CG-Assignment-1

Question-1

Consider the unit sphere centered at the origin as shown in the figure below. An implicit equation of the sphere is $x^2+y^2+z^2-1=0$

Give the matrix S that scales the sphere to the ellipsoid whose radii along the x-, y-, and z-axes are 3, 0.5, and 0.3, respectively. What is the inverse of S?

$$S=\left[\begin{array}{cccc}3& 0& 0& 0\\ 0& 0.5& 0& 0\\ 0& 0& 0.3& 0\\ 0& 0& 0& 1\end{array}\right]S^{-1}=\left[\begin{array}{cccc}0.33& 0& 0& 0\\ 0& 2& 0& 0\\ 0& 0& 3.33& 0\\ 0& 0& 0& 1\end{array}\right]$$

Given the matrix R that rotates the ellipsoid -45 degrees about the z-axis. What is the inverse of R?

$$R=\left[\begin{array}{cccc}\frac{\sqrt{2}}{2}& -\frac{\sqrt{2}}{2}& 0& 0\\ \frac{\sqrt{2}}{2}& \frac{\sqrt{2}}{2}& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]R^{-1}=\left[\begin{array}{cccc}\frac{\sqrt{2}}{2}& \frac{\sqrt{2}}{2}& 0& 0\\ -\frac{\sqrt{2}}{2}& \frac{\sqrt{2}}{2}& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right]$$

Give the matrix T that translates the rotated ellipsoid to (10, 4, 2). What is the inverse of T?

$$T=\left[\begin{array}{cccc}1& 0& 0& -10\\ 0& 1& 0& -4\\ 0& 0& 1& -2\\ 0& 0& 0& 1\end{array}\right]T^{-1}=\left[\begin{array}{cccc}1& 0& 0& 10\\ 0& 1& 0& 4\\ 0& 0& 1& 2\\ 0& 0& 0& 1\end{array}\right]$$

Let M be the matrix representing the overall transformations described in (b), (c), and (d). Give a formula for M in terms of S, R, and T.

$$M=T\cdot R\cdot S$$$$M=\left[\begin{array}{cccc}\frac{3\sqrt{2}}{2}& -\frac{\sqrt{2}}{4}& 0& -10\\ \frac{3\sqrt{2}}{2}& \frac{\sqrt{2}}{4}& 0& -4\\ 0& 0& 0.3& -2\\ 0& 0& 0& 1\end{array}\right]$$

Give a formula for $M^{-1}$

$$M^{-1}=S^{-1}\cdot R^{-1}\cdot T^{-1}$$

Let the equation of the ellipsoid by $M$ from the sphere be $P^T{Q}^{\prime }P=0$. Derive a formula for $Q^{\prime }$.

$$Q^{\prime }=M^T\cdot Q\cdot M$$

Brief Solution:

The original sphere has the quadratic form:

$$P^{T}QP=0$$

After applying the transformation matrix $M$, the sphere becomes an ellipsoid, and its new quadratic form becomes:

$$P^T{Q}^{\prime }P=0$$

To derive $Q^{\prime }$, substitute the transformation $P^{\prime }=M\cdot P$ into the above. The transformed equation is:

$$(M\cdot P{)}^{T}Q(M\cdot P)=0$$

This expands to:

$$P^T{M}^{T}QMP=0$$

Thus, the new quadratic form matrix is:

$$Q^{\prime }=M^{T}QM$$

Question-2

The following shows a perspective projection where the eye is at the origin, the viewing direction is the opposite of the z-axis, and the projection plane is $z=-1$.

A point at $(x,y,z)$ in the viewing frustum is projected on $(x^{\prime },y^{\prime },z^{\prime })$. Give the formula to form $x^{\prime },y^{\prime },z^{\prime }$.

$$x^{\prime }=\frac{x}{-z}$$$$y^{\prime }=\frac{y}{-z}$$$$z^{\prime }=-1$$

Give the $4\times 4$ matrix that represents the projection.

$$M_{\text{perspective projection}}=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& \frac{1}{d}& 0\end{array}\right]$$

Since $d=1$, the matrix is:

$$M_{\text{perspective projection}}=\left[\begin{array}{cccc}1& 0& 0& 0\\ 0& 1& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 1& 0\end{array}\right]$$

Question-3

1. Write down the steps and the composite matrix for rotating 30 degrees about point $(1,2)$.

$$M_{composite}=T(1,2)\cdot R(\alpha )\cdot T(-1,-2)$$$$M_{composite}=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 2\\ 0& 0& 1\end{array}\right]\cdot \left[\begin{array}{ccc}\frac{\sqrt{2}}{2}& -0.5& 0\\ 0.5& \frac{\sqrt{2}}{2}& 0\\ 0& 0& 1\end{array}\right]\cdot \left[\begin{array}{ccc}1& 0& -1\\ 0& 1& -2\\ 0& 0& 1\end{array}\right]$$

2. Write down the composite matrix for rotating 30 degrees about z-axis, then rotating 60 degrees about y-axis.

$$M_{composite}=R_y({60}^{\circ })\cdot R_z({30}^{\circ })$$$$M_{composite}=\left(\begin{array}{cccc}0.5& 0& \frac{\sqrt{3}}{2}& 0\\ 0& 1& 0& 0\\ -\frac{\sqrt{3}}{2}& 0& 0.5& 0\\ 0& 0& 0& 1\end{array}\right)\cdot \left(\begin{array}{cccc}\frac{\sqrt{3}}{2}& -0.5& 0& 0\\ 0.5& \frac{\sqrt{3}}{2}& 0& 0\\ 0& 0& 1& 0\\ 0& 0& 0& 1\end{array}\right)$$

3. Write down the rotation matrix for rotating 30 degrees about the axis $(1,1,1)$. Note that rotation by default is counter-clock wise according to right hand rule.

According to Rodrigues' Rotation Formula

$$R=I+\sin \theta [K]+(1-\cos \theta )K^2$$

where:

• $I$ is the identity matrix,
• $\theta ={30}^{\circ }=\pi /6$,
• $K$ is the skew-symmetric matrix of the unit vector $\mathbf{v}=\frac{1}{\sqrt{3}}(1,1,1)$.

Compute the Skew-Symmetric Matrix $K$

The unit vector along $(1,1,1)$ is:

$$\mathbf{v}=\frac{1}{\sqrt{3}}\left[\begin{array}{c}1\\ 1\\ 1\end{array}\right]$$

The skew-symmetric matrix $K$ is:

$$K=\left[\begin{array}{ccc}0& -\frac{1}{\sqrt{3}}& \frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{3}}& 0& -\frac{1}{\sqrt{3}}\\ -\frac{1}{\sqrt{3}}& \frac{1}{\sqrt{3}}& 0\end{array}\right]$$

Compute $K^2$

$$K^2=\left[\begin{array}{ccc}-\frac{2}{3}& \frac{1}{3}& \frac{1}{3}\\ \frac{1}{3}& -\frac{2}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{1}{3}& -\frac{2}{3}\end{array}\right]$$

Compute the Rotation Matrix

Using $\sin ({30}^{\circ })=\frac{1}{2}$ and $\cos ({30}^{\circ })=\frac{\sqrt{3}}{2}$, we substitute:

$$R=I+\sin ({30}^{\circ })K+(1-\cos ({30}^{\circ }))K^2$$

After computation, the rotation matrix is:

$$M_{composite}=\left[\begin{array}{ccc}\frac{2+\sqrt{3}}{3}& \frac{1-\sqrt{3}}{3}& \frac{1}{3}\\ \frac{1}{3}& \frac{2+\sqrt{3}}{3}& \frac{1-\sqrt{3}}{3}\\ \frac{1-\sqrt{3}}{3}& \frac{1}{3}& \frac{2+\sqrt{3}}{3}\end{array}\right]$$

4. Given eye point $(0,-2,2)$, center point $(0,0,0)$, up vector $(0,1,1)$, find the camera frame, the transformation matrix from world frame to camera frame. (ref: code assign0).

$a=P_{eye}-P_{look}=(0,-2,2)$$r=up\times a=(4,0,0)$$u=a\times r=(0,8,8)$

$\hat{a}=\frac{a}{||a||}=(0,-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$

$\hat{r}=\frac{r}{||r||}=(1,0,0)$

$\hat{u}=\frac{u}{||u||}=(0,\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$

Question-4

Please draw $a$, $up$, $r$, $u$ vector on the below pictures, and specify which up you choose.