Find Equation of Plane Perpendicular to Line $l(t)$

[evinda]

Hello! (Wave)

If we want to find the equation of the plane that is perpendicular to the line $l(t)=(-1,-2,3)t+(0,7,1)$ and passes through $(2,4,-1)$.

If (x,y,z) is a point of the line then it is of the form (-t,-2t+7,3t+1).

We get that $t=-x=\frac{7-y}{2}=\frac{z-1}{3} \Rightarrow -2x=7-y=\frac{2z-2}{3} \Rightarrow -6x=21-3y=2z-2 $

Can we find in that way the equation of the plane?

Also, is the following way also right to find the desired equation?

For $t=1$ , $a=(-1,5,4)$ where $a \in l(t)$.

Let $(x,y,z)$ be a point of the plane.

Then $(x-2,y-4,z+1)$ belongs to the plane.

Since $l(t)$ is perpendicular to the plane, $a$ is also perpendicular to the plane and since $(x,y,z)$ is a point of the plane, $a$ is perpendicular to $(x,y,z)$. So $a \cdot (x,y,z)=0 \Rightarrow -x+5y+4z=14$

 

[Evgeny.Makarov]

For $t=1$ , $a=(-1,5,4)$ where $a \in l(t)$.

Let $(x,y,z)$ be a point of the plane.

Then $(x-2,y-4,z+1)$ belongs to the plane.

Why?

Since $l(t)$ is perpendicular to the plane, $a$ is also perpendicular to the plane

If you write $a \in l(t)$, it means that $a$ is a point, and a point cannot be perpendicular to anything.

since $(x,y,z)$ is a point of the plane, $a$ is perpendicular to $(x,y,z)$.

Again, points can't be perpendicular.

It is known that the equation of a plane is $Ax+By+Cz+D=0$ where $(A,B,C)$ are the coordinates of a vector perpendicular to the plane. In your case, this vector is $(−1,−2,3)$. It is left to substitute the coordinates of a point on the plane into the equation to find $D$.