# Area of a triangle using vectors

I have to find the area of a triangle whose vertices have coordinates

O$(0,0,0)$, A$(1,-5,-7)$ and B$(10,10,5)$

I thought that perhaps I should use the dot product to find the angle between the lines $\vec{OA}$ and $\vec{OB}$ and use this angle in the formula:

area $= \frac{1}{2}ab\sin{C}$

These are my steps for doing this:

$\mathbf{a} \cdot \mathbf{b} = \begin{vmatrix} {\mathbf{a}} \end{vmatrix}\begin{vmatrix} {\mathbf{b}} \end{vmatrix} \sin{\theta}$

Let $\mathbf{a} = \begin{pmatrix} 1 \\ -5 \\ -7 \end{pmatrix}$ and let $\mathbf{b} = \begin{pmatrix} 10 \\ 10 \\ 5 \end{pmatrix}$

$\therefore \begin{pmatrix} 1 \\ -5 \\ -7 \end{pmatrix} \cdot \begin{pmatrix} 10 \\ 10 \\ 5 \end{pmatrix} = (5\sqrt{3})(15)\sin{\theta}$

$\therefore \sin{\theta} = -\dfrac{1}{\sqrt{3}}$

If I substitute these values into the general formula:

area $= \frac{1}{2}ab\sin{C}$

I get:

area $= \frac{1}{2}(5\sqrt{3})(15)(-\dfrac{1}{\sqrt{3}})$

$\therefore$ area $= -\dfrac{75}{2}$

However this isn't right, the area should be $\dfrac{75}{\sqrt{2}}$

I feel I'm missing something really obvious but I can't spot it, can anyone help?

Thank you.

• $\mathbf{a}\cdot \mathbf{b}=|\mathbf{a}||\mathbf{b}| \cos \theta$.
– Meow
Feb 7 '14 at 16:54
• Haha, thank you, can't believe I didn't check that. Feb 7 '14 at 16:55
• Use the cross product ... Feb 7 '14 at 17:31
• @MarkBennet that would make it a lot more efficient, thank you :) Feb 7 '14 at 18:05

The correct formula is $\mathbf{a} \cdot \mathbf{b} = \begin{vmatrix} {\mathbf{a}} \end{vmatrix}\begin{vmatrix} {\mathbf{b}} \end{vmatrix} \cos{\theta}$

So what you really have is $\cos{\theta} = \cfrac{-1}{\sqrt{3}}$

Therefore $$\sin{\theta} = \sqrt{1 - \cos^2{\theta}} = \sqrt{1 - \frac{1}{3}} = \sqrt{\frac{2}{3}} = \frac{\sqrt{2}}{\sqrt{3}}$$

Finally, the area of the triangle is:

$$Area = \frac{1}{2} (5 \sqrt{3}) (15) \frac{\sqrt{2}}{\sqrt{3}} = \frac{75 \sqrt{2}}{2}$$

We can just multiply $\frac{\sqrt{2}}{\sqrt{2}}$ to the area, and then we get the answer you posted:

$$Area = \frac{75 \sqrt{2}}{2} \left(\frac{\sqrt{2}}{\sqrt{2}}\right) = \frac{75}{\sqrt{2}}$$

Since your vectors are in $\mathbb{R}^3$, you can find the area of the parallelogram generated by the vectors by computing the magnitude of the cross product. The area of the triangle is half that value: $Area=(1/2) | a \times b |$.

Alternative solution

$$a = |OA| = \sqrt{1^2 + 5^2 + 7^2} = \sqrt{75} = 3 \sqrt{5}$$

$$b = |AB| = \sqrt{(10-1)^2 + (10--5)^2 + (5--7)^2} = \sqrt{450} = 15\sqrt{2}$$

$$c = |BO| = \sqrt{10^2+10^2+5^2} = 15.$$

Now you can calculate the semiperimeter $$s$$ which is just $$\frac{1}{2} \left(3 \sqrt{5} + 15 \sqrt2 + 15 \right)$$, and use Heron's formula to find the area:

$$A = \sqrt{s(s-a)(s-b)(s-c)}$$

• That's a nice contribution but the problem explicitly said vectors are to be used.. and the solutions to the OP were already presented Sep 1 '19 at 13:39
• If the solutions using vectors were already presented, shouldn't I be able to write an alternative solution? I never meant for my answer to be a standalone answer that directly answers the OP's question. Sep 1 '19 at 13:59