--- /dev/null
+\begin{problem*}{3.24}
+Given the vectors $\vect{A}=2.00\ihat+6.00\jhat$ and
+$\vect{B}=3.00\ihat-2.00\jhat$, \Part{a} draw the vector sum
+$\vect{C}=\vect{A}+\vect{B}$ and the vector difference
+$\vect{D}=\vect{A}-\vect{B}$. \Part{b} Calculate $\vect{C}$ and
+$\vect{D}$, in terms of unit vectors. \Part{c} Calculate $\vect{C}$
+and $\vect{D}$ in terms of polar coordinates, with angles measured
+with respect to the positive $x$ axis.
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{3.28}
+In a game of American football, a quarterback takes the ball from the
+line of scrimmage, runs backward a distance of $10.0\U{yards}$, and
+then runs sideways parallel to the line of scrimmage for
+$15.0\U{yards}$. At this point, he throws a forward pass downfield
+$50.0\U{yards}$ perpendicular to the line of scrimmage. What is the
+magnitude of the football's resultant displacement?
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{3.29}
+The helicopter view in Fig.~P3.29 shows two people pulling on a
+stubborn mule. The person on the right pulls with a force
+$\vect{F}_1$ of magnitude $120\U{N}$ and direction
+$\theta_1=60.0\deg$. The person on the left pulls with a force
+$\vect{F}_2$ of magnitude $80.0\U{N}$ and direction of
+$\theta_2=75.0\deg$. Find \Part{a} the single force that is
+equivalent to the two forces shown and \Part{b} the force that a third
+person would have to exert on the mule to make the resultant force
+equal to zero. The forces are measured in units of newtons
+(symbolized $\bareU{N}$).
+\begin{center}
+\begin{asy}
+draw((0,0)--(1,1));
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{3.36}
+Three displacement vectors of a croquet ball are shown in Figure
+P3.36, where $|\vect{A}| = 20.0\U{units}$, $|\vect{B}| =
+40.0\U{units}$, and $|\vect{C}| = 30.0\U{units}$. Find \Part{a} the
+resultant in unit-vector notation and \Part{b} the magnitude and
+direction of the resultant displacement.
+\begin{center}
+\begin{asy}
+import graph;
+import Mechanics;
+
+real u = 0.1cm;
+
+Vector a = Vector((0,0), mag=20u, dir=90, "$\vect{A}$");
+a.draw();
+Vector b = Vector((0,0), mag=40u, dir=45, "$\vect{B}$");
+b.draw();
+Vector c = Vector((0,0), mag=30u, dir=-45, "$\vect{C}$");
+c.draw();
+
+xaxis("$x$");
+yaxis("$y$");
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{3.43}
+You are standing on the ground at the origin of a coordinate system.
+An airplane flies over you with constant velocity parallel to the $x$
+axis at a fixed height of $7.60\E{3}\U{m}$. At time $t=0$, the
+airplane is directly above you so that the vector leading from you to
+it is $\vect{P}_0=7.60\E{3}\jhat\U{m}$. At $t=30.0\U{s}$, the
+position vector leading from you to the airplane is
+$\vect{P}_{30}=(8.04\E{3}\ihat+7.60\E{3}\jhat)\U{m}$ as suggested in
+Figure P3.43. Determine the magnitude and orientation of the
+airplane's position vector at $t=45.0\U{s}$.
+\begin{center}
+\begin{asy}
+import Mechanics;
+
+real u = 1cm;
+real h = 1u;
+real d = 8.04/7.6*h;
+real dx = 0.2u;
+
+draw((-dx,0)--(d+dx,0));
+draw((-dx,h)--(d+dx,h));
+
+Vector A = Vector();draw((0,0)--(0,h));
+A.draw();
+Vector B = Vector();((0,0)--(d,h));
+B.draw();
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\end{solution}
--- /dev/null
+\begin{problem*}{3.63}
+A rectangular parallelpiped has dimensions $a$, $b$, and $c$ as shown
+in Figure 3.63. \Part{a} Obtain a vector expression for the face
+diagonal vector $\vect{R}_1$. \Part{b} What is the magnitude of this
+vector? \Part{c} Notice that $\vect{R}_1$, $c\khat$, and $\vect{R}_2$
+make a right triangle. Obtain a vector expression for the body
+diagonal vector $\vect{R}_2$.
+\begin{center}
+\begin{asy}
+import graph3;
+import Mechanics;
+
+real u = 1cm;
+
+real a = 1u;
+real b = 1u;
+real c = 1u;
+
+xaxis3(Label("$x$"), 0, 1.5a);
+yaxis3(Label("$y$"), 0, 1.5b);
+zaxis3(Label("$z$"), 0, 1.5c);
+
+draw(xscale3(a)*yscale3(b)*zscale3(c)*unitcube);
+//Vector A = Vector((0,0,0),
+\end{asy}
+\end{center}
+\end{problem*}
+
+\begin{solution}
+\end{solution}