Question

Find the equation of the plane through the point (1,2,3) and
the line l: x = 2+t, y = 2 + 2t, z= -3t.​

Answer 1

Coordinates Geometry (3D) - Plane

Solution:

The given straight line is

$\quad l:x=2+t,\:y=2+2t,\:z=-3t$

$\Rightarrow \frac{x-2}{1}=\frac{y-2}{2}=\frac{z}{-3}=t$

Let the equation of the plane through the above straight line be

$\quad A(x-2)+B(y-2)+Cz=0$ .....(1)

Since the given straight line lies on plane (1), we write

$\quad A+2B-3C=0$ .....(2)

Again the plane (1) passes through the point (1, 2, 3). Then

$\quad -A+0.B+3C=0$ .....(3)

Now eliminating $A$, $B$ and $C$ from (1), (2), (3), we get

$\quad$$\left|\begin{array}{ccc}x-2&y-2&z\\1&2&-3\\-1&0&3\end{array}\right|=0$

$\Rightarrow (x-2)(6-0)-(y-2)(3-3)+z(0+2)=0$

$\Rightarrow 6x-12+2z=0$

$\Rightarrow 6x+2z=12$

$\Rightarrow 3x+z=6$

This is the equation of the required plane.