Question

Find the vector and parametric equations of the following lines.
(c) The line passing through P(3, −1, 4) and Q(3, −1, 5).
(e) The line passing through P(1, 0, −3) and parallel to the line with parametric
equations x = −1 + 2t, y = 2 − t, z = 3 + 3t.

Answer 1

Hmm, alright. Where can we start with this?

Answer 2

Some starters...

The direction vector is the vector that joins any two points on a line.
A vector lying on the line is any multiple of the direction vector added to a point lying on the line.

Answer 3

@3mar can you help me ?

Answer 4

It is not from honesty to answer a question I do not know.
I wish I could have been more help. Sorry I am not so good at it!!

Answer 5

vector equation can be written as \(\vec r = \vec r_o + t \vec d\)

\(\vec r_o\) is any point on the line, \(\vec d\) is the line's direction vector, t is parameter

Answer 6

so \(\vec r_o\) can be either of (3, −1, 4) or (3, −1, 5) [or any other point on the line but you're given those ones

and direction can be \(\vec d = 3, −1, 4) - (3, −1, 5)\), to start with at least, you can scale it down as long as you keep its direction

Answer 7

To finish this question, let me continue what has been previously said:
\[d = (3,-1,5) - (3,-1,4) = (0,0,1)\]
vector equation is:
\[r(t) = (3,-1,4) + (0,0,1)t\]
parametric equation is:
\[x(t) = 3, y(t) = -1, z(t) = 4+t\]
Part e)
Parallel lines have the same directional vector
---> (2,-1,3)
vector equation:
\[r(t) = (1,0,-3) + (2,-1,3)t\]
parametric equation:
\[x(t) = 1+2t, y(t) = -t, z(t) = -3+3t\]