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The pyramid defined by the coordinates A(0,0,0), B(1,0,0), C(0,1,0) and D(0,0,1) is rotated $45^o$ about the line L that has the direction V = J + K and passing through point C(0,1,0).
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Find the coordinates of the rotated figure.

Solution:

enter image description here

Step 1: Translating line (i.e Point C) to the origin O(0,0,0)

$[T r]=\left[\begin{array}{lll}{1} & {0} & {0} \\ {0} & {1} & {0} \\ {0} & {0} & {1} \\ {t x} & {t y} & {t z}\end{array}\right]$

$\left[\mathrm{T}_{\mathrm{r}}\right]=\left[\begin{array}{cccc}{1} & {0} & {0} & {0} \\ {0} & {1} & {0} & {0} \\ {0} & {0} & {1} & {0} \\ {t x} & {t y} & {t z} & {1}\end{array}\right]=\left[\begin{array}{cccc}{1} & {0} & {0} & {0} \\ {0} & {1} & {0} & {0} \\ {0} & {0} & {1} & {0} \\ {0} & {-1} & {0} & {1}\end{array}\right]$

$\{t x=0, t y=-1, t z=0\}$

Step 2: Aligning $\overline{V}$ with Positive z-axis

$\overline{\mathrm{V}}=\mathrm{J}+\mathrm{K}$

On comparing with

$\mathrm{V} \equiv \mathrm{ai}+\mathrm{bj}+\mathrm{ck}$

$\begin{aligned} a=0, b=1, c=1 \\ \text { also }|\overline V|=\sqrt{1^{2}+1^{2}} &=\sqrt{2} \\ \text { and } =\sqrt{b^{2}+c^{2}}=\lambda=\sqrt{2} \end{aligned}$

Now,

$[A v]=\left[\begin{array}{cccc}{\frac{\lambda}{|\overline V|}} & {0} & {\frac{a}{|\overline V|}} &0 \\ {\frac{-ab}{\lambda|\overline V|}} & {\frac{c}{\lambda}} & {\frac{b}{|\overline V|}} & 0 \\ {\frac{-ac}{\lambda|\overline V|}} & {\frac{-b}{\lambda}} & {\frac{c}{|\overline V|}} & 0\\ {0} & {0} & {0} & {1}\end{array}\right]$

$=\left[\begin{array}{cccc}{1} & {0} & {0} & {0} \\ {0} & {1 / \sqrt{2}} & {1 / \sqrt{2}} & {0} \\ {0} & {-1 / \overline{2}} & {1 / \sqrt{2}} & {0} \\ {0} & {1} & {0} & {1}\end{array}\right]$

Step 3: Rotation of the object about z-axis by an angle of $\theta= 45^o$ in CCW direction

$[\mathrm{Rz}]=\left[\begin{array}{cccc}{\cos \theta} & {\sin \theta} & {0} & {0} \\ {-\sin \theta} & {\cos \theta} & {0} & {0} \\ {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {1}\end{array}\right]=\left[\begin{array}{cccc}{1 / \sqrt{2}} & {1 / 2} & {0} & {0} \\ {-1 / \sqrt{2}} & {1 / \sqrt{2}} & {0} & {0} \\ {0} & {0} & {1} & {0} \\ {0} & {0} & {0} & {1}\end{array}\right]$

Step 4: Aligning Z with V

$[A v]^{-1}=\left[\begin{array}{cccc}{\frac{\lambda}{|\overline V|}} & {\frac{-ab}{\lambda|\overline V|}} & {\frac{-ac}{\lambda|\overline V|}} &0 \\ {0} & {\frac{c}{\lambda}} & {\frac{-b}{\lambda}} & 0 \\ {\frac{a}{|\overline V|}} & {\frac{b}{|\overline V|}} & {\frac{c}{|\overline V|}} & 0\\ {0} & {0} & {0} & {1}\end{array}\right]$

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

Step 5: Inverse Translation

$[\mathrm{Tr}]=\left[\begin{array}{cccc}{1} & {0} & {0} & {0} \\ {0} & {1} & {0} & {0} \\ {0} & {0} & {1} & {0} \\ { t x} & {t y} & {t z} & {0}\end{array}\right]=\left[\begin{array}{cccc}{1} & {0} & {0} & {0} \\ {0} & {1} & {0} & {0} \\ {0} & {0} & {1} & {0} \\ {0} & {1} & {0} & {1}\end{array}\right]$

Now,

The combined transformation matrix

$[\mathrm{T}]=[\mathrm{Tr}][\mathrm{R}][\mathrm{M}]|\mathrm{R}|^{-1}[\mathrm{Tr}]^{-1}$

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

$[\mathrm{T}]=\left[\begin{array}{cccc}{0.707} & {0.5} & {-0.5} & {0} \\ {-0.5} & {0.85} & {0.146} & {0} \\ {0.5} & {0.146} & {0.85} & {0} \\ {0.5} & {0.15} & {-0.146} & {1}\end{array}\right]$

Now, new resultant coordinates are:

$\left[\mathrm{X}^{\prime}\right]=[\mathrm{X}][\mathrm{T}]$

$\left[\begin{array}{c}{A^{\prime}} \\ {B^{\prime}} \\ {C^{\prime}} \\ {D^{\prime}}\end{array}\right]=\left[\begin{array}{cccc}{1} & {0} & {0} & {0} \\ {0} & {1} & {0} & {0} \\ {0} & {0} & {1} & {0} \\ {0} & {1} & {0} & {1}\end{array}\right]\left[\begin{array}{cccc}{0.707} & {0.5} & {-0.5} & {0} \\ {-0.5} & {0.85} & {0.146} & {0} \\ {0.5} & {0.146} & {0.85} & {0} \\ {0.5} & {0.15} & {-0.146} & {1}\end{array}\right]$

$\left[\begin{array}{c}{A^{\prime}} \\ {B^{\prime}} \\ {C^{\prime}} \\ {D^{\prime}}\end{array}\right]=\left[\begin{array}{cccc}{0.5} & {0.15} & {-0.146} & {1} \\ {1.207} & {0.6} & {-0.646} & {1} \\ {0} & {1} & {0} & {1} \\ {1} & {0.296} & {0.704} & {1}\end{array}\right]$

Hence,

$A^{\prime}=(0.5,0.15,-0.146)$

$B^{\prime}=(1.207,0.6,-0.646)$

$C^{\prime}=(0,1,0)$

$D^{\prime}=(1,0.296,0.704)$

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