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# Scalars and Vectors Study Guide (page 3)

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Updated on Sep 26, 2011

#### Example 3

Find the magnitude of the resultant vector of the following vectors: (3,–5,–2) m and (–2,0,8) m, and of the difference of the same vectors.

#### Solution 3

The components of the resultant are:

Rx = Ax + Bx = 3 – 2 = 1 m

Ry = Ay + By = –5 + 0 = –5 m

Rz = Az + Bz = – 2 + 8 = 6 m

Therefore:

R2 = 1 + 25 + 36 = 62 m2

and the magnitude of the resultant vector is:

R = 7.9 m

For the difference:

AB = A + (–B)

and the components of the A + (– B) are:

Rx = Ax + (–Bx) = 3 + [– (–2)] = 5 m

Ry = Ay + (–By) = –5 + 0 = –5 m

Rz = Az + (–Bz) = –2 + (–8) = –10 m

Therefore:

R2 = 25 + 25 + 100 = 150 m2

and the magnitude of the resultant vector is:

R = 12 m

We call the scalar product (or dot product) of two vectors A and B a scalar quantity. A · B is obtained by multiplying the magnitudes of the two vectors and the cosine of the angle made by their directions (see Figure 2.9).

A · B = A · B · cos θ

#### Example 4

What is the angle between two vectors that have a scalar product equal to zero?

#### Solution 4

When the scalar product of two vectors of nonzero magnitude is zero:

A · B = 0

It follows that the cosine of the angle made by their directions should be zero, and therefore, the two vectors must be perpendicular to each other.

cos θ = 0

θ = π/2

For the scalar product, the order of multiplication is not important: A · B = B · A (scalar product is commutative). We apply the previous definition to both sides of the equality:

A · B = A · B · cos θ

and

B · A = B · A · cos (–θ )

But A and B are numbers, so A · B will equal B. A, and the cosine function is symmetric: cos θ = cos (–θ). Hence, both sides yield the same answer.

We call the vector product (or cross product) of two vectors A and B a vector quantity: A × B with a magnitude obtained by multiplying the magnitudes of the two vectors and the sine of the angle made by their directions. The direction of the resultant vector is perpendicular on the plane defined by these two vectors (see Figure 2.10).

C = A × B

C = A · B · sin θ

The direction of the resultant vector is given by the right-hand rule (also known as the corkscrew rule). If you position the right hand around the direction of the angle between the two vectors such that the fingers point in the direction of moving the first vector on top of the second vector, and you extend your thumb, the direction of the resultant vector is given by the thumb. Also, for the corkcrew rule: If we rotate the corkscrew such that the first vector advances on top of the second vector, the direction of advance of the screw is the direction of the resultant vector. The two rules are identical, so whichever you find less weird and are able to apply to your problem should bring you to the same result.

For the vector product, the order of multiplication is important because the result is a vector. Vector product is not commutative:

A × B = –B × A

We apply the above definition to both sides of the equality:

|A × B| = A · B · sin θ

and

|B × A| = B · A · sin (–θ)

But A and B are numbers, so A · B = B · A, and the sine function is antisymmetric: sin θ = –sin (–θ). Hence, although the magnitude is the same, the direction is opposite:

|A × B| = A · B · sin θ

|B × A| = –B · A · sin θ

The vector product is distributive. This means that when having three vectors, the following equality holds true:

(A + B) × C = A ×C + B × C

We should also note the following remarkable identities between the unit vectors i, j, and k:

i · i = j · j = k · k = 1

i · j = j · k = k · i = 0

i × j = k, j × k = i, and k × i = j

Using these identities and the other properties of vectors, we can calculate the scalar and vector product of any two vectors when we know their x, y, and z components.

#### Example 5

Show that i × j = k.

#### Solution 5

First, we will show the magnitudes of the left-hand side and right-hand side to be equal:

|i × j| = 1 · 1 · sin 90° = 1

|k| = 1

For the orientation, we can show through a diagram the orientation of the i × j vector. In Figure 2.11, we see the screw rotating from i unit vector toward the j unit vector in the direction of the angel, and the screw advances upward in the direction of k.

Practice problems of this concept can be found at: Scalars and Vectors Practice Questions

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