Real stuck on this and I'm sure I went wrong somewhere. Here is the question.
Using points $A=(1, -1)$; $B=(2,2)$; $C=(4,0)$ find the area of the triangle.
The book states that the way to find it one way is: $$\frac12\|u\| \,\|v-\operatorname{proj}_u(v)\| $$ So I started out by finding the vectors I used u as AC and v as AB which gave me vertices as: $u=[3, -1]$; $v=[1, 3]$
I then calculated the length of u to be $\sqrt{10}$
I then proceeded to find the length of v and the projection of v onto u.
$$v=[1, 3]$$
$$proj= \left[\frac{27}{10}, \frac{-9}{10}\right]$$
I then minused the v from proj by subtraction rule of dot product so the setup (I think) was: $1-\frac{27}{10}+3-\frac{9}{10}$
Now the length I think should've been $\sqrt{\frac{2^2}{5^2}}$
I then multiplied that result by the $\frac12 \sqrt{10}$
$\frac12\sqrt{10}*\sqrt{\frac{4}{25}}$
$\frac12\sqrt{\frac{8}{5}}$
the answer is $\frac{\sqrt{45}}{2}$
where did I get so lost?
the first mistake you made is in calculating the vectors, since:
$$ A=(1,-1)\\ B=(2,2)\\ C=(4,0)\\ $$
I choose $u=B-A=(1,3)$ ; $v=C-A=(3,1)$
$Area=\frac{1}{2}||v|| *||u-proj_v(u)||$
so we will calculate every term:
$$ ||v|| = \sqrt{3^2+1^2} = \sqrt{10}\\ proj_v(u) = \frac{dot(u,v)}{dot(v,v)}v = \frac{3}{5}(3,1)\\ u-proj_v(u) = (1,3)-\frac{3}{5}(3,1)=\frac{4}{5}(-1,3)\\ ||u-proj_v(u)||=\frac{4}{5}\sqrt{10} $$ now let's substitute: $$ Area=\frac{1}{2}||v|| *||u-proj_v(u)||=\frac{1}{2}\sqrt{10}\frac{4}{5}\sqrt{10}=4 $$
I hope i did not make any mistake, but i believe that the correct answer to the book problem, is the one derived here... EDIT. checked against Heron's formula, same result