Can You Solve This 3D Vector Problem Strictly Through Algebraic Manipulation?

[Perion]

Thought someone might enjoy this:

I'm working through my multivariable calc textbook but got stumped for a bit by some fusing algebra (duh). I finally saw the light this morning when I drew a 2D version on paper. Here's the problem:

We are given three vectors, r, a, and b where
r = <x, y, z>
a = <a1, a2, a3>
b = <b1, b2, b3>

Show that the equation (r - a) . (r - b) = 0 (dot product) defines a sphere. Find its radius and its center coordinates.

OK - that seemed pretty straightforward. But, when I plugged the components into the equation I came up with a rather ugly equation that didn't seem to make sense for a sphere nor did it provide me with much insight into the nature of the two vectors a and b.

When I drew the problem (reducing things down to 2D) it wasn't hard to see how it was possible for a and b to determine a sphere and satisfy the dot product condition. Then it was easy to find the center and radius. A not-as-ugly equation for the sphere was then a cinch but it was still ugly enough (to me, anyway) to conceal any obvious equality to the first one. I ended up cheating and used Mathematica to prove that they really were equal .

Can you solve this and understand the nature of a and b strictly from algebraic manipulation of the ugly version that you get when you insert the vectors' components into the dot product equation or do you have to make a drawing too?

Perion

 

[mathwonk]

well if you realize the problem has nothing to do with where the origin is, which follows from the fact that only differences of vectors appear, you can put the oprigin half way between a and b. then the problem becoems to describe the figure defined by (x-a).(x+a) = 0, i.w. |x|^2 = |a|^2, a sphere of radius a, i.e. half the distance between the original two points, and center equal to the midpoint of the line joining them.

but as usual it helps to draw a picture and find out the answer first. Some peopel say even archimedes implied he found his answers first by mechanical models, and then proved them afterwards by geometry.

 

[dextercioby]

Do you know linear algebra...?The nasty quadratic form which results could be brought to the standard form of a quadric,in this case,a sphere.

Daniel.

 

[HallsofIvy]

I don't see why you would think the equation is "ugly". Just straight forward calculation gives:
(x- a₁)(x-b₁)+ (y-a₂)(x-b₂)+ (z-a₃)(z-b₃)= 0
x²- (a₁+b₁)x+ a₁b₁+ y²- (a₂+b₂)x+ a₂b₂+ z²+(a₃+ b₃)+ a₃b₃= 0

and the only thing you need to do is note that the quadratic part is x²+ y²+ z².

 

[Perion]

Do you know linear algebra...?The nasty quadratic form which results could be brought to the standard form of a quadric,in this case,a sphere.

Daniel.

Could you (or someone) show me how to do that for this particular problem?

Thanks
Perion

 

[matt grime]

Move the origin.

Let t = r - (a+b)/2, then the equations are

(t-(a-b)/2) . (t + (a-b)/2) = 0

 

[Perion]

I don't see why you would think the equation is "ugly". Just straight forward calculation gives:
(x- a₁)(x-b₁)+ (y-a₂)(x-b₂)+ (z-a₃)(z-b₃)= 0
x²- (a₁+b₁)x+ a₁b₁+ y²- (a₂+b₂)x+ a₂b₂+ z²+(a₃+ b₃)+ a₃b₃= 0

and the only thing you need to do is note that the quadratic part is x²+ y²+ z².

Straightforward for you (and most everyone else no doubt). For me, it wasn't, as I attempted to force-fit the eq. into the standard form for a sphere. I know this is elementary - I just had better success in visualizing the implications for a and b of <r-a> and <r-b> being orthogonal and also able to completely define a sphere.

Enjoy,
Perion

 

[HallsofIvy]

Good heavens, yes. It just occurred to me! In the plane, a triangle inscribed in a circle so that two vertices are at the ends of a diameter MUST be a right triangle: conversely, any right triangle's circumscribing circle MUST have the hypotenuse of the triangle as diameter.

In 3 dimensions, If the vectors from two given points to a point on a surface are ALWAYS perpendicular, then the three points form a right triangle, so the surface MUST be a sphere and the two points MUST be end points of a diameter. Gosh, that was easy!

 

[philosophking]

Wow! That's REALLY cool halls of ivy. What a great way to look at it.

You are "super mentor" for a reason :)

 

[Perion]

Good heavens, yes. It just occurred to me! In the plane, a triangle inscribed in a circle so that two vertices are at the ends of a diameter MUST be a right triangle: conversely, any right triangle's circumscribing circle MUST have the hypotenuse of the triangle as diameter.

In 3 dimensions, If the vectors from two given points to a point on a surface are ALWAYS perpendicular, then the three points form a right triangle, so the surface MUST be a sphere and the two points MUST be end points of a diameter. Gosh, that was easy!

Yeah! Now you got it! That was what occurred to me when I "drew-down" the situation to 2D and "saw" the geometrical implications of <r-a> and <r-b> having to be perpendicular - and why I thought someone else might get a kick out of the problem. 'Twas a much more satisfying result then my original mechanical (boring) algebraic manipulations of expanded vector components and trying to force-fit a bunch of terms into a form of the general equation for a sphere. One other thing - transforming the origin (e.g. to the center) and so on is fine but seems (to me) to assume some prior understanding of what a and b are all about. Or not? Suppose we weren't given the fact that this scalar product should define a sphere?

Perion

 

[bignum]

Interesting, I just drew it and I see an optical illusion