Answer:
The equation of the plane is [tex]-4x+z=0[/tex].
Step-by-step explanation:
An equation of the plane containing the point [tex](x_0,y_0,z_0)[/tex] with normal vector [tex]\vec{N}=\langle A,B,C \rangle[/tex] is
[tex]A(x-x_0)+B(y-y_0)+C(z-z_0)=0[/tex]
The equation of any plane can be expressed as
[tex]ax+by+cz=d[/tex]
This is called the standard form of the equation of a plane.
Given the points
[tex]P=(1,0,4)[/tex]
[tex]Q=(-1,1,-4)[/tex]
[tex]R=(0,0,0)[/tex]
Two vectors in the plane are
[tex]\overrightarrow{PQ}=\begin{pmatrix}-1&1&-4\end{pmatrix}-\begin{pmatrix}1&0&4\end{pmatrix}=\begin{pmatrix}-2&1&-8\end{pmatrix}\\\\\overrightarrow{PR}=\begin{pmatrix}0&0&0\end{pmatrix}-\begin{pmatrix}1&0&4\end{pmatrix}=\begin{pmatrix}-1&0&-4\end{pmatrix}[/tex]
A normal vector to the plane is
[tex]\overrightarrow{PQ}\times \overrightarrow{PR}=\begin{pmatrix}-2&1&-8\end{pmatrix}\times \begin{pmatrix}-1&0&-4\end{pmatrix}[/tex]
[tex]\left(u_1,\:u_2,\:u_3\right)\times \left(v_1,\:v_2,\:v_3\right)=\left(u_2v_3-u_3v_2,\:u_3v_1-u_1v_3,\:u_1v_2-u_2v_1\right)\\\\\begin{pmatrix}1\cdot \left(-4\right)-\left(-8\cdot \:0\right)&-8\left(-1\right)-\left(-2\left(-4\right)\right)&-2\cdot \:0-1\cdot \left(-1\right)\end{pmatrix}\\\\\begin{pmatrix}-4&0&1\end{pmatrix}[/tex]
Thus, an equation of the plane is
[tex]-4(x-1)+0(y-0)+1(z-4)=0\\-4x+z=0[/tex]
We can check our results with the graph of the plane.
