MHB E.12.4 - Find plane through given point and normal to given vector

[karush]

$\textsf{write an equation for}$

$\textsf{ The plane through the point (3, 2, -5) and perpendicular to the x-axis}$
4ok I know this goes thru $3$ on the axis and it is \parallel to the $yz$ plane

so is it just $x=3$.

 

[MarkFL]

Re: E.12.4

If the plane is perpendicular to the line, the normal vector of the plane is equal to the direction vector of the line (convince yourself of this). A plane equation with normal vector $$\langle a,b,c\rangle$$ and passing through the point $\left(x_0,y_0,z_0\right)$ is:

$$a\left(x-x_0\right)+b\left(y-y_0\right)+c\left(z-z_0\right)=0$$

The family of vectors parallel to the $x$-axis is:

$$k\langle 1,0,0 \rangle$$ where $0\ne k$

And so the plane through the point $\left(x_0,y_0,z_0\right)$ and normal to the $x$-axis is then:

$$k\left(x-x_0\right)+0\left(y-y_0\right)+0\left(z-z_0\right)=0$$

$$k\left(x-x_0\right)=0$$

$$x=x_0$$

We are given:

$$x_0=3$$

And so the plane is:

$$x=3\quad\checkmark$$

 

[HallsofIvy]

Re: E.12.4

If the problem had said "in the yz-plane" then you would have known that x= 0, by definition of "yz-plane". The fact that it is parallel to yz-plane means that every point has the same distance from the yz-plane- and that distance is the xcoordinate.