Question

maths need help for question 6 c)

Answer 1

http://prntscr.com/ly7mjm

Answer 2

Mmmmmmmm ok ok ok i think i get it.
so what are you stuck on? just the whole thing-a-mu-bob?

Answer 3

from the start X'D
Just need to know where I should start

Answer 4

So you have three points floating around in space somewhere.

Answer 5

If you take two vectors in the plane,
and take their cross product,
it will give you a vector NORMAL to the plane (perpendicular).

Answer 6

If our line r is in the plane, then it also will be perpendicular to the normal line.

Answer 7

One way to check for perpendicular lines is by taking a dot product.
Dot product of perpendicular lines is 0.

Answer 8

So we need to set up some vectors,
take their cross product to find a normal line N.

Then take the dot product of r . N to see whether or not it gives us 0.

Answer 9

To find the vector AB, we take each component of A and subtract it from B, ya?\[\large\rm \vec{AB}=<-7-2,~1-5,~4-6>\]

Answer 10

yep ^

Answer 11

You have to be careful how you choose your vectors tho:
You want them to start form the same location (like I illustrated in the picture).

If you were to take AB x BC then you might run into some ... issues..
Err it maybe doesn't matter too much, it just messes up the sign of the result I suppose.

Answer 12

But anyway, it's better to cross AB with AC since the vectors both go out from A.

Answer 13

So if you do some calculations you should get:
AB = <-9, -4, -2>
AC = <4, -7, -15>

Answer 14

And then taking their cross product is going to give you some pretty ugly numbers.

Answer 15

You wanna try working it out on paper maybe? :O

Answer 16

umm,I did some cheat X'D
By using online calculator X'D
I got
http://prntscr.com/ly7s5j

Answer 17

You dirty rotten slacker -_-
Ok good, ya those are the same values I came up with.

Answer 18

>.<

Answer 19

So we'll dot this with ummm, not r, but the direction of r, the slope of r.
If that... makes sense >.<

r = a + bt

We want to see if \(\rm b\cdot N\) is zero.

Answer 20

\(\large\rm <22, 1, -11>\cdot<46,-143,79>\)

and if I didn't miss anything, I think you should end up with = 0.

Answer 21

Oh oh oh, i made a boo boo.
That doesn't quite finish it for us.

That simply tells us that r is PARALLEL to the plane.

Answer 22

Mmmm sec thinking :x

Answer 23

http://prntscr.com/ly7tm2

Answer 24

\(\color{#0cbb34}{\text{Originally Posted by}}\) @MARC
http://prntscr.com/ly7tm2
\(\color{#0cbb34}{\text{End of Quote}}\)
using calculator,we got 0 X'D

Answer 25

We've determined that r is parallel to our plane.

To figure out whether or not it is in the plane,
we'll take the vector connecting any point in the plane,
to any point on our line r,
and making sure that the resulting vector is still perpendicular to our normal line.

Answer 26

So in this illustration, I'm constructing a vector from A to some point on r,
let's say t=0, that will be the easiest point to work with.

Answer 27

\[\large\rm \vec{Ar_o}=<-3-2,~-6-5,~-11-6>\]

Answer 28

And we need to determine if this is perpendicular to our normal line
(dot product again)

Answer 29

\[\large\rm \vec{Ar_o}\cdot \vec{N}=<-5,-11,-17>\cdot<46,-143,79>\]

Answer 30

Now you've got me cheating also -_-
Ok online calculator says = 0

Answer 31

http://prntscr.com/ly7via

Answer 32

hahaha XD

Answer 33

Ok good

Answer 34

yay!
There,the line lies on the plane. :D

Answer 35

So ummm, ya it's in the plane.
I think that's the proper way to do it? 0_O

Seemed kinda long, but maybe that's right.

Answer 36

Therefore**

Answer 37

Answer from the book is the same as yours
XD

Answer 38

Yayyy team! Good job

Answer 39

Thank u @Zepdrix ^