Question

Vector help:
Find an equation of the plane parallel to 2x + y − z = 1 and passing through the point
(0, 2, −2). I don't need an answer, I just need you to guide me through the process--actually, don't give me the answer, just guide me on the way... Thanks!

Answer 1

Planes are specified by their normal vectors in combination with specification of a point in the plane. Figure out how you could create a vector nomral to the plane. 2x + y -z = 1 is parallel to this unknown plane, meaning that 2x + y - z = 1 will never intersect the plane.

Hint: you'll need to find the cross product of two vectors which are themselves in or parallel to the plane.

Answer 2

Welcome to OpenStudy!

Answer 3

Thanks :)

Answer 4

@mathmale To find vectors on the plane/parallel to the plane, do I substitute numbers into x, y, and z?

Answer 5

No.

You are given the equation of a line parallel to the plane. From this equation you can find the x-, y- and z-components of a vector that is parallel to the plane.

Now let (x, y, z) be a point lying in the plane. Find the vector that goes from this point to the given point in the plane.

Take the cross product of these two vectors. Voila! You'll have the equation of the plane.

Answer 6

@mathmale Uh, a few points I don't understand (sorry~)
1. "From this equation you can find the x-, y- and z-components of a vector that is parallel to the plane." So, like there is a vector lying on the give plane which is parallel to the plane that we want? Also, how can I derive a vector from a an equation of the plane?
2. "Now let (x, y, z) be a point lying in the plane. Find the vector that goes from this point to the given point in the plane." Does (x,y,z) lie on the plane that we want, or from the given plane? (Also, to make sure I got the facts right, the vector between the points can be calculated by x,y,z values of vector1-vector2?)

Thank you for being so patient, I've just started learning this and can be a bit...dense. :)

Answer 7

@Keigh2015
@hakunamatata34
@SeanStewart16

Answer 8

yes

Answer 9

Hey, I'm not sure what to do next for this <3

Answer 10

im sorry idk this one:(

Answer 11

@mathmale

Answer 12

2x + y − z = 1 is a given line. What is its direction vector? We are told that it is parallel to the plane. As I said before, this line will never intersect the plane. You could imagine moving this line up or down in a direction perpendicular to the plane until the line actually lies IN the plane.

Next, you create a new vector, a vector that extends from a point (x, y, z) in the plane to the given point in the plane. One way of creating such a vector is represented by the following:\[<x-x _{1},y-y _{1},z-z _{1}>, \]

Answer 13

....
where \[(x _{1},y _{1,}z _{1})\]

... is the given point, the point that lies in the plane whose equation you wish to find.

Answer 14

@mathmale so, I cross the product between the direction vector (how do I find that?) and the vector \[<x-x _{1},y-y _{1},z-z _{1}>, \] (Would any x,y, and z be already okay? Or would they have to be specified?) :)

Thanks~

Answer 15

Wait, is the direction vector the given one? (2x+y-z=1)?

Answer 16

So I cross the vectors (x-0, y-2, z+2) and 2x+y-z=1? (Okai I'm confused) The second one is an equation, right? Or do I find a vector on the given plane?

Answer 17

if you are given the equation of a plane, do you know how to find a vector normal to the plane?

Answer 18

@HexiFlexiDexi

Answer 19

@baru Nope.

Answer 20

are you familiar withthis:
cross product of two vectors gives us a third vector which is perpendicular to the plane containing the original two vectors

Answer 21

Yea :)

Answer 22

if i give you the co-ordinates of two points A and B
do you know how to find a vector that connects A and B?

Answer 23

Yes :) (x1-x2, y1-y2, z1-z2)

Answer 24

awesome,

now we have the plane 2x+y-z=1

if we could some how find three points A,B and C lying on this plane,
we can find two vectors AB and AC lying on the plane.

followed so far?

Answer 25

Yep!

Answer 26

any ideas on how to find three random points on the plane 2x+y-z=1
?

Answer 27

hint:
just substitute random easy numbers for any two variables and solve for the third

i'll do one,
let x=0 and y=0
then 0 +0-z=1
z=-1

so (0,0,-1) is a point on the plane

Answer 28

can you find two more?

Answer 29

Yea~ (0.5,0,0) and (0,1,0) :)

Answer 30

awesome!
lets call those points A, B and C

can you find the two vectors AB and AC ?

Answer 31

AB: (-0.5,1,0)
AC: (-0.5,0,-1)?

Answer 32

yep that looks right

now we have two vectors lying on the given plane,
so their cross product should give us a normal vector to the plane.

are u following? can you do that?

Answer 33

Yep :) (I cross the two vectors?)

Answer 34

yep

Answer 35

Kay :) (Lemme quickly do this on paper)

Answer 36

like this?

Answer 37

its right...
but you have made a sign mistake, the half in the bottom right corner is negative,

Answer 38

so if you correct that, you get
N=<-1, -0.5 , 0.5>

Answer 39

Ah, right! (Fixed it, to lazy to upload again)

Answer 40

now this next part is a bit tricky,

two planes are parallel if the have the same normal vector.

and a vector is normal(N) to a plane, if ALL vectors lying on the plane are perpendicular to N

Answer 41

so let the given point be P=(0 , 2 ,-2)
and let X be an unknown point on the required plane
X=( x , y , z)

Answer 42

can you find the vector PX ?
(it will have variables in them)

Answer 43

(x,y-2,x+2)?

Answer 44

yep... (the x in the end should be z, typo? )

now the vector PX is vector lying on the plane we are trying to find.
and we know that this plane has the same normal vector N we just found.
so PX is perpendicular to N

we know that the dot product of two perpendicular vectors is zero.

so can you do the dot product of PX and N ?

Answer 45

Yea, typo XD (lemme do it on "paper" again)...

Answer 46

Wait, is n the thing we just found using the cross product?

Answer 47

yep

Answer 48

Like this (sorry for terrible handwriting XD)

Answer 49

thats right,
now equate that expression to zero (because dot product of two perpendicular vectors is zero)

rearrange it and what you get is the equation of the required plane :)

Answer 50

:)

Answer 51

awesome!! :)

now observe closely and compare the answer you found and the given equation.
theres a neat shortcut there :)

Answer 52

Only the constant part changes? :)

Answer 53

@baru :)

Answer 54

thats right :)

Answer 55

@HexiFlexiDexi

Answer 56

2x+y-z=1 is the given equation

N=<-1, -0.5, 0.5 > this is the normal vector you found by crossing two vectors on the plane

Answer 57

we know that if we multiply a vector with a scalar, it orientation remains the same.

so multiply N by "-2"

N=<2 , 1 , -1 >