Question

Vector question,See Attachment.

Answer 1

I got answer of (i) as -2i+j-k

Answer 2

I have to review this stuff, hold one >.<

Answer 3

on*

Answer 4

Sure,I'm also sorta reviewing it too .-.

Answer 5

Just to see if we are on the same page, did you find the equation of the line \(\ell\) to be\[\vec r(t)=\langle1+t,-t,2t\rangle\]?

Answer 6

wait i messed up

Answer 7

(1+3t,-t,1+2t)

Answer 8

yep yep

Answer 9

yeah right.

Answer 10

I don't even remember how to do part (ii) lol

Answer 11

I'm just trying to visualize right now,helps.

Answer 12

i think i can do part (ii) better than part (i)

Answer 13

ok to remind me, how did you get the answer to part (i) :P

Answer 14

Well i just found r=a+tb where b=OB-OA and then i put those points i got into the plane x+3y-2z=3 and got t=-1 and then i just put t=-1 into r,got position vector.

Answer 15

Now tell me part (ii)

Answer 16

hm well the normal vector to the plane is (2,3,-2)
let me actually make sure I can do it first :P

dang I have forgotten everything haha, sorry i may be a while in reviewing

Answer 17

Yeah sure take your time

Answer 18

a visual may help

Answer 19

if we cross l and n, that should give us the normal vector of the plane that contains l, don't you agree?

Answer 20

Agreed

Answer 21

well then the normal of our plane should be\[\langle3,-1,2\rangle\times\langle1,3,-2\rangle\]

Answer 22

which i get to be\[\langle-2,8,10\rangle\]is that what you got?

Answer 23

gimme a sec

Answer 24

oh no i messed up again lol

Answer 25

-4?

Answer 26

\[\vec n_\ell=\langle-4,8,10\rangle\]

Answer 27

for the first component?

Answer 28

Okay great :)

Answer 29

Yes

Answer 30

sweet, so then we can just use the formula\[\vec n_\ell\cdot (\vec r-\vec r_0)=0\]