Find the parametric vector and Cartesian equations for the following planes:

a. The plane thru point $(2,1,-2)$ perpendicular to vector $(-1,1,2)$.

b. The plane thru the three points $(2,2,-2)$, $(-1,1,2)$ and $(2,3,1)$.

Please help. Studying for an exam and I attempted this problem several times, but my answers do not match the answer doc. Thank you.

  • $\begingroup$ What did you try, which level are you, what are you suppose to know already? Please give information in order for us to help you UNDERSTAND, and not give you answers that you will just copy/paste and not gain anything by it. $\endgroup$ – Martigan Jun 10 '15 at 7:24
  • $\begingroup$ I'm in Year 3 Calculus. I tried to plug the numbers into an equation given by my lecturer where x(t) = a + lamda (b-a) +mu (c-a) and tried to modify it since for (a) I was only given the point and vector. Am I using the wrong equation? I would like to know the equations you think need to be used. $\endgroup$ – user239904 Jun 10 '15 at 7:32
  • $\begingroup$ Welcome to math.SE! As a heads up, this site is not a solution manual, instead it emphasizes collaboration and human interaction. :) $\endgroup$ – Alp Uzman Jun 10 '15 at 7:32


Let $P=(x,y,z)$ a generic point on the plane and $P_0=(x_0,y_0,z_0)$ the point through it passes, than a vector parallel to the plane is $P-P_0=(x-x_0,y-y_0,z-z_0)^T$. This vector is orthogonal to a ''normal'' $\vec n=(a_n,a_n,a_n)^T$ if the dot product between them is null: $$ (P-P_0) \cdot \vec n=0 \iff (x-x_0,y-y_0,z-z_0)(x_n,y_n,z_n)^T=0 $$ Calculate the dot product and you have the equation of the plane.

For a plane through three given points you can simultaneously solve the three equation obtained by the general equation of a plane $ ax+by+cz+d=0$ whan substitute the coordinate of the three points or $(x,y,z)$.

You can see here.

  • $\begingroup$ Hello. Thank you for your help. Will this equation produce a vector output? My professor gave us the answer for (a): (2,1,-2) +s(-2,2,2) +t(3,3,0). First, all the equations I've tried to use do not give me vector outputs like the answer. I thought that because the vector in (a) was perpendicular to the point, it would equal n⃗ but I don't understand the steps from that point. $\endgroup$ – user239904 Jun 10 '15 at 7:39

Fiest be aware that you can define a plan by different means.

One of them is through parametric equations such as

$(1)$ $x=a_1+\lambda u_1 +\mu v_1$, $y=a_2+\lambda u_2 +\mu v_2$ and $y=a_3+\lambda u_3 +\mu v_3$ with $(a_1,a_2,a_3)$ a point of the plane and $(u_1,u_2,u_3)$, $(v_1,v_2,v_3)$ two non colinear vectors.

An other one is through a cartesian equation such as

$(2)$ $ax+by+cz+d=0$

For your question $(a)$, your professor gave you a parametric equation, which is different from the statement of your problem (give the cartesian equation).

In any case, let's find a parametric equation for the problem $(a)$.

Since you already have a point of the plane $(2,1,-2)$, you just need to find two non-colinear vectors of the plane. Each of them should be perpendicular to the normal vector, that is, the dot product should be $0$.

One of the vector could be very easily $(1,1,0)$, whom dot product with $(1,-1,2)$ is clearly $0$.

The second one could be $(0,2,1)$. Since it is not colinear with $(1,1,0)$, this works.

Hence an equation $(2,1,-2)+\lambda (1,1,0)+ \mu (0,2,1)$

This is different from the answer of your professor, but this is OK, there are an infinity of possible parametric equations for the same plan. Bear in mind that a method to find the last vector, instead of finding our one fitting one, could be to find a vector which is normal to both the vector normal to the plane and the one you already found, with the vector cross product. It would be in that case $(1,-1,2) \times (1,1,0)$. It is probably what your professor did.

In the second case, you can take the first point $A$ as your defining point, and the vectors $AB$ and $AC$ as vectors for the parametrization of the plane.

Or you can also (for cartesian equation) find out the three unknowns by resolving this set of equations:

$ax_1+by_1+cz_1+1=0$, $ax_2+by_2+cz_2+1=0$, $ax_3+by_3+cz_3+1=0$


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