Question

Find parametric equation for the line through the point (0,1,2) that is perpendicular to the line x=1+t, y=1-t, z=2t and intersects this line

Answer 1

Is this calculus BC?

Answer 2

What's the difference?

Answer 3

Wow, you must be the smartest person in your highschool!

Answer 4

What is paramedrics mean?

Answer 5

@terenzreignz

Answer 6

HALP Terence!

Answer 7

Let's see what I can do...

Answer 8

We need a point (already have one) and a direction vector from what I can remember.
Your thoughts?

Answer 9

Yep. parametric equations give us the vector, \(\vec v\) and \( P_0\).

Answer 10

I think I might know how to do it, but Im not confident enough, haha.

Answer 11

I... don't.

Jhanny's forcing me to study... and that's just SICK D:

Teach her a lesson for me @Psymon >:D

Answer 12

Lmao what...

Answer 13

\[\large \vec v = <1,-1,2>\] i found it but how the hell is it parallel to the line L.... |:<

Answer 14

Are you all in paramedrics?

Answer 15

Well....I think we need a point on the line, too. Because if we're going to get something perpendicular to the given line, we need a normal vector. And to get a normal vector out of this we would have to cross two vectors. We have the direction vector given by the line, but we need to pick a point on that line and then make a second vector with the two points we have.

Answer 16

oh, and \(P_0 = (1,1,0) \therefore \vec r_0 = <1,1,0>\)

Answer 17

Yeah, when t = 0 you can say that is a point on the line, absolutely. So now use that point and the given point to make another vector.

Answer 18

Alright.
P= (0,1,2)
\(P_0 = (1,1,0)\)
\[\vec {P_{0}P} = <1-0,1-1,0-2>= <1,0,-2>\]

Answer 19

Right. Now cross that vector with the direction vector given by the line.

Answer 20

Or wait... did i do it backwards?... ?_?

Answer 21

But why are we trying to find a normal vector?

Answer 22

Because a normal vector is always going to be perpendicular to the vectors you're crossing.

Answer 23

How would the normal vector help us find the intersection...

And cross producting 2 vectors GIVES us the normal vector. So then is the normal vector perpendicular to all other vectors in the plane?

Answer 24

Yeah, a normal vector is perpendicular to all vectors in the plane.

Answer 25

But ok. Normal vector