Problem 11 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Answer: The equation of the plane is x + y - z - 4 = 0.

Step 1: Identify the components of the normal vector and a point on the plane

The normal vector is given to be: $$\mathbf{n}=\langle 1,1,-1\rangle$$ Thus, \(A=1\), \(B=1\), and \(C=-1\). The given point on the plane is: $$P_{0}(0,2,-2)$$ Thus, \(x_{0}=0\), \(y_{0}=2\), and \(z_{0}=-2\).

Step 2: Substitute the components of the normal vector and the point into the point-normal form

Using the point-normal form equation, $$A(x-x_{0})+B(y-y_{0})+C(z-z_{0})=0$$, we substitute the components of the normal vector and the point. $$1(x-0)+1(y-2)-1(z-(-2))=0$$

Step 3: Simplify the equation

Now, let's simplify the equation: $$x+(y-2)-(z+2)=0$$ $$x+y-2-z-2=0$$ $$x+y-z-4=0$$

Step 4: Write the final equation

So, the equation of the plane that passes through the point \(P_{0}(0,2,-2)\) with a normal vector \(\mathbf{n}=\langle 1,1,-1\rangle\) is: $$x+y-z-4=0$$