Question

\[give the parametric equestion for the following line.a) the line through (2,9,0) and (-1,5,1) and b)the line through (3,4,5) parallel tp the line x(t)=2-2t , y=4+3t and z=-1-2t

Answer 1

lets make A=(2,9,0) and B= (-1,5,1) and O=the origin
OA= <2,9,0) and AB=<-3,-4,1>
<x,y,z>= <2,9,0> + t<-3,-4,1>
so x=2-3t, y=9-4t, and z=t

Answer 2

and for x(t)=2-2t , y=4+3t and z=-1-2t
you would find the direction vector of this which is <-2,3,-2>

Answer 3

and do the same thing i did for the first problem
<x,y,z>= <3,4,5> + t<-2,3-2?

Answer 4

t<-2,3-2>*

Answer 5

so x=3-2t
y=4+3t
z=5-2t

Answer 6

questions? :/

Answer 7

thank u very much i almost give up but u make my day for c)the line through (1,0,1) perpendicular to the plane 2x-3y+z=9 where does this line intersect the plane

Answer 8

well you would first find the perpendicular direction vector of 2x-3y+z=9

Answer 9

is (2,-3,0)

Answer 10

it that right

Answer 11

well what's the direction vector of 2x-3y+z=9?

Answer 12

hint* there are and infinite amount of ways to find the perpendicular vector of a plane because it is 3d

Answer 13

am still didnt get the right answer i try it many time but i cant answer it right any help please