Question

Does the triangle in 3 space with vertices (-1,2,3), (2,-2,0) and (3,1,-4) have an obtuse angle?

How do I find the angles?
How do I justify the answer?

Answer 1

If\[c^2 > a^2 + b^2\]where \(c\) is the longest side and \(a,b\) are two others, then the triangle HAS an obtuse angle.

Answer 2

So first I'll have to find all the distances?

Answer 3

Yes!

Answer 4

But this question is in my 3 Dimensional Space/Vectors chapter, I was wondering can there be any other method to solve the same question using vectors/matrix something like that?

Answer 5

Finding the magnitude of a vector and finding the distance between two points is essentially the same thing.

Answer 6

You can represent the vector which joins \(\left(-1,2,3\right)\) and \((2,-2,0)\) in component form as \(\langle 3,-4,3\rangle\) or \(3{\bf i} - 4{\bf j} + 3{\bf k}\). Find the magnitude of that.

Answer 7

That thing would give you the length of one side. Similarly, represent the vector between \((2,-2,0)\) and \((3,1,-4)\) and then find the magnitude to get the length of the second side.

Answer 8

Are you following me? @Zeerak

Answer 9

Yes! Just worked the problem out!

Answer 10

Greattt! Thanks for the thorough and prompt reply!

Answer 11

Nice! So, what are the values of \(a,b,c\)?

Answer 12

why not just draw the triangle

Answer 13

lol yeah, but that wouldn't qualify as "work" :-)

Answer 14

after you draw it , it is easy to see the sides can be found using Pythagorus

Answer 15

@UnkleRhaukus Do you mean Law of Cosines? It's not a right triangle.

Answer 16

In this case, we can't use the Law of Cosines either.

Answer 17

We don't know the measure of all sides. :-|

Answer 18

But a simple approach, as I said, is to check if \(c^2 > a^2 + b^2\)

Answer 19

oh, wait we are the 3space, (drawing is not going to be easy_

Answer 20

a= rt34
b= rt66
c= rt26

Answer 21

Remember that \(c\) should be \(\sqrt{66}\) since it's the LONGEST side.

Answer 22

Oh yes, wrote b on that side

Answer 23

Check if\[\left(\sqrt{66}\right)^2 > \left(\sqrt{34}\right)^2 + \left(\sqrt{26}\right)^2\]

Answer 24

Yes 66 > 60

Answer 25

It means it has an obtuse angle

Answer 26

So it does have an obtuse triangle.\(\qquad \qquad \qquad \qquad \boxed{}\)

Answer 27

Does the question ask for the angles as well?

Answer 28

One last thing Parth, I tried to do it first by finding magnitude of (2,-2,0) and (-1,2,3) and then used costheta = u.v / ||u|| ||v|| but it gave me wrong answers

Answer 29

another way to find the angles is using the dot product

Answer 30

I prefer the Law of Cosines but yeah.

Answer 31

No, does not ask for the angles, but if I had to find the angles too, what should i do?

Answer 32

You can use the dot product or the Law of Cosines... I think I like the Law of Cosines a bit too much lol :-P

Answer 33

yeah i would use this formula \[\cos\theta = \frac{u\cdot v}{ \|u\| \|v\|}\]

Answer 34

Yeah, I'd use that formula too :-)

Answer 35

Unkle, it gave me wrong answers

Answer 36

you have already worked out ||u|| and ||v||

Answer 37

i can't see your mistake if i can't see your working

Answer 38

Optional:\[c^2 = a^2 + b^2 - 2ab\cos\gamma \]\[b^2 = a^2 + c^2 - 2ab\cos\beta\]\[a^2 = b^2 + c^2 - 2ab\cos \alpha\]So,\[66 =60 - 2\sqrt{34 \times 26}\cos\gamma \]\[34 = 92 - 2\sqrt{66 \times 26}\cos \beta \]\[26 = 100 - 2\sqrt{66 \times 34}\cos\alpha\]

Answer 39

I'm being too annoying right now with that one. :-|

Answer 40

That's absolutely fine!

If I had to use he dot product method, without finding the lengths of the sides can that be done?

What I did earlier:

||u|| = \[\sqrt{4+4}\] = rt8 = 2rt2
||v|| = \[\sqrt{1+4=9}\] = rt14

Answer 41

Because u = (2,-2,0)
v = (-1,2,3)

Answer 42

Well, the magnitude of the vectors are just the length of the sides in the triangle...

Answer 43

Remember that \(u\) and \(v\) are vectors, not points.

Answer 44

But Im not getting the same answers by two different methods? Why is that?

Answer 45

Oh yes! u and v are vectors I just cant take the magnitude of two point and treat them as vectors!

Answer 46

Oh Thank you Parth!

Answer 47

You have to find the angle between two vectors.

Do you remember the three vectors you got? I gave you one, \(\langle 3,-4,3\rangle \)

Answer 48

You can't treat a point as a vector...

Answer 49

How did you get the lengths of the three sides of the triangle?