How to find the plane which contains a point and a line

Question

I know that $\Pi$ contains the point $(2,0,5)$ and the line $\frac{x-10}3 = \frac{y-3}2 = \frac{z-7}2$. How would I find the minimum vector connecting the point and the line so I can then work out the cross product and thus the normal and equation of plane $\Pi$.

Answer 1

The point $A = (2,0,5)$ is in the plane,

If we set $x = 10$ into $\frac{x-10}3 = \frac{y-3}2 = \frac{z-7}2$ we get $B = (10,3,7)$ is in the plane.

If we set $x = 13$ into $\frac{x-10}3 = \frac{y-3}2 = \frac{z-7}2$ we get $C = (13,5,9)$ is in the plane.

$\quad \vec u =B - A = (8,3,2)$

$\quad \vec v =C - A = (11,5,4)$

Taking $\vec u \times \vec v$ we get that the equation has the form $-2x + 10y -7z = d$.

Plugging in the point $A$ we get the $d = -39$, so the answer is

$$\tag 1 -2x + 10y -7z = -39$$

Answer 2

Take a point on the line say$$P_0(10,3,7)$$ then we get one direction vector of our plane $$\vec{a}=\vec{P_0P}$$. Another vector is the direction vector of our line $$\vec{b}=[3,2,2]$$ so our plane is given by $$\vec{x}=\vec{OP_0}+s\vec{P_0P}+t[3;2;2]$$ where $s$ and $t$ are reals.

Answer 3

$(3,2,2)$ is a direction vector of the line and hence is parallel to the plane.

$(2,0,5)$ and $(10,3,7)$ are two points on the plane. So, $(10,3,7)-(2,0,5)=(8,3,2)$ is a vector parallel to the plane.

$(8,3,2)\times(3,2,2)=(2,-10,7)$ is a normal vector to the plane.

The equation of the plane is $2(x-2)-10(y-0)+7(z-5)=0$, i.e., $2x-10+7z-39=0$.

Answer 4

Other answers show that you don’t need the nearest point on the line in order to compute the normal of the plane that you’re supposed to find. However, it’s easy enough to compute this point via a plane-line intersection: the shortest distance between a point $P$ and a line is measured perpendicularly to the line, so the nearest point on the line to $P$ is the intersection of the line with the plane through $P$ that is perpendicular to the line.

As noted elsewhere, a direction vector of your line is $\mathbf n=(3,2,2)$, so an equation of the perpendicular plane through $P=(2,0,5)$ is $\mathbf n\cdot\mathbf x=\mathbf n\cdot P$, or $3x+2y+2z=16$. Solving the resulting system of equations produces $(4,-1,3)$.