Question

find an equation of the plane passing through the point(-2,3,4) and is perpendicular to the line passing through the points (4,-2,5) and (0,2,4)

Answer 1

to define a plane we need a normal vector\[\vec n\cdot(\vec r-\vec r_o)=0\]and a point in the plane
since we are given that the line passing through (4,-2,5) and (0,2,4) must be perpendicular to the plane we will use that to find our normal vector
so first step, what is that vector?

Answer 2

@bigboss still there?

Answer 3

the dont provide a vector. i think u need to solve my previous questions first?

Answer 4

1. find the plane that passes through the point(2,4,-3) and is parallel to the plane -2x+4y-5z+6=0

Answer 5

2. find the line that passes through the point (2,5,3) and is perpendicular to the plane 2x-3y+4z+7 = 0

Answer 6

you don't need any more info to solve this problem
which one do you want to do first?

Answer 7

can we start with 1, 2 and then the equation question?

Answer 8

1. find the plane that passes through the point(2,4,-3) and is parallel to the plane -2x+4y-5z+6=0

^this is what we are going to do first then....
did you see the hint I gave you about finding the normal vector when given the equation of a tangent plane on your other post?

Answer 9

Two planes are parallel if their normal vectors are perpendicular, so we need the normal vector \(\vec n\) for the plane \(-2x+4y-5z+6=0\) if we are to construct a parallel plane

Answer 10

*...if their normal vectors are parallel

Answer 11

so what is the normal vector of \(-2x+4y-5z+6=0\)?
hint: for plane equations of the form\[ax+by+cz=d\]the normal vector is given by\[\vec n=<a,b,c>\]

Answer 12

ok im with you

Answer 13

great,
so then you know we need \(\vec n\) from \(-2x+4y-5z+6=0\)
what is it?

Answer 14

n = <-2,4,5> ?

Answer 15

careful\[\vec n=<-2,4,-5>\]

Answer 16

correct, sorry

Answer 17

let \(\vec r=<x,y,z>\) be the position vector for any point in the plane
let the position vector for our specific known point be denoted by\[\vec r_0=<2,4,-3>\]if you think about it, this means that \(\vec r-\vec r_0\) will be a vector inside our plane.
Now, remembering that the dot product between two perpendicular vectors will be zero, we have that\[\vec n\cdot(\vec r-\vec r_0)=0\implies \vec n\cdot\vec r=\vec n\cdot\vec r_0\]do out the dot product above and you will get the equation of the plane

Answer 18

can you do that for me? im a lil lost here

Answer 19

I think you need to understand the derivation of the formula for a plane, which is well-described here:
http://tutorial.math.lamar.edu/Classes/CalcIII/EqnsOfPlanes.aspx
please do read that link when we are through here, but I will help you more with this one for now....

Answer 20

thanks alot

Answer 21

we agree that\[\vec n=<-2,4,-5>\]right?
and do you agree that any other plane that has that same normal vector will be parallel to this one?

Answer 22

yes

Answer 23

I now am going to define two position vectors:
(those are vectors that point from the origin to the point in question)
\(\vec r=<x,y,z,>\) will be what I call the vector that points to any point \(P\) in the plane. We will have to leave the components as variables for now.

The other position vector points to the specific point \(P_0\) we know the plane must pass through. Let's call it \(\vec r_0=<2,4,-3>\).
(please look at the picture in the link so you know what I mean by these position vectors)
-let me know when you have gotten all this....

Answer 24

okay im listening

Answer 25

so can you see from the picture in the link that \(\vec r-\vec r_0\) is in the plane?

Answer 26

(it is red in the picture)

Answer 27

yes i see it

Answer 28

now since the normal vector \(\vec n\) is perpendicular to the plane, and because the dot product of two perpendicular vectors is zero, we know that\[\vec n\cdot(\vec r-\vec r_0)=0\]which means that\[\vec n\cdot\vec r=\vec n\cdot\vec r_0\]still with me?
(fear not, we are almost done)

Answer 29

yess with u

Answer 30

so now just do those dot products above, and out pops the equation you are looking for
you know what \(\vec n, \vec r,\) and \(\vec r_0\) are
so we can compute\[\vec n\cdot\vec r=\vec n\cdot \vec r_0\]just plug in and do the dot products
think you can do that?
(it's just a plug-in basically)

Answer 31

(-2+4+-5)*(2+4+-3) ?

Answer 32

that almost looks like the right hand side of what we need, but why did you put + signs between all the components?
you should write them with commas
(-2,4,-5)*(2,4,-3)=?
what is the result of this dot product?

Answer 33

(-2,4,-5)*(2,4,-3)

Answer 34

do you not know how do do dot products, i.e. scalar multiplication of vectors?

Answer 35

how to do*

Answer 36

if you do not know that dot products work like this\[<a,b,c>\cdot<x,y,z>=ax+by+cz\]then you missed something very important, and I do not see how you can do these problems...

Answer 37

...but even if you have never worked with dot-products before, since I have given the formula for how to do them above you should now be able to tell me\[<-2,4,-5>\cdot<2,4,-3>=?\]

Answer 38

(-2)(2)+(4)(4)+(-5)(-3)

Answer 39

great, which equals...?

Answer 40

27

Answer 41

lovely
so that is the RHS of our equation
just to run through it again quickly, we found that\[\vec n\cdot\vec r=\vec n\cdot\vec r_0\]\[\vec n\cdot\vec r=<-2,4,-5>\cdot<2,4,-3>\]\[\vec n\cdot\vec r=27\]now what about the LHS ?
what are those vectors and their dot-product?

Answer 42

bro, i have to go now,thanks so much for ur help, can you please leave the answer for number 2 and 3? will look at it when im back

Answer 43

nope, sorry

Answer 44

LHS = 27 as well

Answer 45

no it does not, I will show you this one, please do not leave

Answer 46

ok plz show me

Answer 47

\[\vec n\cdot\vec r=<-2,4,-5>\cdot<x,y,z>=-2x+4y-5z\]if the LHS was 27 we would have just gotten 27=27, which is definitely not the equation of a plane
so in total we have...

Answer 48

\[\vec n\cdot\vec r=\vec n\cdot\vec r_0\]\[<-2,4,-5>\cdot<x,y,z>=<-2,4,-5>\cdot<2,4,-3>\]\[-2x+4y-5y=27\]

Answer 49

ok i get u nw

Answer 50

so thats the final answer?

Answer 51

...which is the equation of a plane, as opposed to 27=27
which is just a tautology

so the steps are
1) find the normal vector to the given plane \(\vec n\)
2) define position vectors for a general point \(\vec r\), and a specific point in the plane \(\vec r_o\)
3)apply the formula \(\vec n\cdot\vec r=\vec n\cdot\vec r_0\)

Answer 52

thank u :)

Answer 53

can u help me wit 2 and 3 quick?

Answer 54

and yes, the final answer is\[-2x+4y-5y=27\]not to be rough, but you really should be able to recognize that fact by the time we arrive at it
more important is that you understand what we did here so I don't have to explain every step in the other problems, because I will \(not\) just give the answer
so yeah, let's do 2 and 3, but please post a separate thread, this one is getting a bit long

Answer 55

ok hold on, ill post a new thread