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Chapter 13: Problem 33

Find an equation of the plane parallel to the plane \(Q\) passing through the point \(P_{0}\). $$Q: 4 x+3 y-2 z=12 ; P_{0}(1,-1,3)$$

Short Answer

Expert verified
Answer: The equation of the parallel plane passing through point \(P_0(1, -1, 3)\) is: \(4x + 3y - 2z = -5\).

Step by step solution

01

Identify the normal vector

Since plane Q is given by the equation \(4x + 3y - 2z = 12\), the coefficients of x, y, and z are 4, 3, and -2, respectively. Thus, the normal vector (NV) for plane Q is: \(NV = \langle 4, 3, -2 \rangle\).
02

Substitute point P0 into the equation for the parallel plane

Since the desired plane is parallel to plane Q, it also has the same normal vector. So, the equation of the parallel plane passing through the point \(P_0(1, -1, 3)\) will be in the form: \(4x + 3y - 2z = d\). Now we need to find the value of d. Substitute the coordinates of point \(P_0\) into the equation: \(4(1) + 3(-1) - 2(3) = d \Rightarrow 4 - 3 - 6 = d \Rightarrow -5 = d\).
03

Write the equation of the parallel plane

Now that we have found the value of d, we can write the equation of the parallel plane using the normal vector and the constant term. The equation of the parallel plane passing through point \(P_0(1, -1, 3)\) is: $$4x + 3y - 2z = -5$$.

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