Vector Cross Products And Position Vectors

I just realised that i've made a silly mistake on the past few practice exam papers, so I would really appreciate it if you could take a look at how i'm solving this kind of problem so that I can be sure I have sorted it. Essentially, the mistake I had made was solving the problems with the position vectors I had found, rather than the vectors from one vertices to another - It's made me doubt that my method is correct...

(a) The points A, B, C are (2,1,-1), (3,4,-2) and (5,-1,2). Write down the position vectors of A, B and C with respect to a ﬁxed origin O. Using a vector method, calculate the size of angle ABC. (b) Using a vector method, ﬁnd the area of the triangle ABC in part (a).

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

$\vec{OB} = 3\underline{i} + 4\underline{j} - 2\underline{k}$

$\vec{OC} = 5\underline{i} - \underline{j} + 2\underline{k}$

To find the angle $A\hat{B}C$, I need the vectors $\vec{BA}$ and $\vec{BC}$

$\vec{BA} = -\underline{i} - 3\underline{j} + \underline{k}$

$\vec{BC} = 2\underline{i} - 5\underline{j} + 4\underline{k}$

$\vec{BA}.\vec{BC} = |\vec{BA}||\vec{BC}|\sin\theta$

$|\vec{BA}| = \sqrt{(-1)^{2} + (-3)^2 + (1)^2} = \sqrt{11}$

$|\vec{BC}| = \sqrt{(2)^2 + (-5)^2 + (4)^2} = 3\sqrt{5}$

$\vec{BA}.\vec{BC} = (-\underline{i} - 3\underline{j} + \underline{k}).(2\underline{i} - 5\underline{j} + 4\underline{k}) = -2 + 15 + 4 = 17$

$\sin\theta = \frac{17}{\sqrt{11} 3\sqrt{5}}$

$\theta = 49.83\deg$

That is part A done. I know this is quite long, but I really appreciate it if you have read this far!

$Area = \frac{1}{2}|\vec{BA} \times \vec{BC}| = \frac{1}{2} \left| \begin{array}{ccc} \underline{i} & \underline{i} & \underline{k} \\ -1 & -3 & 1 \\ 2 & -5 & 4 \end{array} \right| = \frac{1}{2}|(-12+5)\underline{i} - (-4-2)\underline{j} + (5+6)\underline{k}| = \frac{1}{2} \sqrt{(-7)^2 + (6)^2 + (11)^2} = 7.18 (2.dp)$

I know this is long but I really would appreciate it if you could take a look, so I can be sure that my method is correct.

Thanks so much!

• Bump? I don't expect anyone to go through my working, but I would really appreciate it if someone could look at the mechanics of how i've solved the question. Jun 3 '14 at 11:28

$\vec{BA}\cdot \vec{BC}$ is equal to $|\vec{BA}| |\vec{BC}|\cos\theta$, not $|\vec{BA}| |\vec{BC}|\sin\theta$.
I haven't done the calculation for part B, but your method is correct. If you don't want to go through the determinant-calculations, there's an alternative way. Find $\theta$ from part A and then plug it in the following equation to get the area.
$\text{Area}=\frac{1}{2}\times |\vec{BA}| |\vec{BC|}\sin\theta$.