Part 2: The Angle Between 2 Vectors

Next, let's use the standard inner product to define an angle formed by two vectors. Again, we motivate this definition by considering 2 dimensional vectors. If u and v are two dimensional vectors, then there are angles a and b such that

u = á || u|| cos(a), ||u|| sin(a) ñ        and        v = á ||v|| cos(b),  ||v|| sin(b) ñ

As a result, the inner product of u and v is

u·v = || u|| cos( a)·|| v|| cos( b) +|| u|| sin( a) ·|| v|| sin( b) = || u||  || v|| cos(a) ·cos( b) +|| u|| || v|| sin( a) sin( b) = || u||  || v||   ( cos( a) cos( b) +sin( a)sin( b) )

The difference of the angles identity for the cosine thus leads us to

u·v = || u||  || v||  cos( b-a)

However, q = b - a is the angle formed when u and v share the same initial point:

This leads us define the angle between any two nonzero vectors using the inner product:       

Definition 1.3: Any two non-zero vectors u and v determine an angle q which satisfies

u·v = || u||  || v||  cos(q)

By convention, we often choose q to satisfy 0 < q < 2p.           

EXAMPLE 3    Find the angle in degrees between the vectors u = á -2,2,1 ñ and v = á2,3,6 ñ .       

Solution: To begin with,

|| u|| = 4+4+1 = 3,         || v|| =  4+9+36   = 7

Definition 1.3 implies that q satisfies the equation

cos( q) = u·v ||u||  ||v||

(1)

so that u·v = -2·2+2·3+1·6 = 8 implies that

cos( q) =  8 3·7  =  8 21

A calculator or computer can be used to obtain

q = cos-1 8 21 = 1.17997  radians

Multiplication of q = 1. 17997 by 180^°/p then yields q = 67.61^°.

Maple/Javaview Figure

       

Definition 2.2 also implies that if two vectors u and v form a right angle (q = p/2 ), then their inner product is 0. In particular, we say that u and v are orthogonal if they form a right angle, so that definition 2.2 leads to the following:

Theorem 2.3: The vectors u and v are orthogonal if and only if u·v = 0.

       

Sometimes we write u ^ v to indicate that u and v are orthogonal, so that Theorem 2.3 can be interpreted

u · v = 0       Û     u ^ v

where Û means ïf and only if."       

EXAMPLE 4    Find a number k such that u = á 2,3,4 ñ is orthogonal to v = ák,3,-7 ñ .       

Solution: To do so, we compute u·v and set it equal to 0:

u·v  =  0 2·k+3·3-4·7  =  0 2k-19  =  0 k  =  19 2  = 9.5

Thus, v = á 9.5,3,-7 ñ is orthogonal to u = á 2,3,4 ñ .

Maple/Javaview Figure

       

EXAMPLE 5    Show that i, j, and k are mutually orthogonal.            

Solution: Since i = á 1,0,0 ñ and j = á 0,1,0 ñ , their inner product satisfies

i·j = á 1,0,0 ñ · á 0,1,0 ñ = 1·0+0·1+0·0 = 0

Since i · j = 0, the vectors i and j are orthogonal. Likewise, j · k = 0 and k · i = 0, showing that j ^ k and k ^ i.  

       

In fact, the vectors i, j, and k satisfy the following:

i·i = j·j = k·k = 1 (2) i·j = j·k = k·i = 0 (3)

Any set of 3 vectors that satisfies (2) and (3) is known as an orthonormal basis for the vector space of 3-dimensional vectors. Thus, i, j, and k form an orthonormal basis.

LiveGraphics3d Applet Click and drag red dot to see i, j, k orthonormal basis.

In chapter 3, we will encounter a different orthonormal basis for the vector space of 3-dimensional vectors.

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Check your Reading: For what value of k is v = á -7, k ñ  orthogonal to u = á 3, 7 ñ ?