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Chapter 1: Vectors, Lines and Planes

Section 1.4: Cross Product

Example 1.4.1

For the vectors $A = 3 i - 2 j + 4 k$ and $B = 5 i + 7 j - 6 k$ , and the scalar $c$ ,

a)	Obtain $A \times B$ .

b)	Obtain $θ$ , the angle between A and B.

c)	Verify computationally that $A \times A = 0$ .

d)	Verify computationally that $∥A \times B∥ = ∥A ∥∥ B∥ \sin (θ)$ .

e)	Verify computationally that $B \times A = - A \times B$ .

f)	Verify that both A and B are perpendicular to $A \times B$ .

g)	Verify that ${(A \cdot B)}^{2} + {∥A \times B∥}^{2}$ = ${(∥A ∥∥ B∥)}^{2}$ .

h)	Verify that ${(A \cdot B)}^{2} - {∥A \times B∥}^{2}$ = ${(∥A ∥∥ B∥)}^{2} \cos (2 θ)$ .

i)	Show that $A \times (A \times B) \neq (A \times A) \times B = 0$ .

j)	Verify that $(c A) \times B = c (A \times B) = A \times (c B)$ .

k)	Verify that $(A - B) \times (A + B) = 2 (A \times B)$ .

Solution

Mathematical Solution

Part (a)

Obtain $A \times B$ .

$A \times B$ = $\| \begin{array}{c} i & j & k \\ 3 & - 2 & 4 \\ 5 & 7 & - 6 \end{array} \|$	= $i \|\begin{array}{c} - 2 & 4 \\ 7 & - 6 \end{array}\| - j \| \begin{array}{c} 3 & 4 \\ 5 & - 6 \end{array} \| + k \| \begin{array}{c} 3 & - 2 \\ 5 & 7 \end{array} \|$
	$= i (12 - 28) - j (- 18 - 20) + k (21 + 10)$
	$= - 16 i + 38 j + 31 k$

Part (b)

Obtain $θ$ , the angle between A and B.

The angle between two vectors A and B can be obtained from the following expression.

$\cos (θ) = \frac{A \cdot B}{∥A ∥∥ B∥} = \frac{A \cdot B}{\sqrt{A \cdot A} \sqrt{B \cdot B}}$

The necessary numbers are then

$A \cdot B = [\begin{array}{c} 3 \\ - 2 \\ 4 \end{array}] \cdot [\begin{array}{c} 5 \\ 7 \\ - 6 \end{array}]$	= $15 - 14 - 24 = - 23$
$A \cdot A = [\begin{array}{c} 3 \\ - 2 \\ 4 \end{array}] \cdot [\begin{array}{c} 3 \\ - 2 \\ 4 \end{array}]$	= $9 + 4 + 16 = 29$
$B \cdot B = [\begin{array}{c} 5 \\ 7 \\ - 6 \end{array}] \cdot [\begin{array}{c} 5 \\ 7 \\ - 6 \end{array}]$	= $25 + 49 + 36 = 110$

Finally, the angle $θ$ is given by

$θ = \cos^{- 1} (\frac{- 23}{\sqrt{29} \sqrt{110}})$ = $π - \cos^{- 1} (\frac{23}{\sqrt{29} \sqrt{110}})$ $≐ 1.99$ radians

Part (c)

Verify computationally that $A \times A = 0$ .

$A \times A = [\begin{array}{c} i & j & k \\ 3 & - 2 & 4 \\ 3 & - 2 & 4 \end{array}]$ = $[\begin{array}{c} - 2 \cdot 4 - 4 \cdot (- 2) \\ - (3 \cdot 4 - 4 \cdot 3) \\ 3 \cdot (- 2) - (- 2) \cdot 3 \end{array}]$ = $[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}] = 0$

Part (d)

Verify computationally that $∥A \times B∥ = ∥A ∥∥ B∥ \sin (θ)$ .

The left-hand side:

$∥A \times B∥ = \sqrt{{(- 16)}^{2} + {(38)}^{2} + {(31)}^{2}} = \sqrt{2661}$

The right-hand side:

$∥A∥$ $∥B∥$ $\cos (θ)$	$= \sqrt{29} \sqrt{110} \sin (\cos^{- 1} (\frac{- 23}{\sqrt{29} \sqrt{110}}))$
	$= \sqrt{29} \sqrt{110} (\frac{\sqrt{2661}}{\sqrt{29} \sqrt{110}})$
	$= \sqrt{2661}$

•	With appropriate care for quadrants, the evaluation of $x = \sin (\cos^{- 1} (a / b))$ is effected by setting $y = \cos^{- 1} (a / b)$ so that $\cos (y) = a / b$ and $x = \sin (y)$ .

•	Then, in the right triangle drawn in Figure 1.4.1(a), $x = \sin (y) = \sqrt{b^{2} - a^{2}} / b$ .

Figure 1.4.1(a) $\sin (\cos^{- 1} (a / b))$

Part (e)

Verify computationally that $B \times A = - A \times B$ .

Interchanging two adjacent rows in a determinant negates the value of the determinant. However, the relevant computation is shown below.

$B \times A$ = $| \begin{array}{c} i & j & k \\ 5 & 7 & - 6 \\ 3 & - 2 & 4 \end{array} |$ = $[\begin{array}{c} 7 \cdot 4 - (- 6) \cdot (- 2) \\ - (5 \cdot 4 - (- 6) \cdot 3) \\ 5 \cdot (- 2) - 7 \cdot 3 \end{array}]$ = $[\begin{array}{c} 16 \\ - 38 \\ - 31 \end{array}]$ = $- [\begin{array}{c} - 16 \\ 38 \\ 31 \end{array}]$ = $- A \times B$

Part (f)

Verify that both A and B are perpendicular to $A \times B$ .

If the dot product of two vectors is zero, the vectors are perpendicular.

$A \cdot (A \times B)$ = $[\begin{array}{c} 3 \\ - 2 \\ 4 \end{array}] \cdot [\begin{array}{c} - 16 \\ 38 \\ 31 \end{array}]$ = $- 48 - 76 + 124 = 0$

$B \cdot (A \times B)$ = $[\begin{array}{c} 5 \\ 7 \\ - 6 \end{array}] \cdot [\begin{array}{c} - 16 \\ 38 \\ 31 \end{array}]$ = $- 80 + 266 - 186 = 0$

Part (g)

Verify that ${(A \cdot B)}^{2} + {∥A \times B∥}^{2}$ = ${(∥A ∥∥ B∥)}^{2}$ .

The relevant calculations are summarized below.

Left-hand Side	${(A \cdot B)}^{2} + {∥A \times B∥}^{2}$	= ${(- 23)}^{2} + {(\sqrt{2661})}^{2} = 529 + 2661 = 3190$
Right-hand Side	${(∥A ∥∥ B∥)}^{2}$	= ${(\sqrt{29} \sqrt{110})}^{2} = 29 \cdot 110 = 3190$

Part (h)

Verify that ${(A \cdot B)}^{2} - {∥A \times B∥}^{2}$ = ${(∥A ∥∥ B∥)}^{2} \cos (2 θ)$ .

The relevant calculations are summarized below.

Left-hand Side	${(A \cdot B)}^{2} - {∥A \times B∥}^{2}$	= ${(- 23)}^{2} - {(\sqrt{2661})}^{2} = 529 - 2661 = - 2132$
Right-hand Side	${(∥A ∥∥ B∥)}^{2} \cos (2 θ)$	= ${(\sqrt{29} \sqrt{110})}^{2}$ $(- \frac{1066}{1595}) = - 2132$

Since $\cos (θ) = - \frac{23}{\sqrt{29} \sqrt{110}}$ , the double-angle formula $\cos (2 θ) = 2 \cos^{2} (θ) - 1$ is used to obtain $\cos (2 θ) = - \frac{1066}{1595}$ .

Part (i)

Show that $A \times (A \times B) \neq (A \times A) \times B = 0$ .

The relevant calculations are shown below.

$A \times (A \times B)$ = $| \begin{array}{c} i & j & k \\ 3 & - 2 & 4 \\ - 16 & 38 & 31 \end{array} |$ = $[\begin{array}{c} - 2 \cdot 31 - 4 \cdot 38 \\ - (3 \cdot 31 - 4 \cdot (- 16)) \\ 3 \cdot 38 - (- 2) \cdot (- 16) \end{array}]$ = $[\begin{array}{c} - 214 \\ - 157 \\ 82 \end{array}]$

Using the result of Part (c),

$(A \times A) \times B$ = $| \begin{array}{c} i & j & k \\ 0 & 0 & 0 \\ 5 & 7 & - 6 \end{array} |$ = $[\begin{array}{c} 0 \cdot (- 6) - 0 \cdot 7 \\ - (0 \cdot (- 6) - 0 \cdot 5) \\ 0 \cdot 7 - 0 \cdot 5 \end{array}]$ = $[\begin{array}{c} 0 \\ 0 \\ 0 \end{array}]$ = 0

Part (j)

Verify that $(c A) \times B = c (A \times B) = A \times (c B)$ .

The relevant calculations are summarized below.

$(c A) \times B$	$= \| \begin{array}{c} i & j & k \\ 3 c & - 2 c & 4 c \\ 5 & 7 & - 6 \end{array} \|$ = $[\begin{array}{c} (- 2 c) \cdot (- 6) - 4 c \cdot 7 \\ - (3 c \cdot (- 6) - 4 c \cdot 5) \\ 3 c \cdot 7 - (- 2 c) \cdot 5 \end{array}]$ = $[\begin{array}{c} - 16 c \\ 38 c \\ 31 c \end{array}]$ = $c [\begin{array}{c} - 16 \\ 38 \\ 31 \end{array}]$ = $c (A \times B)$
$c (A \times B)$	$= c [\begin{array}{c} - 16 \\ 38 \\ 31 \end{array}]$
$A \times (c B)$	$= \| \begin{array}{c} i & j & k \\ 3 & - 2 & 4 \\ 5 c & 7 c & - 6 c \end{array} \|$ = $[\begin{array}{c} - 2 \cdot (- 6 c) - 4 \cdot (7 c) \\ - (3 \cdot (- 6 c) - 4 \cdot (5 c)) \\ 3 \cdot (7 c) - (- 2) \cdot (5 c) \end{array}]$ = $[\begin{array}{c} - 16 c \\ 38 c \\ 31 c \end{array}]$ = $c [\begin{array}{c} - 16 \\ 38 \\ 31 \end{array}]$ = $c (A \times B)$

Part (k)

Verify that $(A - B) \times (A + B) = 2 (A \times B)$ .

The relevant calculations are summarized below.

$(A - B) \times (A + B)$	$= (A - B) \times A + (A - B) \times B$	(First Distributive law, Table 1.4.1)
	$= A \times A - B \times A + A \times B - B \times B$	(Second Distributive Law, Table 1.4.1)
	$= 0 - B \times A + A \times B - 0$	$A \times A = B \times B = 0$
	$= - (- A \times B) + A \times B$	Anti-commutation rule
	$= 2 A \times B$	Arithmetic

Maple Solution - Interactive

Initialization

•	Tools≻Load Package: Student Multivariate Calculus

Loading Student:-MultivariateCalculus

•	Enter A as per Table 1.1.1.

•	Context Panel: Assign to a Name≻A

$⟨3, - 2, 4⟩$ $\overset{assign to a name}{\to}$ $A$

•	Enter B as per Table 1.1.1.

•	Context Panel: Assign to a Name≻B

$⟨5, 7, - 6⟩$ $\overset{assign to a name}{\to}$ $B$

Part (a)

Obtain $A \times B$ .

•	Common Symbols palette: Cross-product operator

•	Context Panel: Evaluate and Display Inline

$A \times B$ =

The task template returns the cross product, and a graph of the three vectors A, B, and $A \times B$ . Since the vectors A and B have already been defined, these names for the vectors can be used in the task template.

Tools≻Tasks≻Browse: Linear Algebra≻Visualizations≻Cross-Product Plot

Cross Product

A =

B =

As the legend explains, the red and blue vectors are A and B, respectively, and their cross product is the black vector. The image in the graph can be rotated, thereby emphasizing the orthogonality of the three vectors.

Part (b)

Obtain $θ$ , the angle between A and B

•	Write the sequence of two vectors and press the Enter key.

•	Context Panel: Student Multivariate Calculus≻Lines & Planes≻Angle

•	Context Panel: Assign to a Name≻T

$A, B$ = $\overset{angle}{\to}$ $π - \arccos (\frac{23}{3190} \sqrt{29} \sqrt{110})$ $\overset{assign to a name}{\to}$ $T$

Obtain an approximate value for $θ$

•

Write T.

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Approximate≻5 (digits)

$T$ = $π - \arccos (\frac{23}{3190} \sqrt{29} \sqrt{110})$ $\overset{at 5 digits}{\to}$ $1.9902$

Angle $θ$ has been named T so that it can be referenced in subsequent parts of this example.

Part (c)

Verify computationally that $A \times A = 0$

•	Common Symbols palette: Cross-product operator

•	Context Panel: Evaluate and Display Inline

$A \times A$ =

Part (d)

Verify computationally that $∥A \times B∥ = ∥A ∥∥ B∥ \sin (θ)$ .

•	Enter the notation for the Euclidean norm of the cross product.

•	Context Panel: Evaluate and Display Inline

$∥A \times B∥$ = $\sqrt{2661}$

•	Write the notation for the right-hand side of the identity. Context Panel: Evaluate and Display Inline

•	Context Panel: Simplify≻Simplify

$∥A∥ \cdot &par; B &par; \cdot \sin (T)$ = $\frac{1}{3190} \sqrt{29} \sqrt{110} \sqrt{8488590}$ $\overset{simplify}{=}$ $\sqrt{2661}$

Part (e)

Verify computationally that $B \times A = - A \times B$ .

•	Common Symbols palette: Cross-product operator

•	Context Panel: Evaluate and Display Inline

$A \times B$ =

•	Common Symbols palette: Cross-product operator

•	Context Panel: Evaluate and Display Inline

$B \times A$ =

The vectors $A \times B$ and $B \times A$ differ in sign, that is, $B \times A = - A \times B$ .

Part (f)

Verify that both A and B are perpendicular to $A \times B$ .

•	Common Symbols palette: Dot-product and cross-product operators

•	Context Panel: Evaluate and Display Inline

$A \cdot (A \times B)$ = $0$

•	Common Symbols palette: Dot-product and cross-product operators

•	Context Panel: Evaluate and Display Inline

$B \cdot (A \times B)$ = $0$

The dot products of A and B with the cross product are both zero. Hence, each of A and B is orthogonal to the cross product vector.

Part (g)

Verify that ${(A \cdot B)}^{2} + {∥A \times B∥}^{2}$ = ${(∥A ∥∥ B∥)}^{2}$ .

•	Common Symbols palette: Dot-product and cross-product operators

•	Context Panel: Evaluate and Display Inline

Left-hand Side

Right-hand Side

${(A \cdot B)}^{2} + {∥A \times B∥}^{2}$ = $3190$

$(&par; A &par; \cdot ∥B∥)^{2}$ = $3190$

Part (h)

Verify that ${(A \cdot B)}^{2} - {∥A \times B∥}^{2}$ = ${(∥A ∥∥ B∥)}^{2} \cos (2 θ)$ .

The angle $θ$ was determined and named $T$ in Part (b).

•	Common Symbols palette: Dot product and cross product operators

•	Context Panel: Evaluate and Display Inline

•	Press the Enter key.

•	Context Panel: Simplify≻Simplify

Left-hand Side

Right-hand Side

${(A \cdot B)}^{2} - {∥A \times B∥}_{2}^{2}$ = $- 2132$

$(&par; A &par; \cdot ∥B∥)^{2} \cdot \cos (2 T)$

$3190 \cos (2 \arccos (\frac{23}{3190} \sqrt{29} \sqrt{110}))$

$\overset{simplify}{=}$ $- 2132$

Part (i)

Show that $A \times (A \times B) \neq (A \times A) \times B = 0$ .

The expression on the right is necessarily the zero vector because the cross product of A with itself is the cross product of two collinear vectors. Because of the length property in Table 1.4.1, this is necessarily zero because the angle between A and itself is zero.

•	Common Symbols palette: Cross-product operator

•	Context Panel: Evaluate and Display Inline

Left-hand Side

Right-hand Side

$A \times (A \times B)$ =

$(A \times A) \times B$ =

Part (j)

Verify that $(c A) \times B = c (A \times B) = A \times (c B)$ .

•	Common Symbols palette: Cross-product operator

•	Context Panel: Evaluate and Display Inline

Left-hand Side

Middle

Right-hand Side

$(c A) \times B$ =

$c (A \times B)$ =

$A \times (c B)$ =

Part (k)

Verify that $(A - B) \times (A + B) = 2 (A \times B)$ .

•	Common Symbols palette: Cross-product operator

•	Context Panel: Evaluate and Display Inline

Left-hand Side

Right-hand Side

$(A - B) \times (A + B)$ =

$2 (A \times B)$ =

Maple Solution - Coded

Initialization

•	Install the Student MultivariateCalculus package.

$with (Student :- MultivariateCalculus) &colon;$

•	Define the vectors A and B.

$A, B ≔ ⟨3, - 2, 4⟩, ⟨5, 7, - 6⟩ &colon;$

Part (a)

Obtain $A \times B$ .

•	Apply the CrossProduct command.

$CrossProduct (A, B)$ =

Part (b)

Obtain $θ$ , the angle between A and B

•	Apply the Angle command.

$T ≔ Angle (A, B)$

$π - \arccos (\frac{23}{3190} \sqrt{29} \sqrt{110})$

•	Apply the evalf command.

$evalf (T, 5)$

$1.9902$

Part (c)

Verify computationally that $A \times A = 0$

•	Apply the CrossProduct command.

$CrossProduct (A, A)$ =

Part (d)

Verify computationally that $∥A \times B∥ = ∥A ∥∥ B∥ \sin (θ)$ .

Apply the Norm, CrossProduct, and simplify commands.

Left-hand Side

Right-hand Side

$Norm (CrossProduct (A, B))$

$\sqrt{2661}$

$simplify (Norm (A) \cdot Norm (B) \cdot \sin (T))$

$\sqrt{2661}$

Part (e)

Verify computationally that $B \times A = - A \times B$ .

Apply the CrossProduct command.

Left-hand Side	Right-hand Side
$CrossProduct (B, A)$ =	$CrossProduct (A, B)$ =

Part (f)

Verify that both A and B are perpendicular to $A \times B$ .

Apply the DotProduct and CrossProduct commands.	$DotProduct (A, CrossProduct (A, B))$ = $0$
Apply the DotProduct and CrossProduct commands.	$DotProduct (B, CrossProduct (A, B))$ = $0$

Part (g)

Verify that ${(A \cdot B)}^{2} + {∥A \times B∥}^{2}$ = ${(∥A ∥∥ B∥)}^{2}$ .

Apply the DotProduct , Norm, and CrossProduct commands.

Left-hand Side	$DotProduct {(A, B)}^{2} + Norm {(CrossProduct (A, B))}^{2}$ = $3190$
Right-hand Side	${(Norm (A) \cdot Norm (B))}^{2}$ = $3190$

Part (h)

Verify that ${(A \cdot B)}^{2} - {∥A \times B∥}^{2}$ = ${(∥A ∥∥ B∥)}^{2} \cos (2 θ)$ .

Apply the DotProduct , Norm, CrossProduct, and simplify commands.

$DotProduct {(A, B)}^{2} - Norm {(CrossProduct (A, B))}^{2}$ = $- 2132$

$simplify ({(Norm (A) \cdot Norm (B))}^{2} \cdot \cos (2 T))$ = $- 2132$

Part (i)

Show that $A \times (A \times B) \neq (A \times A) \times B = 0$ .

Apply the CrossProduct command.

Left-hand Side

Right-hand Side

$CrossProduct (A, CrossProduct (A, B))$

$CrossProduct (CrossProduct (A, A), B)$

Part (j)

Verify that $(c A) \times B = c (A \times B) = A \times (c B)$ .

Apply the CrossProduct command.

Left-hand Side

Middle

Right-hand Side

$CrossProduct (c \cdot A, B)$

$c \cdot CrossProduct (A, B)$

$CrossProduct (A, c \cdot B)$

Part (k)

Verify that $(A - B) \times (A + B) = 2 (A \times B)$ .

Apply the CrossProduct command.

Left-hand Side	Right-hand Side
$CrossProduct (A - B, A + B)$ =	$2 CrossProduct (A, B)$ =

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