Plane in 3 Dimensions

Question

I just learned this topic and I'm having trouble with this homework problem...

Find an equation of the plane through the three points given:

$P = (0, 2, 0)$

$Q = (-4, 6, 2)$

$R = (3, 3, -1)$

The equation must $= 4$

Any help is appreciated! :)

Answer 1

HINT

The general equation of a plane is $ax+by+cz=d$, where $a,b,c,d$ are some numbers, not all zero. You have been told that $d=4$ and so your equation is $ax+by+cz=4$ where you need to find the numbers $a,b,c$.

We know that $(x,y,z)=(0,2,0)$ lies on the plane, so if we substitute $x=0$, $y=2$ and $z=0$ then the equation must hold. If we do that we get $a\times 0 + b \times 2 + c \times 0 =4$, i.e. $b=2$.

You now that $b=2$ and $d=4$. Substitute the $x,$ $y$ and $z$ values of the other two points into $ax+2y+cz=4$ and solve the resulting equations simultaneously to find the numbers $a$ and $c$.

Answer 2

HINT

The formula for a plane $\pi$ is $\pi:A(x-x_{0})+B(y-y_0)+C(z-z_0)=0$ where vector $[A,B,C]\perp \pi$ and point $(x_0,y_0,z_0)\in\pi$

One of the possible solutions would be to take two vectors: $ \vec{PQ}$ and $\vec{PR}$ and find a normal vector of our $\pi$ plane.

To do this, simply take the cross product of vectors $\vec{PQ}$ and $\vec{PR}$ which equals to the $[A,B,C]$ vector in our formula: $\vec{PQ}\times\vec{PR}=[A,B,C]$

Choose one of the points, it may be $P$ and put all the other numbers into the formula.