Calculus: Analytic Geometry and Calculus, with Vectors

2.1 Vectors in E, 55

10 Prove that if Pi, P2, P,, P4 are the vertices of a square having its center at
C, then

OC =OPi -I- OP2 + OP3-I- OP4 4

Hint: For each k we can write OPk = OC + CPk and notice that something can

be said about CPk and CP, when Pk and P, are opposite vertices of the square.

11 Prove that if PI, P2, , P8 are the vertices of a cube having its center

at C, then

OC=OPi+OP2-I- +OP8

8

12 Figure 2 196 shows eight vectors i, j, -i, -j, i', etcetera which are unit

vectors (vectors one unit long) having their tails at the origin of an xy plane and

having their tips on the unit circle with center at the origin.

Discover reasons why the first pair of equations

2

is valid, and then solve the first pair to obtain the second

pair.

-i

i

i

-i

Figure 2.196

13 The vectors u, v, w run in positive (counterclockwise) directions along

three consecutive sides of a regular hexagon. Express w in terms of u and v

Hint: Sketch the hexagon and the line segments from the center that separate it

into six equilateral triangles. Perhaps the simplest observation that can be

made is u + v + w = 2v.

14 From the vertices of a triangle, vectors are drawn to the mid-points of the

opposite sides. Prove that the sum of the three vectors is zero._ _

15 What can be said about the location of Q if PQ = PA + PB + PC + PD.

where 11, B, C, D are the vertices of a square and P is on a side of the square?

16 Abilities to sketch figures and construct formulas involving vectors must

be cultivated. As in Figure 2.197, let u be a unit vector

(vector of unit length) having its tail at 0. Show that

1. Let v be another vector having its tail at the same
   point 0. Show that the vector U defined by U = (v.u)u is
   the vector running from 0 to the projection of the tip of v
   on the line bearing the vector u. Observe that the vector V
   defined by V = v - U or by V = v - (vu)u runs from the

Figure 2.197

tip of U to the tip of v. Verify that V is perpendicular to U by showing that

(vu)u] = 0.

Remark: More opportunities to become familiar with these things will appear

later. Meanwhile, we can note that we have seen the (or a) standard procedure

for resolving a given vector v into vector components parallel and perpendicular

to a given unit vector U.

i+j=Vi' i-(ii'-j')

1

j =-(i'+j')