Find the general equation of the plane containing the point \(( 1,2,-1) \) if the vector \(\mathbf{N=\,}2\mathbf{i}+3\mathbf{j}-4 \mathbf{k}\) is normal to the plane. Then use the intercepts to graph the plane.

Solution The general equation of a plane is \(Ax+By+Cz=D,\) where \( A,\) \(B,\) and \(C\) are the components of a normal vector to the plane. Since the normal is \(\mathbf{N=\,}2\mathbf{i}+3\mathbf{j}-4\mathbf{k,}\) we have \[ 2x+3y-4z=D \]

To find \(D\), we use the point \((1,2,-1)\). Then \( D=2(1) +3(2) -4(-1) =12.\) The general equation of the plane is \[ 2x+3y-4z=12 \]

\(2x+3y-4z=12\)

We find the intercepts by letting two variables equal zero and then solving for the third variable. For example, to find the \(x\)-intercept, we let \(y=0\) and \(z=0.\) Then \(2x=12\) or \(x=6.\) Similarly, the \(y\)-intercept is \(4\) and the \(z\)-intercept is \(-3.\) Figure 54 illustrates how the intercepts are used to graph the plane.