Theorem 10.2 Let “, v, w be arbitrary vectors in R” and let r be an arbitrary

scalar. Then,

(a) ” Y = ” “,

(b) u.(v + w) = u.v + u'w,

(c) " (F-v) = r(u v) = (ru) Y,

(d) u. u 2 0,

(e) ” ” = 0 implies ” = 0, and

(f) (u + v) (u + v) = ” ” + 2(” v) + Y v.

The Euclidean inner product is closely connected to the Euclidean length of a

vecto*. Since

u u = u: + IL; + + u: and ll”ll = J uf + u; + + u*,/z

Ilull = JiG

Consequently, the distance between two vectors ” and v can be written in terms

of the inner product as

II” VII = J(” - v) (u - v).

Any two vccto˜s” and v in R” determine a plane, as illustrated in Figure IO. 17.

In that plane we can measure the angle 0 between” and Y. The inner product yields

an important connection between the lengths of” and Y and the angle 0 between

u and v.

[lo.41 LENGTH ANDlNNER PRODVCTIN R" 215

Figure

The angle between two vecfor.s in R”. 10.17

Theorem 10.3 Let u and v be two vectors in R”. Let B be the angle between

them. Then,

Remark Recall that to measure the twine of an angle 0 = LBAC as in Figure

10.18, draw the perpendicular from R to a point D on the line containingA and C.

Then_ in the right triangle BAD, the cosine of H is the length of adjacent side AD

divided by the length oC hypotenuse AB. See the Appendix of this book for more

details. If 0 is an obruse angle (between 90 dcgrccs and 270 degrees), then Fig-

ure 10,lYistherelevantdiagramand thecosine of 8is therqariveof llil˜ll/l˜˜ll.

A D c

Figure

cos 0 = llADll/llASll. 10.18

0 ˜,,

” A c

Fizure

216 EUCLIDEAN SPACES [lOI

In either case, cosine 0 lies between - 1 and + I since a leg of a right triangle can

never be longer than the hypotenuse. For us, the important properties of cos 0 are:

if 0 is acute,

cos 0 > 0

if 0 is obtuse,

cos fl < 0

if 0 is a right angle.

cos 0 = 0

Proof of Themem 10.3 The following proof is a bit more complex than the other

proofs we have seen. It uses the Pythagorean Theorem again. Without loss of

we can work with u and v as vectors with tails at the origin 0; say

u = $ and v = z.Let 1 be the line through the vector Y, that is, the line

through the points 0 and Q. Draw the perpendicular line segment m from the

point P (the head of u) to the line X, as in Figure 10.20. Let R be the point

where m meets JJ. Since R lies on &, % is a scalar multiple of Y = OQ. Write

z = fv. Since u, tv, and the segment m are the three sides of the right triangle

OPR, we can write m as the vector u - fv. Since u is the hypotenuse of this

right triangle,

On the other hand, by the Pythagorean Theorem and Theorem 10.2, the

square of the length of the hypotenuse IS:

Ilull = lltvl/2 + Ilu fVl12

= r2˜˜\˜˜˜2 + (u /v) (u - fV)

= f2˜lV/12 + ” ” 2u (f”) + (fV) (IV)

= iZII”II? + llull˜ 2/(u VI + f˜llVl12.

It follows that

(5)

Plugging equation (5) into equation (4) yields

L10.41 LENCiH AND ,NNER PRODUCT IN R” 217

Figure

Choose r so rhut v and u - fv arc perpendicular: 10.20

Thefollowingexample illustrates how onecanusethe innerproduct tocompute

angles explicitly.

Exmple 10.3 We will use the inner product to compute the angle between the

diagonal of a cube and one of its sides. Consider a cube in R” with each side of

™ length c. Position this cube in R™ in the most natural manner, i.e., with vertices

˜at O(O,O,O), P,(c,O, 0). Pz(O,c,O), and P˜(0,O.c). as in Figure 10.21. Write ui

for the vector z for i = I, 2.3. Then, the diagonal d is ut + u2 + “3, which

is the vector (L., c, c).

The angle H between u1 and d satisfies

u, d (c,O,O) (C.C.C)

˜“” = jlu,jl [Id11 = (™ J<z + (.? + $

˜ Using a trig table or calculator. one finds that cos 0 = I/fi implies that

˜ H = 54™44™.

Figure

10.21

EIJCLIDEAN SPACES [lo]

218

Rarely do we care to know that the angle between two vectors is 71” OI 3r/7

radians. More often, we are interested in whether the angle is acute, obtuse, or a

right angle. Since cos 0 is positive when 0 is acute, negative when 0 is obtuse, and

zero when 0 is a right angle, the dot product tells us the information we want by

Theorem 10.3