y}{8} = \frac{z - 5}{2}$ .