Finding the parametric and vector forms of the line at intersection of planes $2x – y + 3z – 1 = 0$ and $–x + 3y + z – 4 = 0$.

Question

Find the parametric and vector forms of the line at the intersection of planes $$\begin{align}2x – \phantom{3}y + 3z – 1 = 0 \\ –x + 3y + \phantom{3}z – 4 = 0 \end{align}$$

Solution:

If you solve the matrix you would get below:

Parametric form:

$x = 7/5 -2t$

$y = 9/5 - t$

$z = t$

Vector form:

$[x, y, z] = [7/5, 9/5, 0] + t[-2, -1, 1]$

Would this be correct?

Answer 1

Yes, it is correct, and we can check that both plane - equations are satisfied by the parametric equation, that is

• $2(7/5-2t)-(9/5-t)+3(t)=1 \implies 14/5-4t-9/5+t+3t=1$
• $-(7/5-2t)+3(9/5-t)+t=4\implies -7/5+2t+27/5-3t+t=4$

As a complete check, starting from the plane equations

• $2x-y+3z=1$
• $-x+3y+z=4$

by eliminating we obtain

• $2x-y+3z=1$
• $5y+5z=9$

therefore two solutions are

• $P_1=(-11/5,0,9/5)$
• $P_2=(7/5,9/5,0)$

and the parametric equation is

$$P_2+t(P_1-P_2)=(7/5,9/5,0)+t(-2,-1,1)$$