Question

Let L be the line with parametric equations

x = −4+2t
y = 8+t
z = 8+3t

Find the vector equation for a line that passes through the point P=(−8, −6, 7) and intersects L at a point that is distance 3 from the point Q=(−4, 8, 8). Note that there are two possible correct answers. Use the square root symbol '√' where needed to give an exact value for your answer.

I have no idea how to do this question, please help

Answer 1

First I made sure that P was not on the given line, it isn't

So from Q we need a point on the given line which is 3 units away from
Q, let that point be R(x,y,z) or R(-4+2t, 8+t, 8+3t)
vector QR = <-4+2t + 4, 8+t - 8, 8+3t - 8>
= <2t, t, 3t>

|<2t, t, 3t>| = 3
√(4t^2 + t^2 + 9t^2) = 3
14t^2 = 9
t = ± √14/3

looks like a lot of tedious calculations ahead, would have expected
a "nicer" value for t

now that we have t, we can sub into the x,y, and z of the given line
to find the 2 points of R.

I will do the first one, you do the other one

if t = +√14/3
x = -4 + 2√14/3 = (2√14 - 12)/3
y = 8 + √14/3 = (√14 + 24)/3
z = 8 + 3√14/3 = 8+√14
Then vector PR = < (2√14+12)/3 + 8, (√14+24)/3 + 6, 8+√14-7>
= < (2√14 + 36)/3 , (√14 + 42)/3, √14 + 1>
= < 2√14+36, √14+42, 3√14 + 3>
so we now have the direction numbers and a point P(−8, −6, 7)
so the equation of one of the lines is
<b>x = -8 + (2√14+36)t
y = -6 + (√14 + 42)t
z = 7 + (3√14+3)t</b>

Repeat using t = -√14/3 to find the other point R
find the direction vector PR, as above,
then write the other line, perhaps use s as the parameter

Ohh btw, better check my arithmetic