# Find equation of the plane through the origin with basis <1,2,-1> and <2,3,4>.

Find an equation of the plane through the origin with basis <1,2,-1> and <2,3,4>.

Could I get some advice on how to work this problem? I know how to find the basis given some plane, but not the other way around.

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Hint: Compute the cross product of the two vectors. This vector will be orthogonal to the plane and from there you get the equation.

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Okay,so the cross product would be 11i-6j-k? So is this the actual equation I'm looking for? –  briteId Jul 16 '13 at 0:36
Yes, that's the cross product. No, that's not an equation. Given a normal vector $ai+bj+ck$ to a plane $P$ through the origin, an equation for $P$ is given by $ax+by+cz=0$. –  ՃՃՃ Jul 16 '13 at 0:44
Okay, I'm sorry this is not quicker for me to pick up. I appreciate your comments. So, what should I do now with the cross product? Is that the same thing as my "normal" vector? If so, would I just use 11x-6y-z? I think I'm getting confused on how to get from the cross product to the normal vector. Thanks! –  briteId Jul 16 '13 at 1:48
Yes, the vector you get when you compute the cross product is a normal vector to the plane. Since you got $11i-6j-k$ as normal vector, an equation for the plane is $11x-6y-z=0$. And don't worry, we are all learning. :) –  ՃՃՃ Jul 16 '13 at 1:58
Thanks so much! –  briteId Jul 16 '13 at 2:21

Alternative hint $\renewcommand{\vec}[1]{\mathbf{#1}}$

The vector form for the equation of the plane is $$\begin{bmatrix}x\\y\\z\end{bmatrix}=s\begin{bmatrix}1\\2\\-1\end{bmatrix}+t\begin{bmatrix}2\\3\\-4\end{bmatrix},$$ where $s,t\in{\mathbb R}$.

Given an arbitrary point $P=[x,y,z]^T$ on the plane, there must be an $s$ and $t$ so that $P$ can be expressed using the above equation; i.e. $\vec{p}=s\:\vec{a}+t\:\vec{b}$ is a consistent linear system. What conditions must there be on $P$ for this to hold?

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Or you can use determinates for quick way : $$\begin{vmatrix} x & y & z &1 \\ 1&2 &-1 &1 \\ 2&3 & 4 & 1\\ 0 &0 &0 &1 \end{vmatrix} =0$$

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