Question

Multivariable: Assume there is an opaque ball of radius 1 centered at the origin.
Suppose that you stand at the point (2,3,0) and look in the direction of a point that is not visible because it is behind the ball. You will then be looking at a point on the sphere.

Find the point on the sphere at which you look if you are looking in the direction of (−2,−3,2).

Answer 1

I would first find the line connecting the point you stand and the point you look. Then find the intersection between the point and the sphere.

Answer 2

*the line

Answer 3

So I found the vector from PR, P being (2,3,0) and R from (-2,-3,2). It's just <-4,-6,2> but I'm unsure how to Incorporate that now.

Answer 4

You have found the direction vector of the line. Try find the equation of the line.

Answer 5

I have a line now of (2,3,0) + (-4,-6,2)t.
The equation of the sphere is just (x-1)^2 + (y-1)^2 + (z-1)^2 = 1, correct?

Should I set the x,y,z equations equal to each other to get a parameter t, and plug it in?

Answer 6

Hmmm, how would I go about setting the equations equal to each other?
Parameter the sphere? If so, how would I go about that?

Answer 7

I think the equation of sphere that is centred at origin is \(x^2+y^2+z^2=1\). Then you would write the linear equation as \(x=\text{sth},\,y=\text{sth},\,z=\text{sth}\) and substitute back into the sphere equation.

Answer 8

So, I just substituted the corresponding directions into the sphere equation.

(2-4t)^2+(3-6t)^2+(2t)^2=1

I just put this into Wolfram to get the solutions and I got, 3/7 and 1/2.

I used the smaller one, because that's the one I'll see.

I plugged this back in, and the solutions work!

Thanks so much!

Answer 9

Wolfram is nice, but make sure you can solve it by hand or you will be screwed in an exam.

Answer 10

Hahaha, that's very true.
I hadn't thought of that!

Answer 11

I got over-reliant on Wolfram Alpha and screwed up a mock exam. Glad that that was a mock.