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'normalvectors, and therefore parallel tp vector n, xn,, which is given by:i jv=n, xñ, = 1k| 2-42.%3D6 -2-236 -21,=-2i - 27j-14kThe vector V is parallel to the line of intersection. To find a pointon the line, take any point common to the two planes. Substitutingz = 0 in the plane equations and solving for x and ysimultaneously gives the point (2,3,0). Therefore, the line isgiven by theparametric equations:x = 2-2t y=3-271, z=-14t,18

Answered: Image /qna-images/answer/e4aaf09b-e9f4-4e40-8992-916a7ee8924d.jpg

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

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter11: Topics From Analytic Geometry

Section11.4: Plane Curves And Parametric Equations

Problem 53E

See similar textbooks