# Homework Help: Vector Operations

1. Jun 6, 2006

### masterofthewave124

Q: Given the points A(1,6,-2), B(2,5,3) and C(5,3,2). Use two different vector methods to determine whether ΔABC is a right angle triangle.

hmmm vector methods...i'm not quite sure what to do but i have some ideas. for a first method, could i take the magnitude of each side and see if the trig calculations hold valid (pretty weak method considering theres many assumptions here). i'm thinkign a better method would be dot and cross product. i.e. for dot product, see if any two sides produces a product of zero?

i would just someone to guide me the right way so no time is wasted guessing around.

2. Jun 6, 2006

### arunbg

Hint: Use dot product for 1 method and cross product for the second, longer, method . Start by forming vectors AB, BC and CA .

3. Jun 6, 2006

### HallsofIvy

Trig calculations? Do you mean sine and cosine? Since you don't know any angles I don't see how you could do that. Of course the Pythagorean theorem might help.

Yes, that would work as a second method.

Time is never wasted while "guessing around". The more you do it the more you learn.

4. Jun 6, 2006

### masterofthewave124

for dot product: how do i know which two vectors could potentially have an angle of 90 between them? would i have to try all 3 possibilities?

for cross product: what am i doing here?

HallsofIvy: what i meant for trig calculations was in fact pythag. to see if any two sides produced a third side that was the sum of the squares of the other two (i.e. a^2 + b^2 = c^2)

you're also right about guessing around, that's how we learn i guess.

5. Jun 7, 2006

### Hootenanny

Staff Emeritus
You have basically been given three vector equations in the question;

$$\vec{OA} = 1i + 6j -2k$$

$$\vec{OB} = 2i + 5j -3k$$

$$\vec{OC} = 1i + 6j -2k$$

You now need to obtain three more vector equations; $\vec{AB}$, $\vec{BC}$, $\vec{CA}$. Can you do that?

For the dot product basically, your just going to have to try all three combinations (there are only three).

http://en.wikipedia.org/wiki/Cross_product" [Broken]
http://mathworld.wolfram.com/CrossProduct.html" [Broken]

Last edited by a moderator: May 2, 2017
6. Jun 7, 2006

### arunbg

You can also try Pythagoras theorem using the magnitudes of the vectors as Hallsofivy explained .

7. Jun 7, 2006

### HallsofIvy

You are given A(1,6,-2), B(2,5,3) and C(5,3,2). So $\vec{AB}= (2-1)\vec{i}+ (5-6)\vec{j}+ (-2-3)\vec{k}$, $\vec{AC}= (1-5)\vec{i}+ (6-3)\vec{j}+ (-2-2)\vec{k}$, and $\vec{BC}= (5-2)\vec{i}+ (3-5)\vec{j}+ (2-3)\vec{i}$. Is the dot product of any two of those 0? The cross product isn't applicable here.

8. Jun 7, 2006

### arunbg

Why not ? If triangle ABC is rt angled at B , then ,
$$\vert(\vec{AB}\times\vec{BC})\vert = \vert\vec{AB}\vert \vert\vec{BC}\vert$$.
This is surely a necessary and sufficient condition although a bit long .

9. Jun 7, 2006

### Hootenanny

Staff Emeritus
I believe the cross product gives the vector which is perpendicular to both A and B.

10. Jun 7, 2006

### arunbg

I was only taking the magnitude (modulus). Also sin(B) = 1 .

11. Jun 8, 2006

### masterofthewave124

yeah the cross product would work here...none of dot products produced a value of 0 so i guess its not right-angled. didn't make an arithmetic error did i?

edit:

wait if i used the cross product here, then a x b = |a||b| for any vector combination as arunbg stated. but then the end result would be (x,y,z) = a number; how would i rationalize the equality there?

Last edited: Jun 8, 2006
12. Jun 9, 2006

### arunbg

13. Jun 10, 2006

### masterofthewave124

yeah i noticed HallsOfIvy's mistake when reading it, i figured it was a typo. but even with the corrected values, i can't get it to work.

these are the values im getting:
AB • BC
= 10

AB • CA
= -27

BC • CA
= -14

i think it has something to do with the vectors im forming; i formed AB, BC and CA but HallsofIvy formed AB, BC and AC. im getting confused because dot product is taken from tip to tip anyways.

14. Jun 10, 2006

### Hootenanny

Staff Emeritus
Are you sure you are calculating the dot product correctly? Here's my working for the first one;

$$\vec{AB}= (2-1)\vec{i}+ (5-6)\vec{j}+ (2+3)\vec{k}$$

$$\vec{BC}= (5-2)\vec{i}+ (3-5)\vec{j}+ (2-3)\vec{k}$$

$$\vec{AB}{\mathbf\centerdot}\vec{BC} = 1\times 3 + -1 \times -2 + 5\times -1$$

$$\boxed{\vec{AB}{\mathbf\centerdot}\vec{BC} = 0}$$

Now what does this say about the vectors $\vec{AB}$ and $\vec{BC}$? If you can't see it straight away, note that the cosine of the angle between them is given by;

$$\cos\theta = \frac{\vec{AB}{\mathbf\centerdot}\vec{BC}}{|\vec{AB}||\vec{BC}|}$$

Last edited: Jun 10, 2006
15. Jun 10, 2006

### masterofthewave124

sorry i'm so careless (did 3+2+5 instead of 3+2-5). thanks hootenay (note that you have a mistake in your dot product, the first value should be 1 x 3 but i know it was a typo since your answer is still 0).

16. Jun 10, 2006

### Hootenanny

Staff Emeritus
Ahh yes, careless error, I seem to be prone to them recently. Thank you for the correction. Now, have you thought anymore about my question;

17. Jun 10, 2006

### masterofthewave124

it says that the angle between the vectors AB and BC is 90 and hence a right-angle. is this what you were implying?

18. Jun 10, 2006

### Hootenanny

Staff Emeritus
That's it. So now you've proved it with one vector method...