The vector V is parallel to the line of intersection. To find a point on the line, take any point common to the two planes. Substituting z = 0 in the plane equations and solving for x and y simultaneously gives the point (2,3,0). Therefore, the line is %3D
The vector V is parallel to the line of intersection. To find a point on the line, take any point common to the two planes. Substituting z = 0 in the plane equations and solving for x and y simultaneously gives the point (2,3,0). Therefore, the line is %3D
Elementary Geometry for College Students
6th Edition
ISBN:9781285195698
Author:Daniel C. Alexander, Geralyn M. Koeberlein
Publisher:Daniel C. Alexander, Geralyn M. Koeberlein
Chapter10: Analytic Geometry
Section10.CT: Test
Problem 22CT
Related questions
Question
Please help me understand how the problem gets points (2,3,0) at the end to find the parametric equation. I do not fully understand how to please help me. thank you.
![5. For the following two planes x+2y -4z =8 and 6x -2 y +3 z = 6
(a) Find the angle between the two planes.
(b) Find parametric equations for the line in which the two planes intersect.
(c) Find an equation of the plane that is perpendicular to the two planes and containing the point (5,1,-2).
(b) The line of intersection is perpendicular to the planes'normal
vectors, and therefore parallel tp vector n, xn,, which is given by:
i j
v=n, xñ, = 1
k
| 2
-4
2.
%3D
6 -2
-2
3
6 -2
1,
=-2i - 27j-14k
The vector V is parallel to the line of intersection. To find a point
on the line, take any point common to the two planes. Substituting
z = 0 in the plane equations and solving for x and y
simultaneously gives the point (2,3,0). Therefore, the line is
given by theparametric equations:
x = 2-2t y=3-271, z=-14t,
18<t<o](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F2d7d2d74-d827-467b-8d8e-9fee9720e388%2Fe4aaf09b-e9f4-4e40-8992-916a7ee8924d%2Ffnm33aa_processed.jpeg&w=3840&q=75)
Transcribed Image Text:5. For the following two planes x+2y -4z =8 and 6x -2 y +3 z = 6
(a) Find the angle between the two planes.
(b) Find parametric equations for the line in which the two planes intersect.
(c) Find an equation of the plane that is perpendicular to the two planes and containing the point (5,1,-2).
(b) The line of intersection is perpendicular to the planes'normal
vectors, and therefore parallel tp vector n, xn,, which is given by:
i j
v=n, xñ, = 1
k
| 2
-4
2.
%3D
6 -2
-2
3
6 -2
1,
=-2i - 27j-14k
The vector V is parallel to the line of intersection. To find a point
on the line, take any point common to the two planes. Substituting
z = 0 in the plane equations and solving for x and y
simultaneously gives the point (2,3,0). Therefore, the line is
given by theparametric equations:
x = 2-2t y=3-271, z=-14t,
18<t<o
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