Proove Vector Line Eqn is tangent to a sphere?

[ku1005]

Proove Vector Line Eqn is tangent to a sphere??

Hi, iam hoping sum1 could explain to me how to proove that the line
r = -2i-j-11k + L(3i+4k) touches but doesnlt cut the sphere

Ab Value (r - (3i-j+4k)) = 5

I have been able to find the postion vector of the point where they touch, which equals:

(7i-j+k)

however my first idea was that this position vecot would be perpendicular to the radius of the sphere and then use the dot product rule, ie if perpendicular and thus tangent, should = 0...however this method doesnlt work...any ideas or hints would be apprectiaed!

 

[FunkyDwarf]

well if you think about it if the line is tangent to the sphere it will be at right angles to a radius vector, then you can use the dot product

i think

 

[ku1005]

yeah that's EXACTLY wat i thought ...but it doesn't seem to work...hence this post
lol

 

[HallsofIvy]

Hi, iam hoping sum1 could explain to me how to proove that the line
r = -2i-j-11k + L(3i+4k) touches but doesnlt cut the sphere

Ab Value (r - (3i-j+4k)) = 5

I have been able to find the postion vector of the point where they touch, which equals:

(7i-j+k)

however my first idea was that this position vecot would be perpendicular to the radius of the sphere and then use the dot product rule, ie if perpendicular and thus tangent, should = 0...however this method doesnlt work...any ideas or hints would be apprectiaed!

The equation you derived to find the point of intersection should be a quadratic. If the equation you derived has only one solution (a double root) that is sufficient to show that the line is tangent to the sphere.

Any point on the line has r= (3L-2)i- j+ (4L- 11)k. In order to be on the sphere, we must have
|r- (3i- j+ 4k|= |(3L-5)i+ (4L- 15)|= 20 or, squaring both sides,
(3L-5)²+ (4L- 15)²= 25 or
9L²- 30L+ 25+ 16L²- 120L+ 225= 25 so
25L²- 150L+ 225= 0.
Dividing by 25, L²- 6L+ 9= 0.
That is a quadratic equation and so could have no solutions (line does not intersect sphere), two solutions (line goes through sphere), or one solution (line is tangent to sphere). Since L²- 6L+ 9= (L- 3)², this has one solution,L= 3, and so is tangent to the sphere at (3(3)-2)i- j+(4(3)-11)k= 7i- j+ k exactly as you said.

If you really feel the need for further proof that the line is tangent to the sphere, then a radius vector is r= (7i- j+ k)- (3i-j+4k)= 4i- 3k. The dot product of that with the vector along the direction of the line, 3i- j+ 4k is 4(3)+ 0(-1)- 3(4)= 0. You may have forgotten to subtract the center, 3i-j+4k, from the point of intersection.

 

[ku1005]

well yes yes i did!...thanks very much!:):)