Problem 32 - Calculus Early Transcendentals

Question

Find an equation of the plane parallel to the plane \(Q\) passing through the point \(P_{0}\). $$Q: 2 x+y-z=1 ; P_{0}(0,2,-2)$$

Step-by-Step Solution

Answer

Question: Find the equation of the plane parallel to the plane Q: 2x + y - z = 3 and passing through the point P₀ = (0, 2, -2). Answer: The equation of the plane parallel to plane Q and passing through point P₀ is 2x + (y - 2) + (z + 2) = 0.

Step 1: Find the normal vector of the given plane

The normal vector of a plane can be found by looking at the coefficients of the variables in the given equation. For the plane Q, the normal vector is: $$\vec{n} = (2, 1, -1)$$

Step 2: Write the point-normal form of the equation of a plane

The general point-normal form of the equation of a plane is: $$a(x - x_{0}) + b(y - y_{0}) + c(z - z_{0}) = 0$$ Where (a, b, c) is the normal vector and (x0, y0, z0) is a point on the plane. Substitute the normal vector we found in step 1 and the given point P₀ = (0, 2, -2): $$2(x - 0) + 1(y - 2) - 1(z - (-2)) = 0$$

Step 3: Simplify the equation

After simplifying the equation in step 2, we get the equation of the plane parallel to plane Q and passing through P₀: $$2x + (y - 2) + (z + 2) = 0$$