# Thread: Calculus Vectors URGENT HELP!

1. ## Calculus Vectors URGENT HELP!

This unit is on vector equation of lines. Thank you in advance

1. Find equation of the line through the point (-2,1,-5) that meets the line (x+3/-3) = (y-5/4) = (2-z/1) Find the point of intersection.

2. Find the point(s) of intersection of the line (x,y,z)= (0,5,3) + t(1,-3,-2) with the sphere x^2+y^2+z^2 = 6. Is the segment of the line between the intersection points a diameter of the sphere? Please Explain.

3. Determine the distance between the lines x=t , y= 1+2t, z= 3-t and x= -3, y=-6+2s, z= 3-6s. (yes, you have to show how the lines interact)

2. Originally Posted by xkissesx
1. Find equation of the line through the point (-2,1,-5) that meets the line (x+3/-3) = (y-5/4) = (2-z/1) Find the point of intersection.
I'm not sure what you mean here. Unless I'm badly mistaken there are an infinite number of lines through (-2, 1, -5) that can cross that line.

-Dan

3. Originally Posted by xkissesx
2. Find the point(s) of intersection of the line (x,y,z)= (0,5,3) + t(1,-3,-2) with the sphere x^2+y^2+z^2 = 6. Is the segment of the line between the intersection points a diameter of the sphere? Please Explain.
The line (x, y, z) = (0, 5, 3) + t(1, -3, -2) intersects the sphere at
$\displaystyle 0 + t \cdot 1 = x$

$\displaystyle 5 + t \cdot -3 = y$

$\displaystyle 3 + t \cdot -2 = z$

So
$\displaystyle x^2 + y^2 + z^2 = (t)^2 + (-3t + 5)^2 + (-2t + 3)^2 = 6$

Solve this for t. When you insert the two t values into your line equation you will have two points where the line intersects the sphere. If the distance between those two points is equal to the diameter of the sphere, then it is a diameter.

-Dan

4. Originally Posted by xkissesx
3. Determine the distance between the lines x=t , y= 1+2t, z= 3-t and x= -3, y=-6+2s, z= 3-6s. (yes, you have to show how the lines interact)
Doubtless there is an easier way to do this, but I can't think of it off the top of my head (and I'm too stubborn to look it up. )

The minimum distance between the lines will be the length of the line segment that is perpendicular to both lines (and has endpoints on both lines). So we want to find a vector that is perpendicular to the direction vector of both lines. Call this vector $\displaystyle \vec{n}$.
$\displaystyle \vec{n} = (a, b, c)$
I'm leaving it to you to derive the parametric form of both lines, as well as proving that they don't intersect.

The direction vector of the first line is (1, 2, -1), so by using the dot product we see that
$\displaystyle a + 2b - c = 0$

Doing the same for the second line we get
$\displaystyle 2b - 6c = 0$

I solved these two equations and put the solution in terms of c. a and b were both proportional to c, so...
$\displaystyle \vec{n} = (5c, 3c, c) \to \vec{n} = (5, 3, 1)$

This is the direction vector of a line that is perpendicular to both given lines.

In order to find the points of intersection in terms of t and s notice that the point of intersection of the perpendicular line with the second line can be given by
$\displaystyle (1, 1, 3) + t(1, 2, -1) + p(5, 3, 1)$
That is, start on the first line, go up to the point of intersection of the perpendicular and the first line (parametrized by t), then go down the perpendicular (parametrized by p) to the second line.

This will be the same as the intersection point on the second line
$\displaystyle (-3, -6, 3) + s(0. 2, -6)$

So
$\displaystyle 1 + t + 5p = -3$
$\displaystyle 1 + 2t + 3p = -6 + 2s$
$\displaystyle 3 - t + p = 3 - 6s$

Solve this system for t, s, and p. That will give you the two points of intersection. Then find the distance between these points.

-Dan