I am given two vectors u = (3,1,0) and v = (3,0,1).

I am asked to find an equation for the plane containing u and v.

I have applied cross product to give me the vector perpendicular to both u and v and I want to use this as the point to substitute into my equation with either u or v to ultimately give me (a,b,c)⋅($x$-$x_0$,$y$-$y_0$,$z$-$z_0$)=$0$

Am I correct in finding my ($x$,$y$,$z$) point to substitute in by finding the cross product of both given vectors?


  • 2
    $\begingroup$ Well, you get that the equation of the plane is $(1,-3,-3)\cdot (x-x_0,y-y_0,z-z_0)=0.$ That is, $x-3y-3z+D=0.$ You need to know a point of the plane to determine the plane. Unless it is assumed that the plane contains the origin (that is, it is a vector subspace). In such a case, the equation of the plane is $x-3y-3z=0.$ $\endgroup$ – mfl Sep 9 '15 at 20:56
  • $\begingroup$ @mfl I am only given two vectors in this question. I am not given a point specifically. So perhaps I am meant to assume the plane contains the origin. This is where I got a bit lost as I know the equation requires a point. So in this instance I would presume then that it is in fact (0, 0, 0) $\endgroup$ – Metamorphosis Sep 9 '15 at 21:00
  • $\begingroup$ @mfl I don't think one can say for certain without the wording of the problem as written, but I suspect that it is intended that the plane contains the origin, as it's supposed to contain the given vectors, rather than the corresponding points. $\endgroup$ – Travis Willse Sep 9 '15 at 21:00

For any plane $P$ with equation $ax + by + cz = d$, the vector $(a, b, c)$ is perpendicular to $P$.

  • $\begingroup$ So then it would be correct to use the cross product to find my vector perpendicular to both u and v is enough and assume the origin 0? $\endgroup$ – Metamorphosis Sep 9 '15 at 21:04
  • 1
    $\begingroup$ Yeah, if the cross product is $(a, b, c)$, then your plane (passing through the origin) will be $ax + by + cz = 0$ $\endgroup$ – brick Sep 9 '15 at 21:06

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