Here's a question that has been troubling me for a while.
Suppose you have an equation of a line in three-dimensions. The parametric equation for this line is:
x = 2
y = 2 - t
z = t
Where t is the parameter.
What you need to do is determine the equation for a plane that passes through this line, but also tangets a sphere in exactly one place. The equation for this sphere is:
x^2 + y^2 + z^2 = 4
Thats the question, of which I think there are two answers.
The question was longer, but I managed to boil it down to that. If you would like the full question, just ask. If possible, try not to use derivatives to answer this. They confuse me. Thank you in advance for your help.
In fact, I do need two answers for this problem. Here was the original question:
Find the scalar equation of the plane which passes through the line of intersection of the planes x + y + z - 4 = 0 and satisfies each condition:
a) It is 2 units fron the origin (perpendicular distance)
b) It is 3 units from the point A(5,-3,7) (perpendicular distance)
I have a feeling that if you can do part (a), part (b) will fall into place. So, what I did is first found the parametric equation of the line of intersection of the two planes, which was:
x = 2
y = 2 - t
z = t
Then, since when a plane tangents a sphere it is perpendicular to the spere's centre, I contructed a spere of radius 2 around the origin. Now I'm at a loss of how to find this tangent plane.
If there is a way to solve this equation without using spheres and tangent planes, I will accept those answers gratefully as well.
If it helps, I have the answers for (a). They are:
x + 2y + 2z - 6 = 0
or:
x - 2 = 0
So there are two answers, I just don't know how to get them. Thanks for your help.