Rotate that Cuboid

Attaching a right-handed coordinate frame to the cuboid, as shown in the picture above, we see that the $30$ units edge is along the $x$ -axis, the $20$ units edge is along the $y$ -axis, and the $10$ units edge is along the $z$ axis. We first rotate with respect to the $y$ -axis by an angle of $-30^{\circ}$ , where the sign is negative because rotation is clockwise. The second rotation is about the $x$ -axis by an angle of $30^{\circ}$ . This angle is positive because rotation about the $x$ -axis is counter clockwise. The rotation matrix corresponding to the first rotation about the $y$ -axis is given by:

$R_1 = \begin{bmatrix} \cos \phi_1 && 0 && \sin \phi_1 \\ 0 && 1 && 0 \\ -\sin \phi_1 && 0 && \cos \phi_1 \end{bmatrix}$

and the rotation matrix corresponding to the second rotation about the $x$ -axis is given by:

$R_2 = \begin{bmatrix} 1 && 0 && 0 \\ 0 && \cos \phi_2 && -\sin \phi_2 \\ 0 && \sin \phi_2 && \cos \phi_2 \end{bmatrix}$

If $\mathbf{p}_2$ is the coordinate of a point with respect to the final coordinate frame (of the base) after the two rotations, then its coordinates with respect to the coordinate frame after the first rotation is $\mathbf{p}_1 = R_2 \mathbf{p}_2$ . Similarly, the point coordinates with respect the original coordinate frame is $\mathbf{p}_0 = R_1 \mathbf{p}_1 = R_1 R_2 \mathbf{p}_2$ .

We thus have, with $\phi_1 = -30^{\circ}$ and $\phi_2 = 30^{\circ}$ ,

$R = R_1 R_2 = \begin{bmatrix} \dfrac{\sqrt{3}}{2} && 0 && -\dfrac{1}{2} \\ 0 && 1 && 0 \\ \dfrac{1}{2} && 0 && \dfrac{\sqrt{3}}{2} \end{bmatrix} \begin{bmatrix} 1 && 0 && 0 \\ 0 && \dfrac{\sqrt{3}}{2} && -\dfrac{1}{2} \\ 0 && \dfrac{1}{2} && \dfrac{\sqrt{3}}{2} \end{bmatrix}$

Carrying out the above matrix multiplication, we obtain,

$R = \dfrac{1}{4} \begin{bmatrix} 2 \sqrt{3} && -1 && - \sqrt{3} \\ 0 && 2\sqrt{3} && - 2 \\ 2 && \sqrt{3} && 3 \end{bmatrix}$

The highest point of the rotated cuboid is space diagonally opposite to the origin, so its coordinates are $\mathbf{p}_2 = (30, 20, 10)$ . Hence its coordinates in the absolute frame are:

$\mathbf{p}_0 = R \mathbf{p}_2 = \dfrac{1}{4} \begin{bmatrix} 2 \sqrt{3} && -1 && - \sqrt{3} \\ 0 && 2\sqrt{3} && - 2 \\ 2 && \sqrt{3} && 3 \end{bmatrix} \begin{bmatrix} 30 \\ 20 \\ 10 \end{bmatrix} = \dfrac{1}{4} \begin{bmatrix} 50 \sqrt{3} - 20 \\ 40 \sqrt{3} - 20 \\ 90 + 20 \sqrt{3} \end{bmatrix} = \begin{bmatrix} 16.65 \\ 12.32 \\ 31.16 \end{bmatrix}$

For each rotation, view the axis of rotation straight on and call it $(0,0)$ for convenience. They don't have much to do with 3-d coordinates, but they help visualize the change in height.

The first rotation is shown in the first picture. It shows the height of that back corner becoming $15+5\sqrt{3}$

The second rotation is the tricky one. The back corner doesn't really rise by the height of the red line which is $10+5\sqrt{3}-10=5\sqrt{3}$ because the face of that rectangle is tilted 30 degrees from perpendicular. Instead the increase in height is only $\cos{60}=\frac{\sqrt{3}}{2}$ of this.

Total height is then $15+5\sqrt{3}+ \frac{\sqrt{3}}{2}\cdot5\sqrt{3}=\frac{45}{2}+5\sqrt{3}\approx\boxed{31.16}$

Let the lowest corner point of the cube after the two rotations be the origin $O (0,0,0)$ of $(x,y, z)$ coordinate system, where $x$ -, $y$ -, and $z$ -axes are parallel to the respective edges of the cube before the two rotations. Now let the other corner point sharing the adjacent edge with $O$ be another origin $O'$ of $(x',y', z')$ coordinate system, where $x'$ -, $y'$ -, and $z'$ -axes are parallel to the respective edges of the cube after the two rotations.

Then we note that $z=z'\cos 30^\circ + 30\sin 30^\circ = \dfrac {\sqrt 3}2z' + 15$ (see diagram). Since the $z'$ of the highest corner point is $z' = 10 \cos 30^\circ + 20 \sin 30^\circ = 5\sqrt 3+10$ . Therefore, $z = \dfrac {\sqrt 3}2(5\sqrt 3+10) + 15 \approx \boxed{31.16}$ .

Note: $x=x'$ and $y = 15\sqrt 3 - \dfrac {y'}2$ .

Swapping the two rotation steps, that is, rotating the cuboid $30°$ along the $30$ unit edge parallel to the $xz$ -plane and then rotating the result $30°$ parallel to the $yz$ -plane, results in the same altitude, but the calculations are easier since the rotations are both on orthogonal planes.

If we first rotate the cuboid at an angle of $30°$ along the $30$ unit edge, two $30°\text{-}60°\text{-}90°$ triangles are formed with hypotenuses of $10$ and $20$ , giving the highest corner a height of $\frac{1}{2}20 + \frac{\sqrt{3}}{2}10 = 10 + 5\sqrt{3}$ .

If we then rotate the result $30°$ parallel to the $yz$ -plane, another two $30°\text{-}60°\text{-}90°$ triangles are formed with hypotenuses of $30$ and $10 + 5\sqrt{3}$ , giving the highest corner a height of $\frac{1}{2}30 + \frac{\sqrt{3}}{2}(10 + 5\sqrt{3}) = \frac{45}{2} + 5\sqrt{3} \approx \boxed{31.16}$ .

This solution can be expressed in the form of matrices or trig, but the most elegant solution to me is to use quaternions to apply a rotation to a point $p$ in 3D space. Because the quaternion product $qpq^{-1}$ , also called the Hamilton product of two quaternions, gives a rotation $p'$ of vector $p$ by angle $a$ around unit vector $u$ when quaternion $q$ is equal to [ $sin (\frac {a}{2}$ ) + $cos (\frac {a}{2}$ ) * $u$ ], we can calculate these rotations via quaternion products.

It can be seen that the point furthest from the origin will be the one to reach the highest, so we will apply the quaternion operations on that point, being $p=(30,20,10)=30i+20j+10k$ The first rotation is a rotation of 30 degrees around the vector -j (or (0,-1,0)), giving us the quaternion $q=.9659+.2588j$ , of the point $p$ , which gives us the $p'=20.9808i+20j+23.6603k.$ We get the next axis of rotation by rotating the unit vector i around the same axis, giving us the rotation vector $u_{2}=0.866i+0.5k.$ Then to rotate the point $p'$ around $u_{2}$ by 30 degrees we use the quaternion $q_{2}=.9659+.2588 * u_{2}$ By taking this final Hamilton product of $q_{2}p'q_{2}^{-1}$ we get the vector $p''=16.6507i+12.3205j+31.1603k$ . Taking the coefficient for $k$ will give us the answer to the highest point of the rotated shape, being $\boxed{31.160}$ .

A more elegant solution would be to combine these two rotations into a single rotation by multiplying the two quaternions $q$ and $q_{2}$ together before taking the Hamilton product (which I didn't do because I have a dislike for calculators and was dealing with some pretty messy numbers.) The quaternion $q^{-1}$ is the conjugate of the quaternion $q$ and is found by negating all of the coefficients of the imaginary parts of the quaternion, much like a complex conjugate in two dimensions. Much more information on this type of spatial rotation can be found on Wikipedia . This is one of my first solutions, and the first one in which I attempted formatting, so feedback on my presentation (both solution itself and presentation) would definitely be welcome!

In the first rotation about Y axis, the point (30,20,10) shifts to new point (15√3-5, 20, 15+5√3).

In the second rotation about X axis, the above new point shifts to (15√3-5, 15(√3-1)/2, 10+15(1+√3)/2). Thus, the highest point is Z=10+15(1+√3)/2=30.4903.

Thus my answer is 30.4903. It is purely based on the rotation of point (30,20,10) by applying two rotations, first by (-30°) along Y-axis and the second by (30°) along X-axis. The cuboid is irrelevant in this picture.

On second thought, my second rotation has been applied incorrectly on the original edge along the X- axis of the original X-Y face.. The second rotation should be applied along the new edge along the new X-axis along the new X-Y face.

Therefore, my answer 30.4903 is different from the popular answer 31.16 given by others.