Question

find an equation for the plane through point ( p_0 = (5, -2, 1) ) with normal vector ( mathbf{n} = langle -2, -1, 3
angle ):

Explanation:

Step1: Recall the plane equation formula

The general equation of a plane with normal vector \(\mathbf{n}=\langle A,B,C
angle\) passing through point \(P_0=(x_0,y_0,z_0)\) is \(A(x - x_0)+B(y - y_0)+C(z - z_0)=0\).

Step2: Substitute the given values

Here, \(A = - 2\), \(B=-1\), \(C = 3\), \(x_0 = 5\), \(y_0=-2\), \(z_0 = 1\). Substitute these into the formula:
\[
-2(x - 5)-1(y - (-2))+3(z - 1)=0
\]

Step3: Simplify the equation

First, simplify the signs:
\[
-2(x - 5)-1(y + 2)+3(z - 1)=0
\]
Then, distribute the coefficients:
\[
-2x+10 - y - 2+3z - 3=0
\]
Combine like terms:
\[
-2x - y+3z+(10 - 2 - 3)=0
\]
\[
-2x - y+3z + 5=0
\]
We can also multiply both sides by - 1 to get \(2x + y-3z - 5=0\) (either form is correct, but let's present the simplified distributed form).

Answer:

\(-2x - y + 3z+5 = 0\) (or \(2x + y-3z - 5 = 0\))