1. Where the length unit is not specified (in this solution), the unit meter should be understood.
(a) The position vector, according to Eq. 4-1, is r =−5.0ˆ
i+8.0ˆ
j (in meters).
(b) The magnitude is |r|=x2+y2+z2=9.4m.
(c) Many calculators have polar ↔rectangular conversion capabilities which make this computation
more efficient than what is shown below. Noting that the vector lies in the xy plane, we are using
Eq. 3-6:
tan−18.0
−5.0=−58◦or 122◦
where we choose the latter possibility (122◦measured counterclockwise from the +xdirection) since
the signs of the components imply the vector is in the second quadrant.
(d) In the interest of saving space, we omit the sketch. The vector is 32◦counterclockwise from the
+ydirection, where the +ydirection is assumed to be (as is standard) +90◦counterclockwise from
+x,andthe+zdirection would therefore be “out of the paper.”
(e) The displacement is ∆r =r −r where r is given in part (a) and r =3.0ˆ
i. Therefore, ∆r =
8.0ˆ
i−8.0ˆ
j (in meters).
(f) The magnitude of the displacement is |∆r|=82+(−8)2=11m.
(g) The angle for the displacement, using Eq. 3-6, is found from
tan−18.0
−8.0=−45◦or 135◦
where we choose the former possibility (−45◦,whichmeans45
◦measured clockwise from +x,or
315◦counterclockwise from +x) since the signs of the components imply the vector is in the fourth
quadrant.