Question

Given points A(1,-5,2), B(3,5,0), and C(-1,2,-7) in 3-dimension:

a. Find a unit vector orthogonal to the plane containing A, B, and C.

b. Find the equation, in rectangular coordinates, of the plane through A, B, and C.

Answer 1

So part (a) I got n=<104, 14, 34>

Answer 2

Whoa, whoa, whoa, how is a "unit" vector, going to have such massive numbers? :P

Answer 3

and part (b) I got 104(x-1)+14(y+5)+34(z-2)=0

Answer 4

well that is my question. I don't think i did part a right.Lol

Answer 5

Well, recall that a unit vector is found by dividing your vector by its magnitude.

Answer 6

... ugh

Answer 7

i remember you tellling me that a few days ago.

Answer 8

Yep. Well let's just make sure you did the firstpart correctly. So to find a vector orthogonal, what did you do?

Answer 9

v=<2sqrt2/3, 7sqrt2/78, 17sqrt2/8>

Answer 10

Lol, I guess I should just check it instead of asking what you did xDD

Answer 11

you want me to find the normal vector?

Answer 12

I would appreciate that because I don't want to show all my work. Lol

Answer 13

No worries. I was only asking for a verbal explanation is all, dont expect ya to type out all the work xD

Answer 14

Ouch, big numbers.

Answer 15

so?

Answer 16

Your first answer that had large numbers, it looks like you did the same thing I did, but we had some different signs that resulted in different answers. Just checking my cross-product work.

Answer 17

ab=<2,10,2> or <2,10,-2>

Answer 18

at first I had <2,10,-2> but then i made a little error, suppose to be <2,10,2>

Answer 19

No, the very first thing you posted that had <104, 14, 34> Looks like you did some mistakes with signs.

Answer 20

And it should be -2 based on what you typed your points to be.

Answer 21

but it's 0-(-2)=2

Answer 22

Then you typed your point up top wrong.

Answer 23

so it's suppose to be -2?

Answer 24

You typed your point as <1, -5, 2>

Answer 25

So that means it would be (0-2) = -2.

Answer 26

when i am looking for AB you are suppose to take b-a to obtain the vector

Answer 27

Correct.

Answer 28

my given points are right. I just looked them over.

Answer 29

Well, if the point <1, -5, 2> is correct, then it'd be 0 - 2

Answer 30

okay let me fix it..

Answer 31

So the point should be <1, -5,-2> ?

Answer 32

n=<-76, 22, 34>

Answer 33

Bleh, Ill just assume the point is <1. -5. -2> x_x *reworks*

Answer 34

no it's (1,-5, 2)

Answer 35

lol x_x Nvm then. Okay, hang on o.o

Answer 36

Then yes, your normal vector is correct, now you just need to divide it by its magnitude.

Answer 37

kk

Answer 38

<-19sqrt206/309, 11sqrt206/618, 17sqrt206/618>

Answer 39

Alright, awesome.

Answer 40

So now we just need the equation of the plane :P

Answer 41

sorry i lost connection..

Answer 42

okay equation of the plane is -76(x-1)+22(y+5)+34(z-2)=0

Answer 43

Seems good to me :3

Answer 44

it doesn't check though

Answer 45

Oh?

Answer 46

hold on

Answer 47

nvm it does thanks man!

Answer 48

Lol, okay, awesome xD Yeah, you had the right idea, just had to remember how to get the unit vector :P

Answer 49

ive got one more question

Answer 50

kk

Answer 51

#17 find the vector equation of the line containing the points (3,-1,1) and (4,0,2). Determine where, if at all, the line intersects the xy, yz, and xz-planes.



#18 Given points A(1,-5,2), B(3,5,0), and C(-1,2,-7) in 3-dimension:

a. Find a unit vector orthogonal to the plane containing A, B, and C.

b. Find the equation, in rectangular coordinates, of the plane through A, B, and C.



Main question; Determine where the line in #17 intersects the plane in #18; if they do not intersect, state this.

Answer 52

vector equation is v=<3+t, -1-t, 1+t>

Answer 53

i already answered #17 and #18

Answer 54

So only the main question is left?

Answer 55

yes

Answer 56

I havent had a question that asked those specific things, so not exactly sure if Id know how to do it xD But then again I dont know if I do or dont yet. So whats the plane equation since you already did it?

Answer 57

plane equation is -76x+22y+34z=-118

Answer 58

Oh, what we just did./

Answer 59

yeahxD

Answer 60

btw that last question you asked me a few days ago, I can probably help you. I had a problem just like it in my text book.

Answer 61

Yeah, and I dont think ive had a problem like this in mine xDD Ill have to see if I can figure out how to find that intersection.

Answer 62

well the equation of the plane is wrong... I was looking at the key my instructor posted

Answer 63

We still need to know how to check the intersection regardless xD

Answer 64

it was suppose to be those larger numbers... lol

Answer 65

so the-2 was suppose to be a +2

Answer 66

Meh, I dont care at this point, haha. Well if you have the correct plane then we should see how we get the intersection.

Answer 67

correct plane 104x-14y+34z=242

Answer 68

Im not sure how we do a system with a line equation and a plane equation xD

Answer 69

okay lets push this question aside. I've got one more question

Answer 70

Alrighty. Someone else would know of course.

Answer 71

Explain how to find the displacement vector between two points in 3-space, and then find the unit vector in the same direction. Give an example.

Answer 72

Bleh, im getting confused on my own work, lol.

Answer 73

no worries. This material can be quite confusing.

Answer 74

I cant tell when I need to use cross-product and when i don't. I have some problems where it wants parallel planes and lines and uses cross-product and I have some where it wants perpendicular planes and lines and still uses cross-product.

Answer 75

callisto scroll down to the last question

Answer 76

Bleh, I know how to do the work, just dont know when I need cross-product and when I don't -_-

Answer 77

so lets use the points in the previous question

Answer 78

And find where it intersects the plane, right?

Answer 79

no no not that question

Answer 80

This one: Explain how to find the displacement vector between two points in 3-space, and then find the unit vector in the same direction. Give an example.

Answer 81

By displacement vector is that the same as component vector?

Answer 82

i guess

Answer 83

displacement is final minus intital

Answer 84

Right. Final of the corresponding directions. i final minus i initial, etc.

Answer 85

so if we chose two points in space A(1,-5,2) and B(3,5,0)...

Answer 86

we get AB=<2,10,-2>

Answer 87

Pretty easy for this question.
(3-1, 5-(-5), 0-2) = (2, 10, -2)
sqrt(4 + 100 + 4) = sqrt(108) = 6sqrt(3)
<1/3sqrt(3), 5/3sqrt(3), -1/3sqrt(3)>

Answer 88

so the unit vector is taking the vector we obtained from the x product and then dividing it by the magnitude of the vector we found?

Answer 89

It only wanted the unit vector in the same direction as the component vector, so we don't need a cross product for that.

Answer 90

ah okay

Answer 91

So we just took the component vector and divided it by the magnitude.

Answer 92

callisto i can't reply to your email..