Find the equations of the line of intersection of the following planes

Question

Find the equations of the line of intersection of the following planes $2x − 3y + 2z = 5$ and $x + 2y − z = 4$.

So i first put this in the normal vector form

$\langle 2, -3, 2\rangle$

$\langle 1, 2, -1\rangle$

Then i took the cross product which i got

$\langle -1, 4, 7\rangle$

I then made $z = 0$ to solve for $x$ and $y$. Then i subtracted the original equations

$$\begin{matrix}&2x - 3y = 5 \\ - &(x + 2y = 4)\end{matrix}$$ and got

$x-y = 1$

$x = 1+y$

$y = 1-x$

Subbing $x$ and $y$ into the equation of $2x-3y=5$ i got $x= \frac 85$ and $y = -3$

So far i feel as if this is wrong i was trying to solve this question following a youtube video https://www.youtube.com/watch?v=LpardiBTAvU but if this process is correct i do not know how to proceed.

Answer 1

you almost solved it. you have a mistake with the point on the line. equating z to zero and solving a 2x2 linear system will yield the point on the line ${\bf{P_0}}=(\frac{22}{7},\frac{3}{7}, 0)^T$. Hence, the line of intersection is given by its parametric representation as

\begin{equation} {\bf{l}}=(\frac{22}{7},\frac{3}{7}, 0)^T + \alpha (−1,4,7)^T \\ \forall \alpha \in \mathbb{R} \end{equation}

Answer 2

we consider $$2x-3y+2z=5$$ $$x+2y-z=4$$ multiplying the second equation by $$-2$$ and adding to the first one we obtain $-7y+4z=-3$ $$y=\frac{4}{7}z+\frac{3}{7}$$ can you apply this?

Answer 3

solving the system of the equations of the two plane we find the common straight line ( if it exists). $$ \begin{cases} 2x-3y+2z=5\\ x+2y-z=4 \end{cases} \Rightarrow \begin{cases} x=t \;(\forall t \in \mathbb{R})\\ y=13-4x=13-4t\\ z=22-7x=22-7t \end{cases} $$

So the straight line has equation: $$ \begin{bmatrix} x\\y\\z \end{bmatrix} = \begin{bmatrix} 1\\-4\\-7 \end{bmatrix} t+ \begin{bmatrix} 0\\13\\22 \end{bmatrix} $$