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Chapter 13: Problem 31

Find an equation of the plane parallel to the plane \(Q\) passing through the point \(P_{0}\). $$Q:-x+2 y-4 z=1 ; P_{0}(1,0,4)$$

Short Answer

Expert verified
Question: Find the equation of a plane that is parallel to the plane Q: -x + 2y - 4z = 1 and passes through the point P0(1, 0, 4). Answer: The equation of the parallel plane is -x + 2y - 4z = -17.

Step by step solution

01

Determine the normal vector of plane Q

Given the equation of plane Q: -x + 2y - 4z = 1, the coefficients of x, y, and z form the normal vector. So normal vector n = <-1, 2, -4>.
02

Use the point P0 to find the constant in the new plane's equation

P0(1, 0, 4) is a point on the new plane, which means we can plug the coordinates of this point into the equation and solve for the constant. Using the normal vector we just found, the equation of the new plane should have the form: $$-x + 2y - 4z = constant$$ Plugging P0 into the equation: $$-1(1) + 2(0) - 4(4) = constant$$ Solve for the constant: $$-1 - 16 = -17$$ So the constant is -17.
03

Write the equation of the parallel plane

Now that we have found the constant and have the normal vector, we can write the equation of the parallel plane. The equation will be the same as the equation of plane Q, but with a different constant. Therefore, the equation of the parallel plane is: $$-x + 2y - 4z = -17$$

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