Question

Consider the line which passes through the point P(2, -4, 2), and which is parallel to the line x=1+7t,y=2+3t,z=3+7t
Find the point of intersection of this new line with each of the coordinate planes.

Answer 1

first you need the vector of the new line
do you know how to find that?

Answer 2

@TuringTest I don't know how.

Answer 3

consider what this line looks like on a graphthis is the point from which the vector you are given originates
the part with t determines the direction of the line (the vector of its direction)

Answer 4

<1,2,3> + t<7,3,7> ?

Answer 5

is that right?

Answer 6

yeah, sorry I timed out

Answer 7

what you said

Answer 8

so the new line we want should have the same \(\vec v\), right?

Answer 9

im confused why v = <7,3,7>

Answer 10

well a parameterized line is like starting at a point, and then going off in some direction v
you know it as
r(t)=<1,2,3> + t<7,3,7>
for this problem

If we write this generally, this is\[\vec r(t)=P_0+\vec v t\]this is parallel (or antiparallel) to \(\vec v\) because as \(t\) increases, (or gets more negative) \(\vec r(t)\) starts moving away from \(P_0\) in the direction of \(\vec v\)

meditate on that picture for a second because it is hard to visualize at first

Answer 11

These notes are good to help picture what's going on here:
http://tutorial.math.lamar.edu/Classes/CalcII/EqnsOfLines.aspx

Answer 12

so <1,2,3> is a vector but on your graph its a point. I dont get it

Answer 13

argh... this part is hard to explain but I will try
please corroborate what I say with the notes in the link I gave you
btw my internetz is acting funny so sorry if I cut out

Answer 14

ah ok

Answer 15

So <1,2,3> is a "vector" because all points are vectors; that is we can draw a "position vector" from the origin to that point. This is equivalent to when t=0 in our equation

Answer 16

oh ok, i didnt know a position vector is the same as a point

Answer 17

i knew the head was at a point

Answer 18

call this position vector from the origin r(t), then r(0) is the vector that points to P
now as t increases, r starts moving in the direction of v, because we are adding bigger and bigger multiples of v to r

Answer 19

a point and a vector are not strictly the same thing, but to parameterize our curve it is beneficial to treat points as the tips of vectors drawn from the origin, called position vectors and usually labeled "r(t)"

Answer 20

hmm, very confusing to me, not what you said but the way its done

Answer 21

yes, at first it is weird to have to do math with vectors from the originas you can see, when t=1, r(t) is now r(0) + v
When we are thinking about the line itself though, we don't really care much about the direction of the position vector, because that changes as t changes

Answer 22

instead what we care about is the *difference* between two points on the line, say r(0) and r(1), and if you look at this as a vector diagram you can see that r(1)-r(0) is parallel to v

Answer 23

which means our line is parallel to v

Answer 24

hmm, thinking, im really slow sorry

Answer 25

Take your time. The most important part of this stuff is getting past the initial weirdness I think, so absorb it well. Everything else depends on it.

Answer 26

this is equivalent to what I tried to draw above, but pretty http://prntscr.com/5u6xsa