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\section*{Chapter 1 Solutions}
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1.1 We can apply the formula from page 18
$$ \theta = 2 \tan^{-1}\frac{h}{2f},$$
where we have replaced the length of the camera $d$ with the
focal length of the lens $f$. Solving for $f$, we have
$$f=\frac{h}{2}\cot\frac{\theta}{2}.$$
We can use the diagonal of the 24 mm $x$ 36 mm file for $h$.
Solving, we find 
$$f=6\sqrt{13}\cot 45 = 6\sqrt{13}=21.6 mm.$$

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1.3 Suppose that the line segment is between the points $(x_1,y_1)$ and $(x_2,y_2.$
We can use the endpoints of the line segment to determine the
slope and y intercept of a line of which the segment is part,i.e.
$$y=mx+h=\frac{y_2-y_1}{x-2-x_1}y+y_1-\frac{y_2-y_1}{x-2-x_1}x_2.$$
Note that we can deal with horizontal and vertical line segments as special cases.
We can find the intersectiosn with the sides of the window by substituting
$y=y_{max}$, $y=y_{min},$, $x=x_{max}$, and $x=x_{min}$ (the equations
for the sides of the window) into the above equation. We can check the locations
of the points of intersection to detemine if they are on the line segment or
only on the line of whcih it is part.


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1.5 Here we have to intersect the line segment against the polygons that
determine the sides of the window. Each of these polygons is part of a plane.
The equation of each plane is of the form
$$ax+by+cz+d=0,$$
where the coefficients $a$, $b$, $c$, and $d$ can be determined from the
vertices of the parallelepiped. The line segment is either parallel to
a particular plane or intersects it in exactly on place that can be
detemined from the equation of the line. For a line passing through the
two three--dimensional points $(x_1,y_1,z_1)$ and $x_2,y_2,z_2$, we can
write its equation in parametric form as 
$$x(t)=(1-t)x_1+tx_2,$$
$$y(t)=(1-t)y_1+ty_2,$$
$$z(t)=(1-t)z_1+tz_2.$$
We can substitue this equation in the equation for a plane and determine
$t$ for the point of intersection.

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1.7 We have to process 1280 x 1024 x 72 pixels/sec. If we process each
succesively, there is only about 10 nanoseconds to process each. For
a 480 x 640 interlaced display operating at 60 Hz we must process only
480 x 640 x 30 pixels/sec which gives us about 109 nanoseconds to process
each pixel.


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1.9 Good examples for this problem are architecture, mechanical parts design,
electrical circuit design and a paint program.

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1.11 A 1024 x 1280 display has a 4 to 5 aspect ratio. Hence, if the diagonal
is 50 cm and we want square pixels, the screen must be approximately
31 cm x 39 mm. Each pixel is then about 0.3 mm on each side. A smooth display
will require about 3 triads for each pixel, and thus the triads are about
0.1 mm apart. Finally if the shadow mask is
halfway between the screen and electron guns, the shadow mask spacing is
half the triad spacing.
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