How Can Gaussian Elimination Demonstrate a Line of Intersection in 3D Space?

[Muzly]

Hi, first poster here.

I'm stuck on a question in first-year University Maths. I'll put down what I know where I've done it.

Question: Consider these linear equations:

x + 2y - z = 2
2x + y + 3z = 5
x + 5y - 6z = 1

for the planes P1, P2 and P3 respectively.

(a) Using Gaussian elimination, show that there is one free variable, and that the solutions lie along a line L in three dimensions.

Obviously, I'm not going to put down the whole algorithm, but my solution set is as follows:
(S1,S2,S3) = ( (8-7r)/3 , (5r-1)/3 , r )
Where r is our free variable.
Any ideas on how to show (prove) that the equation lies along a line in three dimensions? Or will that be enough?

(b) Write the equation for L in vector form, and find a vector b parallel to the line L.

Would the equation for L in vector form just be [(8-7r)/3]i + [(5r-1)/3]j + rk?

Vector b is parallel to that if you just add a constant, correct?

(c) (The tricky one).
Write down normal vectors n1, n2, n3 for the planes P1, P2, P3 respectively. Obtain a vector c that lies along the line of intersection of P1 and P2, by using the vectors n1 and n2.

Is c parallel to the vector b in (b) above? Should it be?

I have got n1, n2 and n3 through finding the vector equation for P1, P2 and P3, and using cross-multiplication.
The vector forms are:
P1 - 2i + j - 2k
P2 - (2/5)i + (1/5)j + (3/5)k
P3 - i + (1/5)j - (1/6)k

Using X, Y and Z intercepts, and labelling each point A, B, and C respectively, I obtained the following vectors (I'll only show for normal vector for P1, too much to type here!)
A->B = (-2, 1, 0)
A->C = (-2, 0, -2)

Cross multiply these and you'll get:
n1 = 2i + 4j - 2k

Also,
n2 = (3/25)i + (6/25)j + (2/25)k
n3 = (-1/30)i - (1/6)j + (1/5)k

Is anyone able to just check my work (to make sure the response to my problem is correct), and able to explain to me clearly about how to obtain the vector c and if it should be parallel to vector b as in Part B.

All help is GREATLY appreciated!

 

[mcah5]

I'm not quite sure about this, but I think vector c should be the cross product of n1 and n2. I'm having a hard time imagining it in my mind, but I think if a vector is perpedicular to the perpendiculars of two planes, then it should lie parallel to the intersection of the two planes.

 

[Muzly]

Exactly as I thought, but I thought I'd come on here for confirmation.

But, the vector c must lie along the line of intersection of P1 and P2. That's what's stumped me.

And it further asks if vector c (along the line of intersection of P1 and P2) is parallel to the vector b (parallel to the line of intersection of P1 P2 and P3), and should that be so.

If you could cross product n1 and n2 , I had a hint from a tutor that there should be a varying vector attached, which should just be the unit vector of the resultant vector of the cross product of n1 and n2 , with a t stuck outside it. (t is an element of all real numbers)
eg.
(Ai + Bj + Ck) + (Di + Ej + Fk)t

 

[HallsofIvy]

(a) Using Gaussian elimination, show that there is one free variable, and that the solutions lie along a line L in three dimensions.

Obviously, I'm not going to put down the whole algorithm, but my solution set is as follows:
(S1,S2,S3) = ( (8-7r)/3 , (5r-1)/3 , r )
Where r is our free variable.
Any ideas on how to show (prove) that the equation lies along a line in three dimensions? Or will that be enough?

Looks good to me. The "line in three dimensions" referred to is given in parametric equations as x= (8-7r)/3, y= (5r-1)/, z= r. Clearly any point in the intersection lies on that line.

(b) Write the equation for L in vector form, and find a vector b parallel to the line L.

Would the equation for L in vector form just be [(8-7r)/3]i + [(5r-1)/3]j + rk?

Vector b is parallel to that if you just add a constant, correct?

No. any vector parallel to another is a multiple. What you have calculated is a vector function. Since the problem asks for a vector parallel, pick two values for r, calculate the corresponding points on the line and subtract to find the vector. Taking r= 0 and r= 1 should make it very easy!

c) (The tricky one).
Write down normal vectors n1, n2, n3 for the planes P1, P2, P3 respectively. Obtain a vector c that lies along the line of intersection of P1 and P2, by using the vectors n1 and n2.

Is c parallel to the vector b in (b) above? Should it be?

Yes, any vectors "along" the same line will be parallel. Any vector perpendicular to the normal to P1 must lie in P1. Any vector perpendicular to the normal to P2 must lie in P2. The crossproduct of those two normal vectors will be perpendicular to both, therefore lie in both planes, therefore along the line of intersection. Frankly, I don't understand your tutor's suggestion at all! Since the problem asks for a vector, not a vector function, there shouldn't be any variable involved.