5
$\begingroup$

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.

$\endgroup$
4
  • 2
    $\begingroup$ $\mathbf{a}\cdot \mathbf{b}=|\mathbf{a}||\mathbf{b}| \cos \theta$. $\endgroup$
    – Meow
    Feb 7 '14 at 16:54
  • $\begingroup$ Haha, thank you, can't believe I didn't check that. $\endgroup$
    – Elise
    Feb 7 '14 at 16:55
  • 2
    $\begingroup$ Use the cross product ... $\endgroup$ Feb 7 '14 at 17:31
  • $\begingroup$ @MarkBennet that would make it a lot more efficient, thank you :) $\endgroup$
    – Elise
    Feb 7 '14 at 18:05
1
$\begingroup$

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:

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

$\endgroup$
1
$\begingroup$

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: $Area=(1/2) | a \times b |$.

$\endgroup$
1
0
$\begingroup$

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:

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

$\endgroup$
2
  • $\begingroup$ That's a nice contribution but the problem explicitly said vectors are to be used.. and the solutions to the OP were already presented $\endgroup$
    – Sigma
    Sep 1 '19 at 13:39
  • 1
    $\begingroup$ 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. $\endgroup$
    – Toby Mak
    Sep 1 '19 at 13:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.