# Find a vector equation for the line

1. Sep 27, 2005

### dagg3r

1. Find a vector equation for the line that passes through the points
P1 (5,2, -3) and P2 (2, -1, 7 )

how would i do this?
just 5i+2j-3k - (2i -1J + 7k)
3i +3j -10k?
looks a bit simple most likely wrong

2.Find the distance between the plane with equation x + y -4z = -15
and the plane parallel to it, which passes through the point (0, 3 , -4)

what i did i used the equation of the plane x + y -4z = 15
and subed (x=0, y=3, z=-4) and i got d2=-1
now i used the formula
d1-d2 / sqrt(a^2 + b^2 + c^2)
15- -1 / sqrt(18)
= 16 / sqrt(18)
is this how you do it? it just seems like im doing something wrong heh

3. On the curve x=t^4 + 2t - 1
y= e^-2t
find d^2y/dx^2 at the point where t=-1

how do i approach this? thanks

2. Sep 27, 2005

### TD

The vectorial equation of a line through a point P1 and with direction S can be given by $\vec P = \vec P_1 + k\vec S$. When you are given 2 points instead of a direction, the difference of the points gives the direction, so P2-P1 or P1-P2.

3. Sep 27, 2005

### HallsofIvy

Staff Emeritus
Yeah it is a bit too simple- for one thing, it's not an equation! For another there is no variable, t say, to give different points- it's just a vector. The vector equation of a line should look like r= mt+ b where r, m, and b are vectors and t is the scalar parameter.
What you have calculated is the vector pointing from P2 to P1. Since it is showing "direction" think of it as the "slope", m, in the equation of a line. You only need to find b. Hint: what is r when t= 0?

WHY did you use the equation x+ y- 4z= 15?? That clearly does not contain the point (0, 3, -4) because 0+ 3- 4(-4)= 3+16= 19, not 15! I have no idea how you got "d2= -1" because you didn't say what d2 meant!
You are correct that any plane parallel to x+ y- 4z= -15 must be of the form x+ y- 4z= something but that something is not 15.

Chain rule: dy/dx= (dy/dt)/(dx/dt) and then d2y/dx2= (d(dy/dx)/dt)/(dx/dt).