Yahoo Answers is shutting down on May 4th, 2021 (Eastern Time) and beginning April 20th, 2021 (Eastern Time) the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Trending News
Find the equation of the plane in xyz-space through point P and perpendicular to vector n?
Find the equation of the plane in xyz-space through the point P = (5, 2, 2) and perpendicular to the vector n = (2, -1, -3).
before the answer box is says z=_______ but i didn't think you were solving for z in this problem. I got an answer of 5x+2y+2z-2=0 and the online homework engine said that was wrong. I also solved for z to get (5x+2y-2)/2 but was still told I was wrong.
1 Answer
- ?Lv 78 years agoFavorite Answer
A plane with the equation: Ax + By + Cz + D = 0
has a normal vector n=(A,B,C)
Since n = (2, -1, -3), the equation of the plane is
2x -y - 3z + D = 0
Since P = (5, 2, 2) lies on the plane
2(5) -2 - 3(2) + D = 0
10 - 2 - 6 + D = 0
D = -2
So the equation of the plane is
2x -y - 3z - 2 = 0
The required answer is in the format of an equation with z as the subject. Rearranging the above gives:
z = (2x - y - 2)/3
By the way, this is a maths question not physics!