Vtyjkyuk

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 $vecOA$ and $vecOB$ and use this angle in the formula:

area $= frac12absinC$

These are my steps for doing this:

$mathbfa cdot mathbfb = beginvmatrix mathbfa endvmatrixbeginvmatrix mathbfb endvmatrix sintheta $

Let $mathbfa = beginpmatrix 1 \ -5 \ -7 endpmatrix$ and let $mathbfb = beginpmatrix 10 \ 10 \ 5 endpmatrix$

$therefore beginpmatrix 1 \ -5 \ -7 endpmatrix cdot beginpmatrix 10 \ 10 \ 5 endpmatrix = (5sqrt3)(15)sintheta $

$therefore sintheta = -dfrac1sqrt3$

If I substitute these values into the general formula:

area $= frac12absinC$

I get:

area $= frac12(5sqrt3)(15)(-dfrac1sqrt3)$

$therefore$ area $= -dfrac752$

However this isn't right, the area should be $dfrac75sqrt2$

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

Thank you.

The correct formula is $mathbfa cdot mathbfb = beginvmatrix mathbfa endvmatrixbeginvmatrix mathbfb endvmatrix costheta $

So what you really have is $costheta = cfrac-1sqrt3$

Therefore $$sintheta = sqrt1 - cos^2theta = sqrt1 - frac13 = sqrtfrac23 = fracsqrt2sqrt3$$

Finally, the area of the triangle is:

$$
Area = frac12 (5 sqrt3) (15) fracsqrt2sqrt3 = frac75 sqrt22
$$

We can just multiply $fracsqrt2sqrt2$ to the area, and then we get the answer you posted:

$$
Area = frac75 sqrt22 left(fracsqrt2sqrt2right) = frac75sqrt2
$$

Since your vectors are in $mathbbR^3$, 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:
$Area=(1/2) | a times b |$.