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 bea,bandc

cosa= (|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

cosb= (|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

cosc= (|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?