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Chapter 51: Chapter 51

Problem 893

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Find the equation of the plane passing through the point \((4,-1,1)\) and parallel to the plane \(4 x-2 y+3 z-5=0\)

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The equation of the plane passing through the point \((4,-1,1)\) and parallel to the given plane is \[4x - 2y + 3z - 21 = 0.\]
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TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Find the normal vector

First, we identify the normal vector of the given plane, which is \((4,-2,3)\). Since our required plane is parallel to this given plane, it will have the same normal vector.

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

Recall that the point-normal form of a plane is given by \(a(x-x_0) + b(y-y_0) + c(z-z_0) = 0\), where \((a,b,c)\) is the normal vector of the plane and \((x_0,y_0,z_0)\) is a point on the plane. Now substitute the given point \((4,-1,1)\) and the normal vector \((4,-2,3)\) into the equation: \(4(x-4) - 2(y+1) + 3(z-1) = 0\)

Step 3: Simplify the equation

Simplify the equation to get the final answer: $4(x-4) - 2(y+1) + 3(z-1) = 0 \\ 4x - 16 - 2y - 2 + 3z - 3 = 0\) So, the equation of the plane is: \(4x - 2y + 3z - 21 = 0\)

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