Question

Determine the vector equation of a plane that contains the line with symmetric equation (x+3)/-2=(y+1)/5=(z+2)/4 and the point P(1,3,0).

For this question to find the direction vector you would use subtraction with the points (-3,-1,-2) and (1,3,0). Why would it be (1,3,0)-(-3,-1,-2) instead of (-3,-1,-2)-(1,3,0)?

Answer 1

the line has a direction vector (dx,dy,dz) for the denominators of x,y,andz agreed?

Answer 2

tell me how you got your point for subtraction

Answer 3

and quite frankly; for the given point and a point in the line .. the subtraction makes no difference in determing a vector to cross with

Answer 4

its \[\frac{ x+3 }{ -2 }= \frac{ y+1}{ 5 }=\frac{ z+2 }{ 4 }\]

From that the point is (-3,-2,-1). Initially i did (-3,-2,-1)-(1,3,0)=(-4,-5,-1) and the answer was supposed to be (1,3,0)-(-3,-2,-1)=(4,5,0)

Answer 5

Oh wait

Answer 6

* (-4,-5,0) ok so they will give the same answer

Answer 7

let v = a-b

therfore -v = b-a

the vector obtained is just opposite but it doesnt matter when crossing

Answer 8

so it doesn't matter which I subtract from which right?

Answer 9

as far as the cross is concerned, it doesnt matter.

by convention i think they like vectors with a positive x value, but thats just a matter of scaling and is really immaterial overall

Answer 10

ok thanks!