# How do you find the vector parametrization of the line of intersection of two planes 2x - y - z = 5 and x - y + 3z = 2?

Mar 22, 2015

One answer could be:

$x = t$

$z = \frac{1}{4} t - \frac{3}{4}$

$y = \frac{7}{4} t - \frac{17}{4}$.

Let's solve the system of the two equations, explaining two letters in function of the third:

$2 x - y - z = 5$

$x - y + 3 z = 2$

So:

$y = 2 x - z - 5$

$x - \left(2 x - z - 5\right) + 3 z = 2 \Rightarrow x - 2 x + z + 5 + 3 z = 2 \Rightarrow$

$4 z = x - 3 \Rightarrow z = \frac{1}{4} x - \frac{3}{4}$

so:

$y = 2 x - \left(\frac{1}{4} x - \frac{3}{4}\right) - 5 \Rightarrow y = 2 x - \frac{1}{4} x + \frac{3}{4} - 5$

$y = \frac{7}{4} x - \frac{17}{4}$.

SO:

$z = \frac{1}{4} x - \frac{3}{4}$

$y = \frac{7}{4} x - \frac{17}{4}$

and if we put $x = t$:

$x = t$

$z = \frac{1}{4} t - \frac{3}{4}$

$y = \frac{7}{4} t - \frac{17}{4}$.