Area of a Triangle in $3-$Dimensions After Undergoing Three Linear Transformations There is a triangle in $3-$space with vertices at the three following points:
$A=(1,1,1)$, $B=(1,1,-1)$ and $C=(-1,1,1).$
These points then undergo a sequence of three transformations to give $A'$, $B'$ and $C',$ respectively. First, a rotation of $30°$ about the $x-$axis, second, a rotation of $30°$ about the $z-$axis, and finally a contraction with a factor of $k=1/4.$
What would be the transformation matrix that represents these three transformations combined in the order stated?
What would be the area contained in the triangle whose vertices are the points $A',B',C'$?
Finally, what would be the ratio of the area of the triangle whose vertices are $A', B', C'$ to the area of the triangle whose vertices are $A, B, C$?
 A: For the first question, the transformation matrix is the following: $$\begin{bmatrix}{\sqrt{3}\over8} & {-\sqrt{3}\over16} & {1 \over 16}\\{1 \over 8} & {3 \over 16} & {-\sqrt{3} \over 16}\\ 0 & {1 \over 8} & {\sqrt{3} \over 16}\end{bmatrix}$$
For the second question, when the transformation matrix is multplied by each point $ A,B,C $ we get $A' =(\frac{\sqrt{3}+1}{16},\frac{5-\sqrt{3}}{16},\frac{\sqrt{3}+1}{8})$ $B'=(\frac{\sqrt{3}-1}{16},\frac{5+\sqrt{3}}{16},\frac{-\sqrt{3}+1}{8})$ $C'= (\frac{-3\sqrt{3}+1}{16},\frac{1-\sqrt{3}}{16},\frac{\sqrt{3}+1}{8})$.
We can now find two vectors from these three points: $$\vec{A'B'}=B'-A'= (\frac {-1}8,\frac{\sqrt{3}}{8},\frac{-\sqrt{3}}{4})$$ and $$\vec{A'C'}=C'-A'=(\frac {-\sqrt3}4,\frac{-1}{4},0)$$
The area we are looking for can be found by finding the magnitude of the cross product of these two vectors, then multiplying the result by $\frac 12$.
$$Area△A'B'C'= magnitude(det\begin{bmatrix} \hat i&\hat j &\hat k\\-{1 \over 8} & {\sqrt3 \over8} & -{\sqrt{3} \over 4}\\ -{\sqrt3 \over 4} & -{1 \over 4} & 0\end{bmatrix}) *\frac12$$
$$Area△A'B'C'= \frac18$$
A: Here is a sketch assuming you mean rotation wrt right hand rule. disclaimer: it is the opposite of elegant:
Apply the function:
$$\begin{pmatrix}x\\y\\z \end{pmatrix} \mapsto \begin{pmatrix}\frac{1}{4}&0&0\\0&\frac{1}{4} &0\\0&0&\frac{1}{4}  \end{pmatrix}\begin{pmatrix}\frac{\sqrt{3}}{2}&-\frac{1}{2}&0\\\frac12&\frac{\sqrt{3}}{2}&0\\0&0&1\end{pmatrix}\begin{pmatrix}1 & 0 &0\\0&\frac{\sqrt{3}}{2}& -\frac{1}{2}\\0&\frac{1}{2}&\frac{\sqrt{3}}{2} \end{pmatrix} \begin{pmatrix}x\\y\\z\\ \end{pmatrix}$$
to all three points in order to obtain $A^{\prime}, B^{\prime}, C^{\prime}$. From this, form vectors $v:=A^{\prime}-C^{\prime}$ and $u:=B^{\prime}-C^{\prime}$. Take the projection $\frac{(v \cdot u)}{\|u\|}$, and take $\|v-\frac{(v \cdot u)}{\|u\|}\|$ for the height. Multiply this by $\|u\|$ which is the basis and divide by $1/2$ for the area. Or replace the last step with cross product, if you prefer.
