Calculus: Analytic Geometry and Calculus, with Vectors

(lu) #1
2.1 Vectors in Ea 57

interval 0 < 0 < a, is (as in trigonometry) an angle which the line from 0 to P
makes with the positive x axis, then


(2) V = Ivj cos Oi + lvi sin Oj.

Let w be the vector obtained by rotating the vector v through a right angle in
the positive (counterclockwise) direction, so that (as in trigonometry) 0 + 7r/2
is one of the angles which the vector w makes with the positive x axis. Show that


w = IvIcos(0+2)1+jvj sin (0+)i


w = Ivi (-sin 0)i + IV! (cos 0)j

(5) w = -yi + xj.
Remark: While our present interest lies in vectors, our result is equivalent to
the fact that, whatever x and y may be, if we start at P(x,y), the point P having
coordinates x and y, and run in the positive direction along a quadrant of a circle
having its center at the origin, we will stop at the point Q(-y,x). This fact
implies and is implied by the formulas

(6) cos(0+ 2) = -sin 0, in (0 + 2) = cos 0

which were used to obtain (4) from (3).
20 Sketch some figures and discover the circumstances under which two
nonzero vectors u and v are such that lu + vi = lu - vi. Then prove that

(1) lu + vi2 = (u + v).(u + v) = vv
and
(2)

21 The span of the set of n vectors v1, v2, , vnis the set of vectors v
representable in the form
V = ciVi + c2v2 + + cuV.,
where c1, c2, , cn are scalars. Show that the span of the set of three given
vectors v1i v2, v3 is the same as the span of the set of three vectors u1, u2, U3
defined by the system of equations
ui = V1 + V2
U2 = V2 + Va
ua=V1 +Va.
Hint: The proof consists of two parts. Suppose first that w belongs to the span
of u1, u2, ua and seek an easy way to show that w must belong to the span of
vl, v2, Va. It remains to suppose that w belongs to the span of vi, v2, v3 and then
show that w must belong to the span of u1, u2, u3. As a start, solve the given
system of equations for v1, V2, va. One of the results is
V1 = gul - gut + U3.
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