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 Subject: Vector Equations Category: Science Asked by: rosenmatt-ga List Price: \$10.00 Posted: 27 Sep 2006 20:40 PDT Expires: 27 Oct 2006 20:40 PDT Question ID: 769110
 ```Can someone explain how to answer these two problems? Find an equation for the plane that passes through the point (-1,2,1) and contains the interesection of the the planes x+y-z=2 and 2x-y+3z=1 Find an equation for the plane that passes through (6,0,-2) and contains the line x=4-2t, y=3+5t, z=7+4t I am trying to study for a midterm tomorrow and a reply by tomorrow morning would be much appreciated``` Subject: Re: Vector Equations Answered By: hedgie-ga on 28 Sep 2006 08:29 PDT
 ```Matt 0) Given three points, can you find a plane ehich contains them? (hint : three equations fro thre unknowns a b and c) Both your problems can be reusced to that 1) Add a third (arbitrary) plane to the two you have - and get the common point Do it again (with a different plane) You now have a problem 0 2) select two (artbitrary) values of parameter t and get the two points defined by it Again - you have a problem 0) Good luck tomorow Hedgie``` ```Hi!! I hope that the following links would be helpful to you: http://www.math.uga.edu/~chadm/2601/HW2ans.pdf http://caccmath.cacc.cc.al.us/mth227/solved_problems/pg745.pdf```
 ```Hi, Your subject title is correct. Vectors are the way to solve the problems. Here is the solution to problem 1. 1. Find an equation for the plane that passes through the point Q1 = (-1,2,1) and contains the interesection of the the planes P1: x+y-z=2 and P2: 2x-y+3z=1. First, find the line of intersection of the two planes. Any line in the plane that passes thru the foot (tail) of the normal (perpendicular) vector to the plane is perpendicular to the normal vector. So the line of intersection will be normal to the normal vectors of both planes. Finding the normal vectors is easy. They have the same coefficients as the coefficients of the plane. v1 = <1, 1, -1> v2 = <2, -1, 3> To get the vector normal to v1 and v2, take the cross product. n = v1 cross v2 = 2i -5j -3k = <2, -5, -3> n is the directional vector. Now we need to find a point on the line. Let z = 0. Then, x + y = 2 2x - y = 1 3x = 3 x = 1 y = 1 The point Q2 = (1, 1, 0) is on the line. You can check and see that it is in both planes. The equation of the line then is: Line: <1, 1, 0> + tn = <1, 1, 0> + t<2, -5, -3> Now we have two points in the plane. Q1=(-1,2,1) and Q2=(1, 1, 0). Vector q1q2 = <2, -1, -1> Now we have two vectors in the plane. n = <2, -5, -3> q1q2 = <2, -1, -1> If we take the cross product we get the normal vector to the required plane. n cross q1q2 = <2, -4, 8> Since we only care about direction, not magnitude, we can reduce by dividing by two and get <1, -2, 4>. Taking the normal vector to the plane and one point on the plane, the equation of the plane is: 1(x - 1) -2(y - 1) + 4(z - 0) = 0 or x - 2y + 4z + 1 = 0 Here I plugged in point Q2. If I had plugged in point Q1 I would have gotten the same equation. See below. 1(x + 1) - 2(y - 2) + 4(z - 1) = 0 x - 2y + 4z + 1 = 0 Hope this helps.``` 