# Problem with alternate solution — Equation of plane through point and containing intersection line of two planes [Stewart P $803, 12.5.37$]

$37.$ Find an equation of the plane that passes through the point $(1, -2, 1)$
and contains the line of intersection of the planes $x + y - z = 2$ and $2x - y + 3z = 1$. $\bbox[3px,border:2px solid grey]{\text{ Official solution : }}$My modified picture illustrates that the red line of intersection belongs to both planes. A normal vector of each plane $\perp$ to this red line of intersection. Thus, the direction vector of the red line = $\mathbb{n_1} \times \mathbb{n_2} = (1, 1, -1) \times (2, -1, 3) = \color{#C154C1}{(2, -5, -3)}$

For want of a normal vector to the requested plane, require another direction vector on this requested plane. Thus need the vector containing any point on the red line of intersection to the given point $(-1, 2, 1)$ in the plane. WLOG, set $\color{#FF4F00}{x = 0}$: $\begin{cases} x + y - z = 2 \\ 2x - y + 3z = 1 \end{cases} \implies \begin{cases} y - z = 2 \\ -y + 3z = 1 \end{cases}$ $\implies (\color{#FF4F00}{x}, y, z) = (\color{#FF4F00}{0}, 7/2, 3/2).$
Thus another vector parallel to the plane is $(-1, 2, 1) - (\color{#FF4F00}{0}, 7/2, 3/2) = \color{#C154C1}{(-1, -3/2, -1/2)}$.

In toto, a normal vector to the plane $= \color{#C154C1}{(2, -5, -3) \times (-1, -3/2, -1/2)} = 2(-1, 2, -4).$ $... \blacksquare$

$\bbox[3px,border:2px solid grey]{\text{ My solution : }}$ ♦ Any vector on the line of intersection of the two planes produces one vector $\parallel$ to the requested plane.

♠ The other vector $\parallel$ the requested plane is any vector connecting any point on this line of intersection with the given point $(-1, 2, 1)$ in the plane.

♦ Subtract the two equations of the plane to produce their line of intersection: $3x + 2z = 3$. For a direction vector of this line, subtract any two vectors on it. WLOG, put $z = 0 \implies \color{brown}{(1, 0, 0)}$ and put $z = 3 \implies \color{brown}{(-1, 0, 3)}$. Thus direction vector $= (-1, 0, 3) - (1, 0, 0) = (-2, 0, 3)$.

♠ Observe the given point $(-1, 2, 1)$ isn't on this line (but is given to be on the plane). Thus either $\color{brown}{(1, 0, 0)} - (-1, 2, 1) = (2, -2, -1)$ or $\color{brown}{(-1, 0, 3)} - (-1, 2, 1)$ are on the plane.

In toto, a normal vector to the plane = $(-2, 0, 3) \times (2, -2, -1) = 2(3, 2, 2) \neq k(-1, 2, -4)... \blacksquare$

Since the solution's and my normal vectors differ already, what's wrong with my solution?

Let's call the given point $P$. Then there is a typo in your question: at the beginning you specify $P$ as $(1,-2,1)$ and later you change it to $(-1,2,1)$. Let's assume $P =(-1,2,1)$.
Your mistake happens in the third step where you subtract the equations. The point $(1,0,0)$ is not on the line of intersection of the two planes. To see this note that if it were it would satisfy the equation of the first plane, $x+y-z=2$ but $1+0-0\neq 2$.
My guess of what happened: After you subtracted the equations the variable $y$ got eliminated. You assumed that this meant that $y=0$. But this is not the case: any point $(x,y,z)$ on the red line is specified by three coordinates satisfying both $x+y-z=2$ and $2x-y+3z =1$.
• Yep. $3x + 2z = 3$ in $\mathbb R^3$ defines a plane containing the red line rather than the line itself. – epimorphic May 21 '15 at 19:29