Geometric vectors theory question

[iamsmooth]

Geometric vectors theory question :(

Homework Statement

Last question of the night:

A=(-2,1,-2), B=(-3,-5,-7) and C=(1,-1,-3) are the vertices of a triangle. Which of the following points is true?

A. ||AB||²+||BC||²= ||AC||²
B. ||CA||²+||AB||²=||CB||²
C. The triangle is a right triangle
D. The triangle has a right angle at A
E. The triangle has a right angle at B

Homework Equations

c² = a² + b²

The Attempt at a Solution

Well, since I can't plot the points on a graph, I can't imagine what the triangle looks like. Not sure if it's a right triangle. To solve this problem, would I have to plot all the points on the graph properly first and examine the 3 dimensional triangle?

A and B seem like it's the Pythagoras Theorem which means it has to be a right triangle, since all the other options seem to point to the same thing, right? So I think C has to be true.

To find out whether A or B is true, I think I have to figure out if ||AC|| or ||CB|| is the hypoteneus, which I don't know how to do without plotting.

Basically, do I have to plot this question to figure it out? What other way could I figure it out?

Thanks

 

[Dick]

No, you don't have to plot it. You are given the vectors. Can't you figure out the distance between two points if you know the coordinates? Can't you figure out the angle between two vectors using the dot product? Can't you at least try?

 

[iamsmooth]

Oh ok, sorry if it seems like I didn't make an effort. I just don't know how to approach the question since vectors seems to be a big change from matrices and I'm jumping in with no prior knowledge. What you said helps a lot though. Lemme give it a try.

 

[Dick]

Trying is a great idea! Then we can help you if you have problems.

 

[iamsmooth]

Okay, I figured out that the only one correct is C: it is a right triangle.

I did the following:

The vectors,
AB = (-1,-6,-5)
BC = (4,4,4)
CA = (-3,2,1)

Now plugging them into get the norm, as:
\|AB\|=\sqrt{-1^2-6^2-5^2} = \sqrt{62}

\|BC\|=\sqrt{4^2+4^2+4^2}=\sqrt{48}

\|CA\|=\sqrt{-3^2+2^2+1^2}=\sqrt{14}

Now, AB is obviously the biggest being sqrt{62}, which means it must be the hypoteneus, therefore points A and B are proven to be false.

To get the angle, we use the formula:

<br /> \cos\Theta=\frac{u \cdot v}{\|u\|\|v\|}<br />

Which yields

\frac{15}{\sqrt{9}\sqrt{83}} for angle between AB

\frac{23}{\sqrt{83}\sqrt{11}} for angle between BC

\frac{3}{\sqrt{9}\sqrt{11}} for angle between AB

These are all in terms of cos. However, we know cos(90) = 0, and none of those answers are 0, therefore none of them are right angles.

This only leaves the question about whether it is a right triangle or not. A right triangle has an angle of 90 degrees, but none of the sides were proven to be 90 degrees. Why then is the answer "the triangle is a right triangle" true?

Edit: Under closer examination, it asks for the angle of A, and B, not AB, hmmm...

 

[iamsmooth]

Oh wait, I get it, I did the angles wrong. I used the points of u and v instead of the vectors of u and v, so I got a screwy number. Now that I plugged in the proper numbers:

vector AC * CB = AC₁*CB₁ + AC₂*CB₂ + AC₃*CB₃ = -3(4) + 2(4) + 1(4) = 0, Since we know cos 90 = 0 (don't need to plug in the square roots at the denominator, because 0 over anything is 0), we know C is 90 degrees, i.e. right triangle, and the other stuff about the hypoteneus being AB still holds. So I just finished the question yay!

Sorry for all the trouble with vectors, but I've never seen them in my life until a week ago, so it's still weird for me. Thanks for pushing me in the right direction, Dick :D

 

[Dick]

Oh wait, I get it, I did the angles wrong. I used the points of u and v instead of the vectors of u and v, so I got a screwy number. Now that I plugged in the proper numbers:

vector AC * CB = AC₁*CB₁ + AC₂*CB₂ + AC₃*CB₃ = -3(4) + 2(4) + 1(4) = 0, Since we know cos 90 = 0 (don't need to plug in the square roots at the denominator, because 0 over anything is 0), we know C is 90 degrees, i.e. right triangle, and the other stuff about the hypoteneus being AB still holds. So I just finished the question yay!

Sorry for all the trouble with vectors, but I've never seen them in my life until a week ago, so it's still weird for me. Thanks for pushing me in the right direction, Dick :D

Your off to a great start if you can not only plug in the numbers but figure out when something is wrong and fix it yourself. Nice work.