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Vector help

Anonymous Mon 09 Nov 2020 00:16:53 No.918775 View View Reply Original Report

Quoted By: >>918946

Plz explain to me how to solve this problem. I know I'm supposed to use the cross-product formula to find the normal vector, but that's about it; so now I want to be able to replicate this process on similar problems.

Thanks kings!

Anonymous

Anonymous Mon 09 Nov 2020 04:35:53 No.918874 Report

Quoted By: >>919132

regarding the filename: yes, math is indeed fun

Summary:

1) Turn both of your line equations into vectors
2) Take the cross product of the two vectors to get a vector normal to the plane (so you had that part right)
3) You'll also need a point on the plane; get it from one of the line equations. Call it P
4) Note that every point Q on the plane will have the property that Q - P will be perpendicular to your normal vector. That is, the dot product of (Q - P) and your normal vector is 0. So that's how you'll get the equation of the plane from just P and your normal vector. I'll do one step per post because I'm scared of the character limit

Anonymous

Anonymous Mon 09 Nov 2020 04:42:35 No.918883 Report

Quoted By: >>918946 >>919132

1) Turn both of your line equations into vectors

Many methods to do this. Maybe just get two points and find the vector between them.

How to get two points on each line? Look at the left equation first. You need two different points - i.e. two different sets of (x, y, z) that satisfy the equation. Not much solving required - just pick some random number and set it equal to all the other ones. Let's pick 0 first because it's easy:

0 = (x-2)/4 = (y-3)/5 = (z+2)/3

That's easy because you just take the negative of the other thing in the numerators, so you get x=2, y=3, z=-2, or (2, 3, -2). Let's do 1 next:

1 = (x-2)/4 = (y-3)/5 = (z+2)/3

1 = (x-2)/4 means 4 = x-2 so 2 = x
1 = (y-3)/5 means 5 = y-3 so 8 = y
1 = (z+2)/3 means 3 = z+2 so 1 = z

So the other point on the line is (2, 8, 1).

Vector between them is (2, 8, 1) - (2, 3, -2), which gives <0, 5, 3> = 0*i + 5*j + 3*k, if i, j, k are the unit vectors.

Same thing with the other line (this crap is so much typing), except I did it in my head because it's so much typing:

0 = x/2 = (y+2)/3 = z+1 yields (0, -2, -1)
1 = x/2 = (y+2)/3 = z+1 yields (2, 1, 0)

Vector between them is (2, 1, 0) - (0, -2, -1) which is <2, 3, 1> or 2i + 3j + 1k.

Anonymous

Anonymous Mon 09 Nov 2020 04:47:15 No.918885 Report

Quoted By: >>919132

2) Take the cross product of the two vectors to get a vector normal to the plane

Cross product of <0, 5, 3> and <2, 3, 1>.

https://en.wikipedia.org/wiki/Cross_product#Matrix_notation

i j k
0 5 3
2 3 1

So:
x-component: (5*1)-(3*3) = 5-9 = -4
y-component: (0*1)-(3*2) = 0-6 = -6
z-component: (0*3)-(5*2) = 0-10 = -10

So the cross product is <-4, -6, -10>. Recall this is a normal vector to a plane, so we can do whatever crap with it we want so long as we don't change its direction. I choose to reduce by a factor of 2 and take the negative, resulting in <2, 3, 5>.

Anonymous

Anonymous Mon 09 Nov 2020 04:48:17 No.918886 Report

Quoted By: >>919132

3) You'll also need a point on the plane; get it from one of the line equations. Call it P

From step 1, the second point of the second line (chosen because the numbers look nice) is (2, 1, 0).

Anonymous

Anonymous Mon 09 Nov 2020 04:53:21 No.918890 Report

Quoted By: >>919132

4) Actually get the equation of the plane

Remember that thing I said:
>Note that every point Q on the plane will have the property that Q - P will be perpendicular to your normal vector. That is, the dot product of (Q - P) and your normal vector is 0. So that's how you'll get the equation of the plane from just P and your normal vector.

Key: the dot product of (Q - P) and your normal vector is 0. Recall that the dot product is just x1*x2 + y1*y2 + z1*z2. Let's say point Q is (x, y, z), and we know point P (the chosen point on the plane) is (2, 1, 0).

So <x - 2, y - 1, z - 0> dotted with <2, 3, 5> must be 0:

2*(x-2) + 3*(y-1) + 5*(z-0) = 0
2*x - 4 + 3*y - 3 + 5*z - 0 = 0
2*x + 3*y + 5*z = 7

Hey, 2, 3, 5, 7, the first four primes. Cool.

Let's check the other points we got from those lines because we do things right around here:

(2, 3, -2) plugged in is 2*2 + 3*3 + 5*(-2) = 4 + 9 - 10 = 13-10 = 3, which is not 7. Fuck fuck fuck.

I will check my math

Anonymous

Anonymous Mon 09 Nov 2020 05:00:49 No.918892 Report

Quoted By: >>919132

Oh yeah, the y-component gets negated in the cross product. Been a very long time since I did a cross product...

So the normal vector is <2, -3, 5>. Picking up in step 4:

<x - 2, y - 1, z - 0> dotted with <2, -3, 5> must be 0
2*(x-2) - 3*(y-1) + 5*(z-0) = 0
2*x - 4 - 3*y + 3 + 5*z = 0
2*x - 3*x + 5*z = 1

Plug in points:
(2, 3, -2) plugged in is 2*2 - 3*3 + 5*(-2) = 4 - 9 - 10 = -15

?!?!??!?!??!?!

Anonymous

Anonymous Mon 09 Nov 2020 05:08:03 No.918894 Report

Quoted By: >>919132

Oh I fucked step 1 up, the second point on the first line had a bad x-coordinate; 4 = x-2 means x is 6, not x is 2.

So second point on the left line is (6, 8, 1). Left line's direction vector is (6, 8, 1) - (2, 3, -2) = <4, 5, 3>. Funny thing is, I had a vague bad feeling typing out that the direction vector had a x-component of 0, but didn't know what it was coming from.

Well, anyway, time for another cross product

i j k
4 5 3
2 3 1

5*1 - 3*3 = 5-9 = -4
-(4*1 - 3*2) = -(4-6) = 2
4*3 - 5*2 = 12 - 10 = 2

direction vector <-4, 2, 2> reduces to <-2, 1, 1>

<x - 2, y - 1, z - 0> dotted with <-2, 1, 1> must be 0:

-2*(x-2) + 1*(y-1) + 1*(z-0) = 0
-2*x + 4 + y - 1 + z = 0
-2*x + y + z = -3
2x - y - z = 3

Plug in points again
(2, 3, -2) plugged in is 2*2 - 3 -(-2) is 4 - 3 + 4 = 5

excuse me what the fuck I'm incompetent

Anonymous

Anonymous Mon 09 Nov 2020 05:16:23 No.918895 Report

Quoted By:

I am going stark raving mad, all the example points work for the "<x - 2, y - 1, z - 0> dotted with <-2, 1, 1> must be 0" statement but not for the derived equation of the plane...

(2, 3, -2) plugged in is <2-2, 3-1, -2-0> dot <-2,1,1> = <0,2,-2> dot <-2,1,1> = 0*(-2) + 2*1 + (-2)*1 = 0
(6, 8, 1) plugged in is <6-2, 8-1, 1-0> dot <-2,1,1> = <4,7,1> dot <-2,1,1> = 4*(-2) + 7*1 + 1*1 = -8 + 7 + 1 = 0
(2, 1, 0) plugged in is 0, of course
(0, -2, -1) plugged in is <0-2, -2-1, -1-0> dot <-2,1,1> = <-2,-3,-1> dot <-2,1,1> = 4 - 3 - 1 = 0

!??!???!?!?!??!??!??!?!?!??!?!?!??!?!?!??!?!??!?!??!

Anonymous

Anonymous Mon 09 Nov 2020 05:19:44 No.918896 Report

Quoted By: >>919132

oh I plugged in wrong, yes the plane equation is 2x - y - z = 3

(2, 3, -2) plugged in properly is 2*2 - 3 -(-2) is 4 - 3 + 2 = 3

and the other ones follow, I'm too tired for the rest of this shit

Anonymous

Anonymous Mon 09 Nov 2020 08:07:17 No.918946 Report

Quoted By: >>919132

>>918775
Step 1: Compute three points not on the same line, say as >>918883

P = (2,3,-2)
but he made an error in the second point, it should be
Q = (6,8,1),
and the third one from the second line,
R = (0,-2,-1).

2) Check that P3 does not lie on the first line:
(0-2)/4 = -1/2 not equal to (-2-4)/5 = -1

3) Insert the three points in a three-points-formula for planes, say for example
det[X-P;Q-P;R-P] = 0
|(x-2) 4 -2|
|(y-3) 5 -5| = 0
|(z+2) 3 1|

(x-2)(5+15) - (y-3)(4+6) + (z+2)(-20+10) = 0
20(x-2) -10(y-3) -10(z+2) = 0
2(x-2) -(y-3) -(z+2) = 0
2x - y - z -3 = 0

Anonymous

View Same Google ImgOps iqdb SauceNAO a09.png, 387KiB, 680x708

Anonymous Mon 09 Nov 2020 17:41:27 No.919132 Report

Quoted By:

>>918874
>>918883
>>918885
>>918886
>>918890
>>918892
>>918894
>>918896
>>918946

WOW SHIT THIS IS REALLY THOROUGH.

You guys are kings and I give you my utmost thanks for enhancing my learning process.

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