Finding plane equation by using parametric equation

Question

Find the plane equation passing through $(4,-2,6)$

$x=3-2t$, $y=t$ and $z=5+2t$

How was the solution of these type of questions..Thanks

Answer 1

Let $P(4,-2,6)$ and line $AB$: $X_\ell=A+t\,(B-A)=(3,0,5)+t\,(-2,1,2)$, then the plane equation will be $n\cdot (X-A)=0$ where $n=[(P-A)\times(B-A)]$ and $X=(x,y,z)$ an arbitrary point on the plane.
As the mixed product corresponds to determinant, we have $$\left| \begin{array}{ccc} x-3&y&z-5\\ 4-3&-2-0&6-5\\ -2&1&2 \end{array} \right|=0$$ $$-5 x - 4 y - 3 z + 30=0$$ We can test now that $P$ and all the line $AB$ lies in the plane: WA results for the point and the line although it's not needed.

Answer 2

We are given one point in the plane $(4,-2,6)$. Two other points can be obtained by taking parameter $t=0$ and $t=1$ on the line: $(3,0,5)$ and $(1,1,7)$. Three non-collinear points determine a plane. One way to find it is to take the cross product of two difference vectors: $(1,-1,1)\times(3,-3,-1)=(5,4,3)$ is a normal vector to the plane. Can you take it from here?

Answer 3

The generic equation of a plane is $\pi : ax + by + cz + d = 0$. This generic plane $\pi$ is parallel to the plane $ W_{\pi} : ax + by + cz = 0 $.

You can write $W_{\pi}$ as $Span\{(1,-2, 1), (-2, 1, 2)\}$, where $(1,-2, 1)$ is the vector $\vec{PQ}$, where $P(4, -2, 6)$ is the given point and $Q(3, 0, 5)$ is a point in the given line (obtained by taking $t = 0$), and $(-2,1,2)$ is the direction of the line.

We can write $Span\{(1,-2, 1), (-2, 1, 2)\}$ as:

$$\begin{cases} x = t - 2s \\ y = -2t + s \\ z = t + 2s \end{cases} $$

after some steps you'll find the cartesian equation $5x+4y+3z = 0$, which corresponds to $ax+by+cz = 0$. To find $d$ we solve $5x+4y+3z + d = 0$ substituting to $x$, $y$ and $z$ the coordinates of $P$. This way you can find the equation $5x+4y+3z -30 = 0$ of the plane $\pi$, parallel to $W_{\pi}$ and containing the point $P$.