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Chapter 13: Problem 14

Find an equation of the plane that passes through the point \(P_{0}\) with a normal vector \(\mathbf{n}\). $$P_{0}(1,2,-3) ; \mathbf{n}=\langle-1,4,-3\rangle$$

Short Answer

Expert verified
Question: Find the equation of a plane that passes through point P_0 (1, 2, -3) with a normal vector 饾惂 = <-1, 4, -3>. Answer: The equation of the plane is -1x + 4y - 3z = 16.

Step by step solution

01

Identify the Normal Vector and Point

The normal vector 饾惂 is given as <-1, 4, -3>, which means the coefficients A, B and C are: A = -1 B = 4 C = -3 The given point P_0 = (1, 2, -3), has the coordinates x_0, y_0, z_0: x_0 = 1 y_0 = 2 z_0 = -3
02

Obtain the Equation of the Plane

We know that the equation of a plane in 3D space can be given by: \(Ax + By + Cz = D\) Plugging in the values of A, B, and C, we get: \(-1x + 4y - 3z = D\) Now, we need to find D, and to do this we will use the point P_0.
03

Calculate the Value of D

Plug the given point P_0 (1, 2, -3) into the equation: \(-1(1) + 4(2) - 3(-3) = D\) Solving for D, we obtain: D = -1 + 8 + 9 = 16
04

Write the Final Equation

With the value of D, we can now write the final equation of the plane: \(-1x + 4y - 3z = 16\) This is the equation of the plane that passes through the point (1, 2, -3) with a normal vector of <-1, 4, -3>.

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