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Q: Vector Equations ( Answered,   2 Comments )
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

 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

Subject: Re: Vector Equations
From: edithcen-ga on 28 Sep 2006 02:26 PDT

I hope that the following links would be helpful to you:
Subject: Re: Vector Equations
From: ansel001-ga on 29 Sep 2006 00:39 PDT

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
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.

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