Problem 13 - Calculus Early Transcendentals

Question

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

Step-by-Step Solution

Answer

Question: Determine the equation of the plane that passes through the point \(P_{0}(2, 3, 0)\) and has a normal vector \(\mathbf{n} = \langle -1, 2, -3 \rangle\).

Step 1: Write down the given information

We are given the point \(P_{0}(2, 3, 0)\) and the normal vector \(\mathbf{n} = \langle -1, 2, -3 \rangle\).

Step 2: Write down the general form of the equation

We have the general equation of a plane, which is \(Ax + By + Cz = D\).

Step 3: Plug in the coefficients

From the normal vector \(\mathbf{n} = \langle -1, 2, -3 \rangle\), we can say that \(A = -1\), \(B = 2\), and \(C = -3\). Therefore, the equation becomes \(-x + 2y - 3z = D\).

Step 4: Substitute the coordinates of the point \(P_{0}\) to find D

We know that the point \(P_{0}(2, 3, 0)\) lies on the plane; thus, it must satisfy the equation. Substitute its coordinates into the equation as x, y, and z: \(-2 + 2(3) - 3(0) = D\). Solve for D: \(-2 + 6 = 4\).

Step 5: Write the final equation

Now that we have found D, we can write the complete equation of the plane: \(-x + 2y - 3z = 4\).