classmate
Date
Page
Assume
2 vectors
AB = i + } & Bc =
Then a vector of length
5 m
perpendicular to both
both ă & ' is
and
Answers
Answer:
It may not be immediately clear that the question makes sense, but it's not hard to turn it into a question that does. Since vectors have no position, we are as usual free to place vectors wherever we like. If the two vectors are placed tail-to-tail, there is now a reasonable interpretation of the question: we seek the measure of the smallest angle between the two vectors, in the plane in which they lie. Figure 12.3.1 illustrates the situation.
θAB
Figure 12.3.1. The angle between vectors A and B.
Since the angle θ lies in a triangle, we can compute it using a bit of trigonometry, namely, the law of cosines. The lengths of the sides of the triangle in figure 12.3.1 are |A|, |B|, and |A−B|. Let A=⟨a1,a2,a3⟩ and B=⟨b1,b2,b3⟩; then
|A−B|2 =|A|2+|B|2−2|A||B|cosθ 2|A||B|cosθ =|A|2+|B|2−|A−B|2 =a
2
1
+a
2
2
+a
2
3
+b
2
1
+b
2
2
+b
2
3
−(a1−b1)2−(a2−b2)2−(a3−b3)2 =a
2
1
+a
2
2
+a
2
3
+b
2
1
+b
2
2
+b
2
3
−(a
2
1
−2a1b1+b
2
1
)−(a
2
2
−2a2b2+b
2
2
)−(a
2
3
−2a3b3+b
2
3
) =2a1b1+2a2b2+2a3b3 |A||B|cosθ =a1b1+a2b2+a3b3 cosθ =(a1b1+a2b2+a3b3)/(|A||B|)
So a bit of simple arithmetic with the coordinates of A and B allows us to compute the cosine of the angle between them. If necessary we can use the arccosine to get θ, but in many problems cosθ turns out to be all we really need.
The numerator of the fraction that gives us cosθ turns up a lot, so we give it a name and more compact notation: we call it the dot product, and write it as
A⋅B=a1b1+a2b2+a3b3.
Explanation: