Problem 12 - Calculus Early Transcendentals

Question

Find an equation of the plane that passes through the point \(P_{0}\) with a normal vector \(\mathbf{n}\). $$P_{0}(1,0,-3) ; \mathbf{n}=\langle 1,-1,2\rangle$$

Step-by-Step Solution

Answer

The equation of the plane is x - y + 2z + 5 = 0.

Step 1: Identify the coefficients of the normal vector

From the given normal vector \(\mathbf{n}=\langle 1,-1,2\rangle\), we can identify the coefficients as \(a=1, b=-1\), and \(c=2\). These coefficients will be used in the equation of the plane.

Step 2: Substitute the point and coefficients into the equation of a plane

We are given the point \(P_{0}(1,0,-3)\). Substitute the coordinates of the point and the coefficients obtained from the normal vector into the equation of the plane: \(ax + by + cz = d\). Therefore we have: \(1(x-1)-(y-0)+2(z+3)=0\)

Step 3: Simplify the equation

Now, let's simplify the equation: \(x-1-y+2z+6=0\) Combining all the terms, we get: \(x-y+2z+5=0\)

Step 4: Write the final answer

Finally, the equation of the plane that passes through the point \(P_{0}(1,0,-3)\) with a normal vector \(\mathbf{n}=\langle 1,-1,2\rangle\) is: $$x-y+2z+5=0$$