Advanced Vectors

[Wong]

Revisiting maths I did many years ago and have forgotten how to do questions such as this one. Please help.

The point P is at the foot of the perpendicular from point (1,1,9) to the plane 2x+y-z=6. Find the coordinates of P.

Thanks.

[jim360]

Hi,

I hope this is relatively clear and also that other people agree with me.

The vector equation of a line is of the form r.n = a.n, where r = (x,y,z), n is some vector perpendicular to the plane, and a the coordinates of a point in the plane. the . appearing is the Scalar product defined by a.d =(A,B,C).(D,E,F) = AD + BE + CF, just to clear up notation.

The vector equation for a line is r = c + Ld, where c is some point on the line, L is any real number, and d is the direction of the line, i.e. some vector parallel to the line.

So here we know that if the coordinates of P are described by the vector r, then r.n = 6. We also know, reading off from the equation of the plane, that n = (2,1,-1). Furthermore, as P is at the foot of a perpendicular, we must have that the line joining P to (1,1,9) is parallel to (2,1,-1).

Therefore we have the following two equations for the point P described by position vector r:

r.n = 6 (1)

c + Ln = r (2)

Where we choose c = (1,1,9) and n = (2,1,-1). Now what we want to do firstly is to find the value of L. This can be done by "dotting" equation 2 with the vector n, to obtain:

c.n + L n.n = r.n

And then as equation (1) says r.n = 6, we have:

c.n + Ln.n = 6

Since c and N are known we can work out :

c.n = (1,1,9).(2,1,-1) = 2 + 1 -9 = -6 ;
n.n = (2,1,-1) = 6.

hence we have:

c.n + Ln.n = -6 + 6L = 6 => L = 2

Then we have, from equation 2:

r = c + Ln = c + 2n = (1,1,9) + 2*(2,1,-1) = (1,1,9) + (4,2,-2) = (5,3,-7).

To check that this is right take (5,3,-7).(2,1,-1) = 10 + 3 - 7 = 13 -7 = 6.

So anyway the coordinates of P are (5,3,-7).

Notation: here I have used r, n, a, etc as vectors and A, B, C as pure numbers. Usually vectors should be in bold or underlined to avoid confusion with pure numbers. Can't think how to avoid the clash so clearly in general.

[vascop]

Hi Wong and Jim
I did this a different way and got the answer P(5,3,7). So i looked at your solution and I think there is a small error near the end.

//
Then we have, from equation 2:

r = c + Ln = c + 2n = (1,1,9) + 2*(2,1,-1) = (1,1,9) + (4,2,-2) = (5,3,-7).

To check that this is right take (5,3,-7).(2,1,-1) = 10 + 3 - 7 = 13 -7 = 6.

So anyway the coordinates of P are (5,3,-7).
//
should be
Then we have, from equation 2:

r = c + Ln = c + 2n = (1,1,9) + 2*(2,1,-1) = (1,1,9) + (4,2,-2) = (5,3,7).

To check that this is right take (5,3,7).(2,1,-1) = 10 + 3 - 7 = 13 -7 = 6.

So anyway the coordinates of P are (5,3,7).

If anyone is interested in my solution I will post it.

[vascop]

Square of distance between any point (x,y,z) and (1,1,9) is
(x-1)^2+(y-1)^2+(z-9)^2
if (x,y,z) is the point P (the foot of the perpendicular) then this distance will be a minimum.
Substitute in the above for z. Since the point P is on the plane then it must satisfy 2x+y-z=6, and so z=2x+y-6
So square of distance to P is (x-1)^2=(y-1)^2+(2x+y-6-9)^2
If we now calculate the partial derivative with respect to x of the square of the distance and equate it to zero and the partial derivative with respect to y and equate it to zero this will give the conditions for the distance to be a minimum.
Missing out the algebra this gives 2 equations as follows:
5x+2y=31
4x+4y=32
Solving these simultaneous equations gives
x=5 and y=3
Substituting these values in the equation above for z, that is z=2x+y-6 gives z=7
So the point P is (5,3,7).

[jim360]

Vascop - thanks for the correction. 9-2 = -7?? No wonder I stopped doing this sort of number work.

All that happened was that dodgy subtraction - the rest is correct and I think more elegant and simpler than vascop's method, since the algebra he's skipped is far from trivial. Sorry for the confusion.

[sunny-dave]

I think the Ed should be posting this as Thread of the Week ... it certainly brought back some long disused neurons into circulation.

Please don't move on to Fourier Analysis - my brain could well explode.

[Wong]

Thanks very much Vascop & Jim, I will try and work through it again and see if it makes more sense now. In the meantime you can expect a couple more shortly but I can guarantee that I am not revisiting Fourier Analysis.

[vascop]

The algebra I missed out in my post is very simple:
Partial derivative with respect to x of the distance squared is

2(x-1)+4(2x+y-15)

and with respect to y is

2(y-1)+2(2x+y-15)

Setting both these to zero gives the 2 equations in my previous post:

5x+2y=31
4x+4y=32

[vascop]

Having said that I do think Jim's solution is more elegant, but it requires a knowledge of elementary vector analysis.

[jim360]

Yeah when I looked again at it I thought "Oh yeah that's not too bad actually". Thought you were taking the square root for some reason. Of course you don't need to, making it doable, just about, with a first course in calculus + knowing how partial differentiation works.

[Wong]

I have looked through the explanations and, perhaps surpisingly, I found Vascop's much easier to follow. I will persevere though. In the meantime I amposting another question which I hope you will have a look at. I think I know how to do it but definitely need confirmation.
Thanks
.

[factor-fiction]

Is this definitely a vectors problem then? I'm out of touch with this level of problem but I think I'd have just used 3d pythagoras and the equation of the plane

[factor-fiction]

I can see now that vectors is a good way- I have used vectors to solve your latest query about intersecting planes. When I get chance I must try to solve this problem using Pythagoras in 3d though.
It's a shame we don't get more Maths questions like this on here. Thanks