The base $ABCD$ of the figure has area $9$. The point $M$ divides the segment $AB$ on ratio $2$ and the edge $BF$ of length $2$ forms an angle of $60º$. Calculate $[CM,CB,BF]$, knowing that $\mathbb{V}^3$ is oriented by a positive basis.
I did the following:
- I assumed $CB=(3,0,0)$.
- To find the coordinates of $CM$: The $x$ coordinate seems to be $3$ too and the $y$ coordinate seems to be $\frac{3}{2}$, as it's a point in the plane, the $z$ coordinate is $0$, hence: $CM=(3,\frac{3}{2},0)$
Now to find the coordinates of $BF$, the $x$ coordinate seems to be also $3$, to obtain the $y$ coordinate, I've imagined a line starting on $F$, parallel to $GP$, then I'd have a triangle $BFP'$. As the angle $FBP'=60º=\frac{\pi}{3}$, I found the length $BP'$ doing the following:
The slope of $BF$ is $\frac{\pi}{3}$, then I can think about a function $f(x)=\frac{\pi x}{3}$, I guess that $BP'$ can be found taking the distance from $(0,0)$ to $(x,\frac{\pi x}{3})$ such that this distance is equal to $2$ and hence:
$$\sqrt{x^2 + \frac{\pi^2 x^2}{9}}=2 \\ x^2 + \frac{\pi^2 x^2}{9} = 4 \\ x=\pm \frac{6}{\sqrt{9+\pi^2}}$$
Then if the $y$ coordinate of $CM$ is positive and the $y$ coordinate of $CB$ is zero, then the $y$ coordinate of $BP'$ must be negative.
Now to obtain the $z$ coordinate. I know that $x=3, y= -\frac{6}{\sqrt{9+\pi^2}}$, then:
$$a^2+\left(-\frac{6}{\sqrt{9+\pi^2}}\right)^2=2^2 \\ a=\pm \frac{2 \pi }{\sqrt{9+\pi ^2}}$$
And hence: $BF=\left(3,-\frac{6}{\sqrt{9+\pi^2}},\frac{2 \pi }{\sqrt{9+\pi ^2}}\right)$
Finally: $$ \begin{vmatrix} 3 & \frac{3}{2} & 0 \\ 3 & 0 & 0 \\ 3 & -\frac{6}{\sqrt{9+\pi ^2}} & \frac{2 \pi }{\sqrt{9+\pi ^2}} \\ \end{vmatrix}=-\frac{9 \pi }{\sqrt{9+\pi ^2}} $$
As the basis is positive, it should be: $$\frac{9 \pi }{\sqrt{9+\pi ^2}}$$.
The problem is that this result does not match with the answers given by the book. What have I done wrong?
I'll denote $[a\times b]$ or $[a,b]$ the cross product; $(a,b)$ or $a\cdot b$ the dot product and $(a,b,c)=[a\times b]\cdot c$.
We're asked to find $$(CM,CB,BF)=[CM\times CB]\cdot BF=[CB+BM\times CB]\cdot BF= ([CB\times CB]+[BM\times CB],BF)=(0+[BM\times CB],BF)$$ Now, I'd suppose "The point $M$ divides the segment $AB$ on ratio 2" is $\frac{AM}{MB}=2$, as may be seen from the diagram, so $BM=\frac{1}{3}BA$. Then, $$(CM,CB,BF)=([BM\times CB],BF)=([\frac{1}{3}BA\times CB],BF)=\frac{1}{3}([BA,CB],BF)=\frac{1}{3}(-[BA,BC],BF)=\frac{1}{3}([BC,BA],BF)= \frac{1}{3}|[BC,BA]|\cdot |BF|\cdot \sin \frac{\pi}{3}$$ Briefly, why $\sin \frac{\pi}{3}$ : $[BC,BA]$ is the vector with length of $\operatorname{Area}(ABCD)$, orthogonal to plane $ABC$ and pointing upwards or downwards. To determine, upwards or downwards we use that rule:"Direction of $[a\times b]$ is selected such, that the shortest turn from $a$ to $b$, viewed from end of $[a\times b]$ would be counterclockwise" in a positive basis.
The shortest turn from $BC$ to $BA$ is counterclockwise, when viewed from from above of the plane $ABC$, so the vector $[BC\times BA]$ points upwards.
$([BC,BA],BF)=|[BC,BA]|\cdot |BF|\cos \theta$, where $\theta$ is the angle between $[BC,BA]$ and $BF$, but $\theta+$given $\frac{\pi}{3}=\frac{\pi}{2}$, so $\cos \theta = \sin \frac{\pi}{3}$.
Finally, $$(CM,CB,BF)=\frac{1}{3}|[BC,BA]|\cdot |BF|\cdot \sin \frac{\pi}{3} = \frac{1}{3}\cdot 9\cdot 2\cdot \frac{\sqrt{3}}{2}=3\sqrt{3}\hbox{.}$$