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I am having a brain fart with a simple methodology here.

Say I have three points in 3 dimensions: (1,2,0) , (0,0,0) , and (-2,1,0)

The formula to find component form is < t point - i point > but my question is: What if the initial and terminal points are not specified as above? How do we know which points to subtract and add from each correctly?

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It will be clearer if you state the question exactly as is given to you. :) But you are right, a vector is specified by choosing an initial point and a terminal point. If there are 3 points, then we do not know which is initial and which is terminal point. –  Srivatsan Sep 20 '11 at 0:21
    
"The vertices of a triangle are given, determine whether the triangle is acute, right, or obtuse" –  Matt Sep 20 '11 at 0:28
    
Do you know how to calculate the angle between two vectors? –  Srivatsan Sep 20 '11 at 0:34
    
Yes, the trouble I am having is getting the vectors in component form since I need them in that form to continue the problem and find the angles. Can I just choose my initial and terminal points arbitrarily? –  Matt Sep 20 '11 at 0:36
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The vertices of the triangle are given to be $A(1,2,0)$, $B(0,0,0)$ and $C(-2,1,0)$. The question asks us whether this triangle is acute, obtuse or right.

Recall: A triangle is obtuse if any one angle of the triangle is obuse-angled. It is a right triangle if any one angle is a right angle. Finally, it is an acute triangle if all the three angles are acute. So to do the problem, we need to check whether each of the angles is acute or obtuse or right. I will show the steps for one angle. You can do for the other two angles similarly.

First of all, let $\theta$ be some angle in $(0, \pi)$. Then

  • $\theta$ is acute $\iff$ $\theta < \frac{\pi}{2}$ $\iff$ $\cos \theta > 0$.
  • $\theta$ is right $\iff$ $\theta = \frac{\pi}{2}$ $\iff$ $\cos \theta = 0$.
  • $\theta$ is obtuse $\iff$ $\theta > \frac{\pi}{2}$ $\iff$ $\cos \theta < 0$.

Now, to see if (say) angle $A$ of the triangle $ABC$ is acute/right/obtuse, we need to check whether $\cos \angle BAC$ is positive/zero/negative. But what is $\cos \angle BAC$? It is the angle made by the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$. (When you are computing the angle at a particular vertex $v$, you should make sure that both the vectors corresponding to the two adjacent sides have that vertex $v$ as the initial point.) We will first compute these two vectors: $$ \overrightarrow{AB} = (0,0,0) - (1,2,0) = (-1,-2,0) $$ $$ \overrightarrow{AC} = (-2,1,0) - (1,2,0) = (-3,-1,0) $$ Therefore, the angle between these vectors is given by: $$ \cos \angle BAC = \frac{\overrightarrow{AB} \cdot \overrightarrow{AC}}{|\overrightarrow{AB}| |\overrightarrow{AC}|} = \ldots \tag{1} $$ Can you take it from here? From the sign of this value, you should be able to decide if angle $A$ is acute/right/obtuse.

Now, do the same procedure for the remaining two angles $B$ and $C$ as well. That should help you solve the problem.

A shortcut. Since you are not interested in the actual values of the angles, but you need only whether they are acute, obtuse or right, it is enough to compute only the sign of the numerator (the dot product between the vectors) in formula (1). The denominator is always positive.

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"When you are computing the angle at a particular vertex, you should make sure that the both vectors have that vertex as the initial point." I think that is what I was confused on. I will post my results when I have them. thanks :) –  Matt Sep 20 '11 at 1:05
    
@Matt Ok. This is what I meant. Suppose you want to compute angle $C$. Then the two adjacent sides are $AC$ and $BC$. Now, when you are representing these as vectors, you should pick $C$ to be the initial point. So, you will take the vectors as $\overrightarrow{CA}$ and $\overrightarrow{CB}$. (Figure out what to do for $B$ yourself.) Warning: this is an important point; if you do this carelessly then you'll end up with incorrect answers. :) –  Srivatsan Sep 20 '11 at 1:12
    
Ah I understand now! For angle B I would find component forms for vectors BC-> and BA-> and then use the cosine formula. Interestingly, My homework solution website doesn't even show the whole formula, I assume thats what you meant when you said being able to tell the angle from the signs. If dot product on the top of the formula = 0, the angle is right. If it is < 0, it is obtuse. If it is > 0 it is acute. Isn't that right? –  Matt Sep 20 '11 at 1:22
    
PS. Here is a link to my textbook solution website: calcchat.com/book/Calculus-ETF-5e use chapter 11, sect 3, exercise 29 for this problem –  Matt Sep 20 '11 at 1:23
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In geometric problems like this, it's good to have an intuition of what sort of information would actually allow you to solve the problem.

For instance, you've already given one example: if you know the initial and final points on the vector, then there's really only one vector that can go between them. Those initial and final points specify that vector.

I don't know the exact problem you're trying to solve, but very often you don't have the final and initial points. Can you think of any other bits of information could you have that would specify the vector?

Perhaps you've been given the magnitude of the vector, and also another vector in the same direction. In this case, there's only one vector with your given magnitude and pointing in the given direction. So magnitude plus direction also specify that vector.

Maybe there's some way in your question to figure out the magnitude and direction of the vector? If this is true, see if you can figure out how to find the vector, possibly using unit vectors... But before you dive into the maths, make sure you can see that all of your constraints actually specify one unique vector.

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mmm...didn't see the comments on the question till after writing this, sorry. Hope it's still of some use to someone :) –  tom Sep 20 '11 at 0:58
    
Should I delete this answer??? –  tom Sep 20 '11 at 1:00
    
Hi Tom, this is the other context that may help in the problem: "The vertices of a triangle are given, determine whether the triangle is acute, right, or obtuse" I still don't see how that give definitive initial and terminal points though. –  Matt Sep 20 '11 at 1:02
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