Parametric / vector question.

Question

Question 10 [10 points]

Let L be the line with parametric equations

$$ x = −6−3t $$ $$ y = 6+3t $$ $$ z = −8+2t $$

Find the vector equation for a line that passes through the point P=(−1, 2, 3) and intersects L at a point that is distance 2 from the point Q=(−6, 6, −8). Note that there are two possible correct answers.

If anyone already has a similar question up, even just pointing me in the right direction would help a lot!

Answer 1

Let the required line be L1

From the equation of line L: line L contains the point Q (t=0). A distance of 2 means it can happen in two ways. Imagine a circle of radius 2, with the origin set at point Q. Where this circle cuts the line L, will become the points of intersection. (This will happen exactly at 2 points, hence two answers)

1. Find these points of intersection (co-ordinates) 1a. Using distance formula, a distance of 2 from Q. 1b. This should form a quadratic equation in t. Solve it to get two values of t. 1c. Find the coordinates (x,y,z) corresponding to each value of t.
2. Using these coordinates (one by one) and the coordinates of point P, find the equations of line L1.

Answer 2

Hint:

Note the point $Q$ is on $L$, so you just need to find the point $P'$ on $L$ s.t. $|P'Q|=2$, which means you just need to find corresponding $t$. There are two $t$ satisfying the requirement.

After you get the points $P'$, just connect it with $P$ by a line.

Answer 3

This is my question: http://s10.postimg.org/po9b9w5bt/Screenshot_2015_11_13_20_52_38.png

And Here is my answer: http://s19.postimg.org/za2ihshfn/IMG_0356.jpg

Is it correct?