Question

Calc III.
Consider the line which passes through the point P(-3,1,-2), and which is parallel to the line x=1+1t,y=2+5t,z=3+2t find the point of intersection of this new line with each of the coordinate planes:
xy-plane:(x,y,z)
xz-plane:(x,y,z)
yz-plane: (x,y,z)

Answer 1

As a start, write out the equation of the line in parametric form

Answer 2

<1,2,3>+t<1,5,2> would be the parametric form of the equation right?

Answer 3

`<1,2,3>+t<1,5,2> would be the parametric form of the equation right?`
looks good

now find the parallel line

Answer 4

How would I go about that exactly?

Answer 5

btw it's a vector equation not parametric

http://tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfLines.aspx

Answer 6

well you know how the line is pointing so to speak based on this vector `<1,5,2>` after the t

Answer 7

all parallel lines to this line will also point in the same direction

Answer 8

r = <1,2,3>+t<1,5,2>

anything parallel to line r will be in the form

s = P + t<1,5,2>

where P is any point you want s to run through

Answer 9

you can think of "<1,5,2>" as the "slope" of the line

it's not really the slope, but it plays a similar role

Answer 10

is s just a another name for the parallel line

Answer 11

I just made up the label

Answer 12

imagine we're in R2

and we have r = <1,1> + t*<3,4>

so we would start at (1,1) and go up 4, to the right 3

Answer 13

anything parallel to line r in R2 is going to have the same "slope"

so we're going to go up 4, to the right 3

if we want to go through say (2,-3), then we just start there and move along following the vector <3,4>