What's the parametric equation for the plane through a point (x,y,z) perpendicular to (a,b,c)?

Question

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.

Answer 1

Hint:

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.

Answer 2

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:

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