Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 5 down vote accepted

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

share|cite|improve this answer
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?

share|cite|improve this answer

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

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.