3D Equations of Planes

Question

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So the above question is from a Calc III class I'm taking but I'm not sure I understand the solution...
$n_1$ represents the vector normal to the given plane. Therefore taking the cross product of $n_1$ and the vector $P_1P_2$ would output a vector that is parallel to the given plane, not orthogonal as desired. Unless I'm mistaken... Anyone care to elaborate? Are both solutions correct?

Here was my attempted solution:

Let $r$ be the vector $<3,-1,-1>$ and let $M_1$ and $M_2$ be arbitrary points satisfying the equation of the plane given. I chose $M_1 = (3,10,0)$ and $M_2 = (2,6,0)$ therefore let $m$ be the corresponding vector $m = <-1,-4,0>$ then $$r \times m = <-4,1,-13>$$ which gives a final equation of $$ -4(x+2) + y-1 -13(z-4)$$

$m$ is the vector from $P_1$ to $P_2$ lying in the plane given — Anthony Peter, Jul 15 '13 at 04:39
But $P_1$ is $(-2,1,4)$ and $P_2$ is $(1,0,3)$ from the problem statement. Why do you use different points? — Daryl, Jul 15 '13 at 04:42
I should rename them $M_1$, $M_2$ there isn't a correlation between the two. — Anthony Peter, Jul 15 '13 at 04:44
Ok. I understand now. Your new plane must be satisfied by the two points $P_1$ and $P_2$, not arbitrary points on the first plane. With the information givven, of the two vectors which span the new plane, one must be orthogonal to the original plane and the second must be parallel to the vector joining $P_1$ and $P_2$. — Daryl, Jul 15 '13 at 04:50
you're saying that we need to find a vector $n$ such that $$n \times n_1 =0$$ and $$n \times r =0$$ — Anthony Peter, Jul 15 '13 at 05:01

score 1 · Accepted Answer · answered Jul 15 '13 at 05:08

We need to find the plane which is perpendicular to the plane $4x-y+3z=2$ and passing through the points $P_1(-2,1,4)$ and $P_2(1,0,3)$.

The plane can be written as $n\cdot(r-p)=0$ where $n$ is orthogonal to the plane and $p$ is a point on the plane. Given that we know $P_1$ and $P_2$ lie on the plane, we can choose $p$ as either of these points. Arbitrarily choose $p=P_2=(1,0,3)$ (it is just as correct to choose $p=P_1$). We also have $r=(x,y,z)$ as an arbitrary point on the plane.

The plane $4x-y+3z=2$ can also be expressed as $(4,-1,3)\cdot(r-q)=0$ for some point $q$ on the plane. Hence, a normal to the first plane is $n_o=(4,-1,3)$. The requirement now is that $n$ is orthogonal to $n_o$, since $n_o$ is parallel to the given plane, which is orthogonal to the required plane.

Also, since $P_1$ and $P_2$ lie in the plane, the vector joining the two points is parallel to the required plane. Hence, the vector $a=(3,-1,-1)$ is also parallel to the required plane.

This leads to $n$ needing to be orthogonal to $n_o$ and $a$. Thus $n=n_0\times a=(4,13,-1)$.

The required plane, is then $4(x-1)+13(y-0)-(z-3)=0$ or $4x+13y-z=1$.

Oh. I wasn't visualizing that $r$ lies in the plane and therefore we want to find it's orthogonal counterpart. thank you — Anthony Peter, Jul 15 '13 at 05:18

3D Equations of Planes

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