Let A = (a1, a2, . . . , an), B = (b1, b2, . . . , bn), and C = (c1, c2, . . . , cn) be vectors inRn, and letk be a real number. Then
(1) AãB=BãA
(2) Aã(B+C) =AãB+AãC (3) AãkB=kAãB =k(AãB)
(4) AãA≥0, andAãA= 0if and only if A=O
Proof. Here are the proofs of (1), (3), and (4). The proof of (2) is left as an exercise.
(1) AãB = (a1, a2, . . . , an)ã(b1, b2, . . . , bn)
= a1b1+a2b2+ã ã ã+anbn
= b1a1+b2a2+ã ã ã+bnan
= (b1, b2, . . . , bn)ã(a1, a2, . . . , an)
= BãA
(3) AãkB = (a1, a2, . . . , an)ãk(b1, b2, . . . , bn)
= (a1, a2, . . . , an)ã(kb1, kb2, . . . , kbn)
= a1(kb1) +a2(kb2) +ã ã ã+an(kbn)
= k(a1b1) +k(a2b2) +ã ã ã+k(anbn)
= k(a1b1+a2b2+ã ã ã+anbn)
= k(AãB)
The proof thatkAãB=k(AãB) is exactly the same. Habits of Mind Note that in the equation kAãB = k(AãB), the insertion of parentheses changes the object that is being multiplied by k.
On the left side, you are multiplyingkbyA, a vector inRn; on the right side, you are multiplyingk byAãB, a real number.
(4) AãA = (a1, a2, . . . , an)ã(a1, a2, . . . , an)
= a21+a22+ã ã ã+a2n
Now, the sum of squares of real numbers is nonnegative, and such a sum is 0 if and only if eachai= 0.
In the proof of part (4), you see the equation AãA=a21+a22+ã ã ã+a2n
2.2 Dot Product
The right-hand side of that equation should look familiar—you saw it in the definition of the length of a vector,
A=
a21+a22+ã ã ã+a2n
So you can substitute AãAfor a21+a22+ã ã ã+a2n to get a more efficient way to write the length of a vector.
Theorem 2.2
IfA is a vector inRn,A=√ AãA.
Example 5
Problem. Show that ifAand B are vectors inRn,
(A+B)ã(A+B) =AãA+ 2(AãB) +BãB
Solution. The proof of this fact is exactly the same as the proof from elementary algebra that established the identity (a+b)2=a2+ 2ab+b2.
(A+B)ã(A+B) = (A+B)ãA+ (A+B)ãB (Theorem 2.1 (2))
=Aã(A+B) +Bã(A+B) (Theorem 2.1 (1))
=AãA+AãB+BãA+BãB (Theorem 2.1 (2))
=AãA+AãB+AãB+BãB (Theorem 2.1 (1))
=AãA+ 2(AãB) +BãB
Example 6
Problem. LetA= (1,4,0,1). For what values ofciscAãcA= 72?
Solution. Using part (3) of Theorem 2.1, you have
cAãcA=c(AãcA) =c2(AãA) SinceAãA= 18, this becomes 18c2= 72, soc=±2.
Example 7
Problem. LetA andB be vectors inRn, withB =O. Show that ←−
This example will be important in the next lesson.
A−AãB
BãB
B
ãB= 0
Solution. An equation like this can be confusing, since it mixes operations between vectors and numbers. It may help to first read through the equation to check that the operations are working on the right kind of input.
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• AandB are vectors.
• AãB andBãB (each the dot product of two vectors) are both numbers.
• Thus, ABããBB (the quotient of two numbers) is also a number. ←−
Note that since B = O, BãB >0, so division by BãB is okay.
• AãB
BãB
B (a scalar multiple of a vector) is a vector.
• That meansA−AãB
BãB
B (the difference of two vectors) is a vector.
• Finally,
A−AãB
BãB
B
ãB (the dot product of two vectors) is a number.
To see that this number is 0, use Theorem 2.1.
A−AãB
BãB
B
ãB =AãB−AãB
BãB
B
ãB
=AãB−AãB
BãB
(BãB)
=AãB−AãB= 0
The next two examples show how the basic rules for dot product can be applied to geometry.
Example 8
Problem. SupposeAandBare nonzero orthogonal vectors inRnandc1A+c2B=O.
Show thatc1=c2= 0.
Solution. Take the equationc1A+c2B=Oand dot both sides withA:
Aã(c1A+c2B) =AãO Aã(c1A) +Aã(c2B) = 0 c1(AãA) +c2(AãB) = 0
SinceAis orthogonal toB,AãB= 0, so this last equation becomesc1(AãA) = 0. Since A=O,AãA >0. Sincec1(AãA) = 0, it follows thatc1= 0. To provec2= 0, take the equationc1A+c2B=O and dot both sides withB.
Example 9
Problem. Consider the triangle inR3 whose vertices are A= (3,2,5), B = (5,2,1), andC= (2,1,3). Show that the angle formed by−→
CAand−−→
CB is a right angle.
Solution. You want to show that the angle formed by−→
CA and −−→
CB is a right angle.
But −→
CA is equivalent to A−C and −−→
CB is equivalent to B −C. So, you only have to show that A− C = (1,1,2) is orthogonal to B −C = (3,1,−2). And it is:
(A−C)ã(B−C) = (1,1,2)ã(3,1,−2) = 0.
2.2 Dot Product
One of the most beautiful theorems in mathematics is the Pythagorean Theorem. Does it extend toRn?
InR2, ifAandBare nonzero vectors that aren’t scalar multiples of each other, thenA+B and A−B are the two diagonals of the parallelogram whose sides areAandB. So the lengths of the diagonals are A+Band A−B.
Now, from plane geometry, if the diagonals of a parallelogram have the same length, the parallelogram is a rectangle. So, if the parallelogram determined byAandB is a rectangle, thenAis perpendicular toB. Thus, Ais perpendicular toBif and only if the diagonals,A+BandA−B, have the same length. The same fact is true inRn, but the proof is via algebra.
Lemma 2.3
If A and B are vectors in Rn, A is orthogonal to B if and only if
←−A lemma is a result that’s needed to prove another result. In this case, Lemma 2.3 is needed to prove Theorem 2.4. Usu- ally, people discover that they need some fact when they try to prove a theo- rem, so they call that fact a lemma and prove it sep- arately. The German word for lemma is hilfsatz—
“helping statement.” Note the connection between the words “lemma” and
“dilemma.”
A+B=A−B.
Proof. SupposeA+B=A−B. By Theorem 2.2 and squaring both sides, you have (A+B)ã(A+B) = (A−B)ã(A−B). This simplifies to 4(AãB) = 0, so AãB = 0, and A is orthogonal to B. The proof of the converse is just as simple.
In the next figure,A,B, and A−B are three sides of a triangle. Since Ais orthogonal toB, it follows from Lemma 2.3 thatA−B=A+B, so the hypotenuse of the right triangle whose legs areAandB has length A+B.
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