# Vector equation of planes

So I have these two planes:

$$\pi_1: X = (1,0,0) + \lambda(0,1,1) + \gamma(1,2,1)$$ $$\pi_2: X = (0,0,0) + \lambda(0,3,0) + \gamma(-2,-1,-1)$$

I need to find two points $A$ and $B$ of the intersection of the two planes and then determinate the line that passes between these two points. I already know one way to solve this:

I can find the normal vector of the two planes by doing the cross product of its direction vectors (for $\pi_1$ we have $\vec v_1 = (0,1,1), \vec v_2 = (1,2,1)$ and for $\pi_2$ we have $\vec v_3 = (0,3,0), \vec v_4 = (-2,-1-1)$. If I take the cross product of each pair of vectors, I'm gonna have the normal vectors to the planes $\vec n_1, \vec n_2$. Then, I suppose that exists the direction vector of the line that passes by $A$ and $B$, let's call this vector $\vec v$. I then, can take the dot product of this vector with $\vec n_1$ and $\vec n_2$, so I'll find the vector in question. But then, for the line, I still need to find a point of it, so I can create the vector equation. How could I do it?

Also, is there another way of doing this, without determining the line equation from the vectors? (I really need another way too, so I can understand it better)

• Are there any constraints on $A$ or $B$? Do they need to be at the intersection of these planes? On these planes? Any two points $A$, $B$? – Carser May 20 '14 at 3:18
• Sorry, they're the intersection. Gonna edit it. – Marter Js May 20 '14 at 3:19
• No worries, that makes it much clearer :) – Carser May 20 '14 at 3:51

You have two planes, $$\pi_1: X_1 = (1,0,0) + \lambda_1(0,1,1) + \gamma_1(1,2,1)$$ $$\pi_2: X_2 = (0,0,0) + \lambda_2(0,3,0) + \gamma_2(-2,-1,-1)$$ which you could also write as $$\pi_1: z_1 = -\frac{x_1}{2} + \frac{y_1}{2} + \frac{1}{2}$$ $$\pi_2: z_2 = \frac{x_2}{2}$$ You have already found the plane normals where (not unit normals) $$n_1 = \begin{bmatrix} -1 \\ 1 \\ -2 \end{bmatrix}, ~~~~~~~ n_2 = \begin{bmatrix} -3 \\ 0 \\ 6 \end{bmatrix}$$
The planes are not parallel ($\hat{n}_1 \ne \hat{n}_2$) and you can get a vector for this line which is $\hat{v} = \hat{n}_1 \times \hat{n}_2$, as you say. You can use any point that lies on the line of intersection, which you can find by determining an $(x,y,z)$ which satisfies both plane equations above.
For example: $$z_1 = z_2$$ $$-\frac{x}{2} + \frac{y}{2} + \frac{1}{2} = \frac{x}{2} \implies y=1$$ $$\vdots$$
EDIT: You want to find at least one point $A$ by finding $\lambda_1$, $\lambda_2$, $\gamma_1$, and $\gamma_2$ that satisfy $X_1 = X_2$. Once you have $A$, you can choose any $B = A + c v$. So, how to find this $A$? Can you find values to satisfy this: $$\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + \lambda_1 \begin{bmatrix} 0 \\ 1 \\ 1 \end{bmatrix} + \gamma_1 \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} + \lambda_2 \begin{bmatrix} 0 \\ 3 \\ 0 \end{bmatrix} + \gamma_2 \begin{bmatrix} -2 \\ -1 \\ -1 \end{bmatrix}$$ Hint: you should be able to choose $A = \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix}$
• Could you be more clear in how to find $z_1 = ...$? I apreciate so much if possible :). Also, do you have any hint to me, so I can find the points $A$ and $B$? I'm intersted in solving the two possible ways. – Marter Js May 20 '14 at 3:54