# Thread: 3D Vector Geometry Problem

1. ## 3D Vector Geometry Problem

Greeting MathHelpForum,

I'm currently working through grade 12 Calculus and Vectors in an independent-learning course. I've just recently finished the calculus portion of the course, finding it relatively painless on the whole, and so I assumed the vector portion would be a walk in the park, especially since I've done some work with vectors in grade 12 physics. That does not seem to be case however; I've had a few questions now that have stumped me for quite a while, and I really can't seem to wrap my head around this one in particular. I don't have any teachers or instructors for this course than I can go to for help, so if anyone here could point out where I'm going wrong it would be greatly appreciated. The question is as follows:

The vertices of a figure are given by A(-8, 4, 2), B(-6, 3, 5) and C(-10, 5, -9). What type of figure is ABC? Justify your answer. (Hint: Calculate the magnitudes of the sides of the figure.)

and here is my attempt at an answer:

AB = (-8, 4, -2) – (-6, 3, 5) = (-2, 1, -7)
|AB| = √[(-2)2 + (1)2 + (-7)2] = √(54) ≈ 7.35
BC = (-6, 3, 5) – (-10, 5, -9) = (4, -2, 14)
|BC| = √[(4)2 + (-2)2 + (14)2] = √(216) ≈ 14.7
CA = (-10, 5, -9) – (-8, 4, -2) = (-2, 1, -7)
|CA| = √[(-2)2 + (1)2 + (-7)2] = √(54) ≈ 7.35

Can I simply leave my answer here, and state that ABC is an isosceles triangle? I know, at this point, that ABC has three vertices and three sides, two of which are the same, is this enough information? To me, it seemed that I should go a step further, and prove that ABC is a triangle by showing that all internal angles add up to 180. Since I have all three sides, I should be able to use Cosine Law to determine each angle, but when I apply Cosine Law I find the following:

Let the internal angles of this figure be a, b and c
cos a = (|BC|2 + |CA|2 – |AB|2) / 2|BC||CA|
= [(14.7)2 + (7.35)2 – (7.35)2] / 2(14.7)(7.35)
= 216.09 / 216.09 = 1
a = cos-1 (1) = 0

cos b = (|AB|2 + |CA|2 – |BC|2) / 2|AB||CA|
= [(7.35)2 + (7.35)2 – (14.7)2] / 2(7.35)(7.35)
= -108.045 / 108.045 = -1
b = cos-1 (-1) = 180

cos c = (|AB|2 + |BC|2 – |CA|2) / 2|AB||BC|
= [(7.35)2 + (14.7)2 – (7.35)2] / 2(7.35)(14.7)
= 216.09 / 216.09 = 1
a = cos-1 (1) = 0

So, by my calculations, the three angles DO add up to 180, but only because two of the angles are zero... I'm not sure what to make of this result, can anyone enlighten me?

2. ## Re: 3D Vector Geometry Problem

Originally Posted by Lethargic
Greeting MathHelpForum,

I'm currently working through grade 12 Calculus and Vectors in an independent-learning course. I've just recently finished the calculus portion of the course, finding it relatively painless on the whole, and so I assumed the vector portion would be a walk in the park, especially since I've done some work with vectors in grade 12 physics. That does not seem to be case however; I've had a few questions now that have stumped me for quite a while, and I really can't seem to wrap my head around this one in particular. I don't have any teachers or instructors for this course than I can go to for help, so if anyone here could point out where I'm going wrong it would be greatly appreciated. The question is as follows:

The vertices of a figure are given by A(-8, 4, 2), B(-6, 3, 5) and C(-10, 5, -9). What type of figure is ABC? Justify your answer. (Hint: Calculate the magnitudes of the sides of the figure.)

and here is my attempt at an answer:

AB = (-8, 4, -2) – (-6, 3, 5) = (-2, 1, -7)
|AB| = √[(-2)2 + (1)2 + (-7)2] = √(54) ≈ 7.35
A = (-8,4,2) not (-8,4,-2) as you've written when calculating |AB| and |AC|

fix this and redo it.

3. ## Re: 3D Vector Geometry Problem

Originally Posted by Lethargic
The vertices of a figure are given by A(-8, 4, 2), B(-6, 3, 5) and C(-10, 5, -9). What type of figure is ABC? Justify your answer. (Hint: Calculate the magnitudes of the sides of the figure.)
and here is my attempt at an answer:
AB = (-8, 4, -2) – (-6, 3, 5) = (-2, 1, -7)
|AB| = √[(-2)2 + (1)2 + (-7)2] = √(54) ≈ 7.35
BC = (-6, 3, 5) – (-10, 5, -9) = (4, -2, 14)
|BC| = √[(4)2 + (-2)2 + (14)2] = √(216) ≈ 14.7
CA = (-10, 5, -9) – (-8, 4, -2) = (-2, 1, -7)
|CA| = √[(-2)2 + (1)2 + (-7)2] = √(54) ≈ 7.35
Notice that $-2<-2,1,-7>=<4,-1,14>$ that is $-2\overrightarrow{AB} =\overrightarrow {BC}$.
When two vectors are multiples of each other then they are parallel.

In this case ${A,~B,~\&~C}$ are co-linear. So there is no triangle.

You should have seen that $2\sqrt {54} = \sqrt {216}$

4. ## Re: 3D Vector Geometry Problem

In answer to the question 'can I simply leave my answer here'?, the answer is no you can't. Why can't the three points lie on straight line ? In fact they do ! But only because of a typo. Check your co-ordinates for the point A.

5. ## Re: 3D Vector Geometry Problem

Thank you all so much! I can't believe I didn't catch that typo! Everything works out much better with the correct values, it is a triangle after all