Problem 34 - Calculus Early Transcendentals

Question

Find an equation of the plane parallel to the plane \(Q\) passing through the point \(P_{0}\). $$Q: x-5 y-2 z=1 ; P_{0}(1,2,0)$$

Step-by-Step Solution

Answer

Question: Find the equation of a plane that is parallel to the plane Q: (x - 5y - 2z = 1) and passes through the point P₀(1,2,0). Answer: The equation of the plane parallel to plane Q and passing through point P₀(1,2,0) is x - 5y - 2z = -9.

Step 1: Find the normal vector of plane Q

The coefficients of the variables in the equation of plane Q, \((x - 5y - 2z = 1)\), are the components of the normal vector. So, the normal vector \(\vec{n}\) of plane Q is: $$\vec{n} = \langle 1, -5, -2 \rangle$$

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

The point-normal form of the equation of a plane is given by: $$a(x - x_0) + b(y - y_0) + c(z - z_0) = 0$$ Here, \((a, b, c)\) is the normal vector \(\vec{n}\), and \((x_0, y_0, z_0)\) is the given point \(P_0\).

Step 3: Plug in the normal vector and point P_0

Now, plug in the coordinates of the normal vector \((1, -5, -2)\) and point \(P_0 (1,2,0)\) into the point-normal form equation: $$(1)(x - 1) - 5(y - 2) - 2(z - 0) = 0$$

Step 4: Simplify the equation

Finally, simplify the equation of the plane: $$x - 1 - 5y + 10 - 2z = 0$$ Combining constants and rearranging terms, we get: $$x - 5y - 2z = -9$$ This is the equation of the plane parallel to plane Q and passing through point \(P_0\).