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13.4 The Cross Product

Another way vectors multiply

dot product: \(\vec{u} \cdot \vec{v} = |\vec{u}| |\vec{v}| \cos \theta\)

Diagram showing vectors u and v with an angle theta between them.

\(\vec{u} = \langle a, b, c \rangle \quad \vec{v} = \langle d, e, f \rangle\)

\(\vec{u} \cdot \vec{v} = ad + be + cf\) Scalar

the cross product of \(\vec{u}, \vec{v}\) is \(\vec{u} \times \vec{v}\) (or \(\vec{v} \times \vec{u}\))

the result is a vector

magnitude: \(|\vec{u} \times \vec{v}| = |\vec{u}| |\vec{v}| \sin \theta\)

direction: by the right-hand rule

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CURL RIGHT HAND RULE

Right hand rule diagram with thumb as a x b, index as a, and middle finger as b.
3D axes labeled with unit vectors i, j, k corresponding to index, middle, and thumb.

\(\vec{i} \times \vec{j} = \vec{k}\)

\(\vec{j} \times \vec{i} = -\vec{k}\)

Hand gripping a vertical axis with thumb up, illustrating the curl right hand rule for cross products.

Sweep a into b

Two perpendicular vectors labeled a and b.

\(\vec{a} \times \vec{b}\) : out of screen/paper

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\[ \vec{u} \times \vec{v} = -(\vec{v} \times \vec{u}) \quad \text{in cross product, order matters} \]

How to compute \( \vec{u} \times \vec{v} \)?

One way: use \( |\vec{u} \times \vec{v}| = |\vec{u}||\vec{v}| \sin \theta \) to find magnitude, then right-hand rule for direction.

The other (algebraic) way: the determinant of a special matrix.

\( 2 \times 2 \) matrix \( \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} \) determinant is

A 2x2 determinant with arrows showing the product of diagonals: (1 times 4) minus (2 times 3).
\[ = (1)(4) - (2)(3) \]\[ = 4 - 6 = -2 \]
\[ \begin{vmatrix} -3 & 1 \\ 7 & -4 \end{vmatrix} = (-3)(-4) - (1)(7) = 12 - 7 = 5 \]

We need determinant of a special \( 3 \times 3 \) matrix for cross product

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\( \vec{u} = \langle 2, 1, 2 \rangle \quad \vec{v} = \langle 5, 0, 1 \rangle \)

Find \( \vec{u} \times \vec{v} \)

\[ \vec{u} \times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 2 & 1 & 2 \\ 5 & 0 & 1 \end{vmatrix} \]
  • First row: \( \vec{i}, \vec{j}, \vec{k} \)
  • 2nd row: first vector in cross product
  • 3rd row: the second vector

Move along first row (\( \vec{j} \) gets negative)

\[ = \vec{i} \begin{vmatrix} 1 & 2 \\ 0 & 1 \end{vmatrix} - \vec{j} \begin{vmatrix} 2 & 2 \\ 5 & 1 \end{vmatrix} + \vec{k} \begin{vmatrix} 2 & 1 \\ 5 & 0 \end{vmatrix} \]

determinant of leftover after removing row and column of \( \vec{i} \)

determinant of leftover after removing row, column \( \vec{j} \) is in

\[ = \vec{i}(1) - \vec{j}(-8) + \vec{k}(-5) = \vec{i} + 8\vec{j} - 5\vec{k} = \langle 1, 8, -5 \rangle \]
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\[ \vec{u} = \langle 2, 1, 2 \rangle \quad \vec{v} = \langle 5, 0, 1 \rangle \]
\[ \vec{u} \times \vec{v} = \langle 1, 8, -5 \rangle \quad \vec{v} \times \vec{u} = -\langle 1, 8, -5 \rangle = \langle -1, -8, 5 \rangle \]

(check on your own)

notice

\[ (\vec{u} \times \vec{v}) \cdot \vec{u} = \langle 1, 8, -5 \rangle \cdot \langle 2, 1, 2 \rangle = 2 + 8 - 10 = 0 \]
\[ (\vec{u} \times \vec{v}) \cdot \vec{v} = \langle 1, 8, -5 \rangle \cdot \langle 5, 0, 1 \rangle = 5 + 0 - 5 = 0 \]

this means, \( \vec{u} \times \vec{v} \) (or \( \vec{v} \times \vec{u} \)) is perpendicular or orthogonal to BOTH \( \vec{u} \) and \( \vec{v} \)

if \( \vec{u} \) and \( \vec{v} \) form the two sides of a parallelogram sharing a corner, then we can also show that \( |\vec{u} \times \vec{v}| = \text{area of the parallelogram} \)

Diagram of a parallelogram formed by vectors u and v with angle theta between them.
\[ \begin{aligned} \text{area} &= |\vec{u} \times \vec{v}| \\ &= |\vec{u}| |\vec{v}| \sin \theta \end{aligned} \]
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example

Find area of parallelogram with vertices

\( A(-3, 4), B(-1, 7), C(3, 5), D(1, 2) \)

Parallelogram ABCD plotted on a 2D coordinate system with vectors u and v labeled.
\[ \begin{aligned} \vec{u} &= \langle 4, -2 \rangle = \langle 4, -2, 0 \rangle \\ \vec{v} &= \langle 2, 3 \rangle = \langle 2, 3, 0 \rangle \end{aligned} \]

area of p-gram is \( |\vec{u} \times \vec{v}| \) or \( |\vec{v} \times \vec{u}| \)

\[ \vec{u} \times \vec{v} = \begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ 4 & -2 & 0 \\ 2 & 3 & 0 \end{vmatrix} = \vec{i} \begin{vmatrix} -2 & 0 \\ 3 & 0 \end{vmatrix} - \vec{j} \begin{vmatrix} 4 & 0 \\ 2 & 0 \end{vmatrix} + \vec{k} \begin{vmatrix} 4 & -2 \\ 2 & 3 \end{vmatrix} \]
\[ = \vec{i}(0) - \vec{j}(0) + \vec{k}(16) = 16\vec{k} = \langle 0, 0, 16 \rangle \]

\( |\vec{u} \times \vec{v}| = 16 \) area of the p-gram

or \( |\vec{u}| |\vec{v}| \sin \theta \)

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Collinear Vectors and Cross Products

collinear vectors: along the same line

if \( \vec{AB} \parallel \vec{AC} \), then angle between is \( \theta = 0 \) (same direction) or \( 180^\circ \) (opposite direction) and \( \sin \theta = 0 \)

therefore, if \( \vec{AB} \parallel \vec{AC} \), then \( |\vec{AB} \times \vec{AC}| = 0 \)

Points A, B, and C on a single line segment with vectors AB and AC shown.
Two vectors u and v originating from a common point with an angle theta between them.

\( A, B, C, D \) are points on a plane

\[ \vec{AB} \times \vec{AD}, \vec{AD} \times \vec{AB}, \vec{AB} \times \vec{AC}, \vec{AC} \times \vec{AB} \]\[ \vec{AD} \times \vec{AC}, \vec{AC} \times \vec{AD} \]

are ALL orthogonal to the plane

cross product is orthogonal to BOTH parent vectors

A parallelogram in a plane defined by points A, B, C, and D with vectors AB, AC, and AD.
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Physical Applications of Cross Product

physical applications: torque or moment can be found by cross product

\( \vec{F} \): force applied at \( P \) at distance \( r \) from pivot

the torque about pivot is \( \vec{OP} \times \vec{F} \)

Diagram showing a pivot O, a point P at distance r, and a force vector F applied at P.

Magnetism

\[ \vec{F} = q (\vec{v} \times \vec{B}) \]
  • \( q \): charge
  • \( \vec{v} \): flow of current
  • \( \vec{B} \): magnetic field