Question

Find the vector, parametric and scalar equation of the plane passing through A(3, 5, 2), B(0, 5, −1) and C(1, 5, −3)

Answer 1

Find your range and then use your kernel

Answer 2

the vector equation is just "normal" dotted to one of the given points

Answer 3

i cant say that im familiar with the other two forms .....

Answer 4

Amistre are you like 22? :D

Answer 5

pfft, i havent been 22 in ages :)

Answer 6

;)

Answer 7

So I cross two points then dot it with the third?

Answer 8

cross 2 vectors to define the normal vector

Answer 9

you are given 3 points to define vectors with

Answer 10

the ax+by+cz +d = 0 is the scalar format

Answer 11

would parametric be the matrix form?

Answer 12

So it'd be (AxB) . C?

Answer 13

yes

Answer 14

Parametric is when you have its in rows. Yeah somewhat like Matrix form. I know how to get the respect forms I'm just not sure what to do to get started

Answer 15

A cross B, dotted to C

Answer 16

A(3, 5, 2), B(0, 5, −1) C(1, 5, −3)
-5 +1 -5 +1 -5 +1
----------------------------------
3 0 3 0 0 0 1 0 -2

Answer 17

Parametric is just the vector form broken up

Answer 18

x 3 1
y 0 0
z 3 -2

x = 0
y = -(6-3)
z = 0

Answer 19

When I got (AxB) . C I end up with -45

Answer 20

C is a generic vector (x-xo, y-yo, z-zo)

Answer 21

So after I got this -45 how do I get the forms?

Answer 22

um, you dont get -45 to start with ...

Answer 23

Well A cross B is (-15,3,15) dot (1,5,-3) is -45

Answer 24

C = ( \(x-1,y-5,z+3\)~)

Answer 25

your operations are not correct

Answer 26

Okay lemme try that.

Answer 27

Nvm could you please explain

Answer 28

your points given all have y=5; which means that this plane is parallel to y=5

Answer 29

the equation is:

x=x, z=z, y=5

Answer 30

A(3, 5, 2)
B(0, 5, −1)
C(1, 5, −3)
^^
all ys are the same

this is a plane parallel to the zx plane

Answer 31

Okay. So how do I go about getting the vector equation?

Answer 32

your normal is just the (0,1,0) vector dotted to the generic C

Answer 33

When you dot something isn't it a scalar value? How do you dot something with an x,y,z variable in it?

Answer 34

Wouldn't I just end up with y-5?

Answer 35

take generic point (x,y,z), and create a vector to if from C(1,5,-3)

this produces all possible vectors from C to every single point

(x-1,y-5,z+3) defines all vectors from C to a given general point

Answer 36

the dotproduct to a normal defines all vectors from C that are perpendicular (orthogonal) to the normal vector

Answer 37

a plane is defined as all the vectors that are perpendicular to a given vector

Answer 38

the vector equation is : \(\vec n\cdot \vec C\)\[(0,1,0)\cdot (x-1,y-5,z+3)\]

Answer 39

the parametric is the result of that
\[0(x-1)+1(y-5)+0(z+3)=0\]

Answer 40

or is that the scalar?

Answer 41

the parametric might be:

x = 0 + 1x + 0z
y = 5 + 0x + 0z
z = 0 + 0x + 1z

Answer 42

Umm this is what I did. Please let me know if it makes sense.

a(x-xo) + b(y-y0) + c(z-z0)

vector 1 = AB = (-3,0,-3)
vector 2 = AC = (-2,0,-5)

v1 x v2 = (0,-9,0)

0(x-1) -9(y-5) + 0(z+3) = 0
-9y-45 = 0
-9y = 45
y = -5

And that would be the equation of the plane. Does that make sense?

Answer 43

it almost makes sense ... y = 5 should be the outcome so you most likely ignored a negative along the way

Answer 44

-9(-5) = +45

Answer 45

Ahh. Yeah. Makes sense now.