## LECTURE 3: DOT PRODUCT CONTINUED AND THE CROSS PRODUCT

LECTURE 3: DOT PRODUCT CONTINUED AND THE CROSS PRODUCT DAGAN KARP A BSTRACT. In this lecture we’ll conclude our initial study of the dot product, and ...
Author: Nicholas Walton
LECTURE 3: DOT PRODUCT CONTINUED AND THE CROSS PRODUCT DAGAN KARP A BSTRACT. In this lecture we’ll conclude our initial study of the dot product, and begin our study of the cross product.

1. D OT PRODUCT CONTINUED Last time we discovered the dot product and learned some of its basic properties. Let’s study an example. Example 1. What is the angle between a = (1, 0, 1) and b = (1, 1, 0)? The answer is given in terms of the dot product: a·b . θ = cos−1 |a| |b| So we compute   1+0+0 −1 √ √ θ = cos 2 2   1 = cos−1 2 π = . 3 1.1. Projections. Another convenient use of the dot product is to determine the projection of one vector onto another. The idea is the following: any two vectors determine a plane (we’ll see this in detail soon). So given vectors a and b, we can ask: what is the orthogonal projection of b onto the line (through the origin) with direction a? See Figure 1.1. It is worth reiterating that b is projecting onto the line determined by a and not the vector a itself. Let θ be the angle between a and b. Then the projection proja b of b onto a is given by proja b = |b|| cos θ|. In order to express this otherwise, it is helpful to be acquainted with unit vectors. Definition 2. A unit vector is simply a vector of unit length, i.e. a is a unit vector if and only if |a| = 1. a Remark 3. Let a be a nonzero vector. The unit vector v in the direction of a is v = . |a| To see that this is true, note that v is a scalar multiple of a, and hence parallel to a. Also, |v| = 1. Date: March 12, 2009. 1

b θ

a projab F IGURE 1. The projection of vector b onto the vector a. We now return to our projection. Let’s compute. proja b = |b|| cos θ| |b| |a|| cos θ| |a| a·b = |a|   a·b a = |a| |a|   a·b = a. a·a =

(1)

Here Equation 1 follows from multiplication by the unit vector in the direction of a; this is allowed as proja b is in the direction of a by definition, hence multiplication by the unit vector in the direction of a changes neither the magnitude nor direction of proja b. Example 4. Note that this projection generalizes the coordinate projections. The coordinate projections are maps πi : R3 → R, for i = 1, 2, 3, given by πi (b1 , b2 , b3 ) = bi . Here b = (b1 , b2 , b3 ) is any vector in R3 . We may also compute the projection of (b1 , b2 , b3 ) onto the coordinate axes given our new construction. For example, projection in the direction of the x-axis is given by π1 and also projection onto i. We compute to show agreement.   b·i proji b = i i·i   (b1 , b2 , b3 ) · (1, 0, 0) = · (1, 0, 0) (1, 0, 0) · (1, 0, 0)   b1 = (1, 0, 0) 1 = (b1 , 0, 0) = (π1 (b), 0, 0). Similar agreement holds for the projections π2 and π3 . 2

2. C ROSS PRODUCT Definition 5. Let a = (a1 , a2 , a3 ) and b = (b1 , b2 , b3 ). The cross product of a and b is the vector a × b = (a2 b3 − a3 b2 , a3 b1 − a1 b3 , a1 b2 − a2 b1 ). Remark 6. The cross product is given by the determinant of the following matrix.   i j k a × b = det a1 a2 a3  . b1 b2 b3 Theorem 7. The vector a × b is orthogonal to both a and b. P ROOF. Recall that two vectors are orthogonal if and only if their dot product is zero. Hence to prove this theorem is suffices to show (a × b) · a = (a × b) · bb = 0. We will show the first equality and omit the proof of the second. So let’s compute. (a × b) · a = a1 (a2 b3 − a3 b2 ) + a2 (a3 b1 − a1 b3 ) + a3 (a1 b2 − a2 b1 ) = a1 a2 b3 − a1 a3 b2 + a2 a3 b1 − a2 a1 b3 + a3 a1 b2 − a3 a2 b1 = 0.  Theorem 8. Let 0 6 θ 6 π be the angle between a and b. Then |a × b| = |a| |b| sin θ. P ROOF. We compute the square of what we desire. |a × b|2 = (a2 b3 − a3 b2 )2 + (a3 b1 − a1 b3 )2 + (a1 b2 − a2 b1 )2

= (a21 + a22 + a23 )(b21 + b22 + b23 ) − (a1 b1 + a2 b2 + a3 b3 )2

= |a|2 |b|2 − (a · b)2

= |a|2 |b|2 − |a|2 |b|2 cos2 θ = |a|2 |b|2 (1 − cos2 θ) = |a|2 |b|2 sin2 θ. Now note that sin θ > 0 as 0 6 θ 6 π. Therefore we may take square roots and |a × b| = |a| |b| sin θ.



Corollary 9. Two nonzero vectors a and b are parallel if and only if a × b = 0. P ROOF. Nonzero vectors a and b are parallel if and only if θ = 0 or θ = π. In either case, sin θ = 0. Hence |a × b| = 0, and thus a × b is the zero vector. Proposition 10. For any two nonzero vectors a and b, the length of the cross product of a and b is equal to the area of the parallelogram determined by a and b. 3

b

|b| sin θ a

F IGURE 2. The length of the cross product is the area of the area of the parallelogram P ROOF. The area of the parallelogram is given by base times height. We may take the base as |a| and the height is then given by |b| sin θ, where θ continues to be the angle between a and b. Then the area A is given by A = |a|(|b| sin θ) = |a| |b| sin θ = |a × b|.



Theorem 11. Let a, b, c be vectors in R3 and k ∈ R be a scalar. Then, (1) (2) (3) (4) (5)

a × b = −b × a (ka) × b = k(a × b) = a × (kb) a × (b + c) = a × b + a × c a · (b × c) = (a × b) · c a × (b × c) = (a · c)b − (a · b)c.

P ROOF. There is nothing to do here but write out each vector in components and use the definition of dot and cross product. I’ll leave that for your enjoyment. As a last remark about the cross product, here’s a geometric interpretation of the product in part (4) above. In fact this product has a special name; it is called the triple product. Proposition 12. The volume V of the parallelepiped determined by a, b, c is given by the magnitude of their tipple product; V = a · (b × c). P ROOF. The volume of the parallelepiped is given by the product of the height and area of the base of the parallelepiped. We may choose any two vectors as the base. Let’s choose b and c for the base. Then the base of this parallelepiped is a parallelogram and has area A = |b × c|. Now the vector b × c is orthogonal to the base of the parallelepiped, which is the parallelogram determined by b and c. So let θ be the angle between a and b × c. Then the height of the parallelepiped is given by h = |a| | cos θ|, where we take absolute value in case θ > π/2. Therefore the volume is given by V = Ah = |b × c| |a| | cos θ| = |a · (b × c)|.

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

b×c θ a c b F IGURE 3. The parallelepiped.

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