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High School Math Calculus PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

The general form of the equation of the plane passes through ( 1 , − 2 , 4 ) and ( 4 , 0 , − 1 ) and is perpendicular to the x z -plane.

The general form of the equation of the plane passes through ( 1 , − 2 , 4 ) and ( 4 , 0 , − 1 ) and is perpendicular to the x z -plane.

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

8th Edition

ISBN: 9781337904285

Author: Larson

Publisher: CENGAGE L

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Question

Chapter 10.4, Problem 34E

To determine

To find: The general form of the equation of the plane passes through (1,−2,4) and (4,0,−1) and is perpendicular to the xz -plane.

Expert Solution & Answer

Answer to Problem 34E

The general form of the equation of the plane passes through (1,−2,4) and (4,0,−1) and is perpendicular to the xz -plane is −5x−3z+17=0 .

Explanation of Solution

Given information:

The plane passes through (1,−2,4) and (4,0,−1) and it is perpendicular to the xz -plane.

Calculation:

Since the plane is perpendicular to the xz -plane then the normal vector to the plane will be parallel to the plane that can be obtained by finding the cross product of vectors i and k as shown below.

v=|ijk100001|=i(0−0)−j(1−0)+k(0−0)=−j=〈0,−1,0〉

Calculate the second vector u as follows.

u=〈4−1,0+2,−1−4〉=〈3,2,−5〉v

Calculate the cross product of the vector u and to find the normal vector on the plane.

u×v=|ijk32−50−10|=−5i−3k=〈a,b,c〉

Substitute −5 for a , 0 for b , −3 for c , 1 for x1 , −2 for y1 and 4 for z1 in the equation a(x−x1)+b(y−y1)+c(z−z1)=0 to find the general form of the equation of the plane passes through (1,−2,4) and (4,0,−1) and perpendicular to the xz -plane.

−5(x−1)+0(y+2)−3(z−4)=0−5(x−1)−3(z−4)=0−5x+5−3z+12=0−5x−3z+17=0

Therefore, the general form of the equation of the plane passes through (1,−2,4) and (4,0,−1) and perpendicular to the xz -plane is −5x−3z+17=0 .

Chapter 10.4, Problem 33E

Chapter 10.4, Problem 35E

Chapter 10 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

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