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Chapter 1: Vectors, Lines and Planes

Section 1.3: Dot Product

Example 1.3.2

If $A = 3 i + 2 j + 7 k$ and $B = 4 i - 5 j + 6 k$ ,

a)	Obtain $A \cdot B$

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

c)	Verify that ${∥A∥}^{2} = A \cdot A$

d)	Obtain the scalar projection of B on A

e)	Obtain the vector projection of B on A

f)	Obtain the component of B orthogonal to A

g)	For A, obtain direction cosines, angles, and numbers.

Solution

Mathematical Solution

	Part (a)
	$A \cdot B = [\begin{array}{c} 3 \\ 2 \\ 7 \end{array}] \cdot [\begin{array}{c} 4 \\ - 5 \\ 6 \end{array}]$ = $3 (4) + 2 (- 5) + 7 (6) = 12 - 10 + 42 = 44$

Part (b)

${∥A∥}^{2} = A \cdot A = [\begin{array}{c} 3 \\ 2 \\ 7 \end{array}] \cdot [\begin{array}{c} 3 \\ 2 \\ 7 \end{array}] = 9 + 4 + 49 = 62$ $\Rightarrow ∥A∥ = \sqrt{62}$

${∥B∥}^{2} = B \cdot B = [\begin{array}{c} 4 \\ - 5 \\ 6 \end{array}] \cdot [\begin{array}{c} 4 \\ - 5 \\ 6 \end{array}] = 1 6 + 2 5 + 36 = 7 7$ $\Rightarrow ∥B∥ = \sqrt{77}$

$θ = \cos^{- 1} (\frac{A \cdot B}{∥A ∥∥ B∥}) = \cos^{- 1} (\frac{44}{\sqrt{62} \sqrt{77}}) ≐ 0.88044$ radians

	Part (c)
	In Part (b), the norm of A was found to be $\sqrt{62}$ by computing $\sqrt{A \cdot A}$ . In Section 1.1, the norm was defined as the square root of the sum of the squares of the components. Hence, calculate $∥A∥ = \sqrt{3^{2} + 4^{2} + 7^{2}} = \sqrt{9 + 4 + 49} = \sqrt{62}$

Part (d)

•	The vectors A, B, $B_{A}$ and $B_{⊥ A}$ are drawn in Figure 1.3.2(a).

•	The vectors A and B are drawn in red and green, respectively. The vector projection of B on A, namely, $B_{A}$ , which is collinear with A, is drawn as an arrow having a black outline and residing within the arrow representing A.

•	The scalar projection of B on A is the length of the vector projection, that is

$∥B_{A}∥$ = $\frac{B \cdot A}{∥A∥}$ = $\frac{44}{\sqrt{62}} ≐ 5.588$

Figure 1.3.2(a) Vectors A, B, $B_{A}$ and $B_{⊥ A}$

	Part (e)
	The vector $B_{A}$ , the vector projection of B on A, is shown in Figure 1.3.2(a). It is obtained by the following calculation. $B_{A} = \frac{B \cdot A}{A \cdot A} A$ = $\frac{44}{62} A$ = $\frac{22}{31} [\begin{array}{c} 3 \\ 2 \\ 7 \end{array}]$

	Part (f)
	The component of B orthogonal to A is the vector $B_{⊥ A}$ , shown in blue in Figure 1.3.1(a). It is obtained by the following calculation. $B_{⊥ A} = B - B_{A}$ = $[\begin{array}{c} 4 \\ - 5 \\ 6 \end{array}] - \frac{22}{31} [\begin{array}{c} 3 \\ 2 \\ 7 \end{array}]$ = $\frac{1}{31} [\begin{array}{c} 58 \\ - 199 \\ 32 \end{array}]$

Part (g)

The direction cosines for A are the components of $A / ∥A∥$ = $(3 i + 2 j + 7 k) / \sqrt{62}$ .

Direction Cosines	Direction Angles (in radians)
$\cos (α) = 3 / \sqrt{62}$	$α = \cos^{- 1} (3 / \sqrt{62})$ $≐ 1.18$
$\cos (β) = 2 / \sqrt{62}$	$β = \cos^{- 1} (2 / \sqrt{62}) ≐ 1.31$
$\cos (γ) = 7 / \sqrt{62}$	$g = \cos^{- 1} (7 / \sqrt{62}) ≐ 0.48$

A common multiple of the direction cosines yields a set of direction numbers. Hence, the components of A itself constitute a set of direction numbers.

Maple Solution - Interactive

Part (a)

Data entry

•	Write $A = \dots$ as per Table 1.1.1.

•	Context Panel: Assign Name

$A = ⟨3, 2, 7⟩$ $\overset{assign}{\to}$

•	Write $B = \dots$ as per Table 1.1.1.

•	Context Panel: Assign Name

$B = ⟨4, - 5, 6⟩$ $\overset{assign}{\to}$

Compute the dot product

•	Common Symbols palette: Dot-product operator.

•	Context Panel: Evaluate and Display Inline

$A \cdot B$ = $44$

Although Figure 1.3.2(a) could be constructed interactively, its construction is tedious. Below, in the section where a coded solution is given, Maple commands for drawing the figure more efficiently are given.

Part (b)

Begin by installing the Student MultivariateCalculus package since it contains an Angle command, which then becomes available in the Context Panel.

•	Tools≻Load Package: Student Multivariate Calculus

Loading Student:-MultivariateCalculus

•	Form a sequence of the names of the two vectors. Context Panel: Evaluate and Display Inline

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

•	Context Panel: Approximate≻5 (digits)

$A, B$ = $\overset{angle}{\to}$ $\arccos (\frac{2}{217} \sqrt{62} \sqrt{77})$ $\overset{at 5 digits}{\to}$ $0.88044$

Part (c)

With the Student MultivariateCalculus package installed, the norm of A can be accessed via the Context Panel or via the notation $∥A∥$ , as per the note on norms in Section 1.1.

Obtain the square of the 2-norm via standard math notation

•	Context Panel: Evaluate and Display Inline

${∥A∥}^{2}$ = $62$

•	Common Symbols palette: Dot-product operator

$A \cdot A$ = $62$

•	Expression palette: Square root template

•	Common Symbols palette: Dot-product operator

•	Context Panel: Evaluate and Display Inline

$\sqrt{A \cdot A}$ = $\sqrt{62}$

Obtain the 2-norm via the Context Panel

•	Write the name A. Context Panel: Evaluate and Display Inline

•	Context Panel: Student Multivariate Calculus≻Norm

$A$ = $\overset{norm}{\to}$ $\sqrt{62}$

There is no Context Panel option for obtaining the square of the norm. Hence, the alternate calculation shows $∥A∥ = \sqrt{A \cdot A}$ instead of ${∥A∥}^{2} = A \cdot A$ .

Part (d)

Implement $∥B_{A}∥$ = $\frac{B \cdot A}{∥A∥}$ using math notation for the norm

•	Common Symbols palette: Dot product operator

•	Subscript norm operator with 2.

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Approximate≻5 (digits)

$\frac{B \cdot A}{{∥A∥}_{2}}$ = $\frac{22}{31} \sqrt{62}$ $\overset{at 5 digits}{\to}$ $5.5880$

Implement $∥B_{A}∥$ = $\frac{B \cdot A}{∥A∥}$ using Context Panel for the norm

•	Write the name A. Context Panel: Evaluate and Display Inline

•	Context Panel: Student Multivariate Calculus≻Norm

$A$ = $\overset{norm}{\to}$ $\sqrt{62}$

•	Common Symbols palette: Dot-product operator

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Approximate≻5 (digits)

$\frac{B \cdot A}{\sqrt{62}}$ = $\frac{22}{31} \sqrt{62}$ $\overset{at 5 digits}{\to}$ $5.5880$

Implement $∥B_{A}∥$ = $\frac{B \cdot A}{∥A∥}$ as $\frac{B \cdot A}{\sqrt{A \cdot A}}$ to avoid norm notation

•	Common Symbols palette: Dot product operator

•	Context Panel: Evaluate and Display Inline

•	Context Panel: Approximate≻5 (digits)

$\frac{B \cdot A}{\sqrt{A \cdot A}}$ = $\frac{22}{31} \sqrt{62}$ $\overset{at 5 digits}{\to}$ $5.5880$

Part (e)

The task template, implemented in Table 1.3.2(a), provides a solution for both $B_{A}$ and $B_{⊥ A}$ , and in addition, draws a graph of B and its two components.

Tools≻Tasks≻Browse: Linear Algebra≻Visualizations≻Projection Plot onto 1-D

Projection onto 1-D Subspace

Vector to be projected

Project onto

Table 1.3.2(a) Projection Plot onto 1-D task template

Table 1.3.2(b) implements the calculation of $B_{A}$ via the recipe listed in Table 1.3.1.

•	Common Symbols palette: Dot product operator

•	Context Panel: Evaluate and Display Inline

$\frac{B \cdot A}{A \cdot A} A$ =

Table 1.3.1(b) Computation of $B_{A}$ via the recipe listed in Table 1.3.1.

Table 1.3.2(c) implements the calculation of $B_{A}$ via the Projection option in the Context Panel, an option available only when the Student MultivariateCalculus package is installed.

•	Write a sequence of the names of the vectors. Context Panel: Evaluate and Display Inline

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

$B, A$ = $\overset{projection}{\to}$

Table 1.3.2(c) Computation of $B_{A}$ via the Projection option in the Context Panel

Part (f)

Calculate $B_{⊥ A} = B - B_{A}$

•	Common Symbols palette: Dot-product operator

•	Context Panel: Evaluate and Display Inline

$B - \frac{B \cdot A}{A \cdot A} A$ =

Compare this result to that provided in Table 1.3.2(a).

Part (g)

•	Type the name A. Context Panel: Evaluate and Display Inline

•	Context Panel: Student Multivariate Calculus≻Normalize

•	Context Panel: Assign to a Name≻T

$A$ = $\overset{normalize}{\to}$ $\overset{assign to a name}{\to}$ $T$

•	Set each direction cosine equal to the appropriate component of T, the normalization of A.

•	Context Panel: Solve≻Solve

•	Context Panel: Select Element≻1

•	Context Panel: Approximate≻5 (digits)

$\cos (α) = T_{1}$ $\overset{solve}{\to}$ $\{α = \arccos (\frac{3}{62} \sqrt{62})\}$ $\to$ $α = \arccos (\frac{3}{62} \sqrt{62})$ $\overset{at 5 digits}{\to}$ $α = 1.1799$

$\cos (β) = T_{2}$ $\overset{solve}{\to}$ $\{β = \arccos (\frac{1}{31} \sqrt{62})\}$ $\to$ $β = \arccos (\frac{1}{31} \sqrt{62})$ $\overset{at 5 digits}{\to}$ $β = 1.3140$

$\cos (γ) = T_{3}$ $\overset{solve}{\to}$ $\{γ = \arccos (\frac{7}{62} \sqrt{62})\}$ $\to$ $γ = \arccos (\frac{7}{62} \sqrt{62})$ $\overset{at 5 digits}{\to}$ $γ = 0.47571$

Note: The symbol $γ$ is a special mathematical quantity, which Maple therefore does not treat the same as it would a plane name. Here, it was converted to an atomic identifier, which freezes it as a name distinct from its meaning as a special symbol.

Maple Solution - Coded

Part (a)

As per Table 1.3.2, the Student MultivariateCalculus package contains a DotProduct command defined over the real numbers.

For completeness and clarity, the Coded solutions begin with the execution of a restart command.

Initialize

•	Execute a restart.

$restart$

•	Install the Student MultivariateCalculus package.

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

•	Define the vectors A and B.

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

Compute $A \cdot B$

•	Apply the DotProduct command.

$DotProduct (A, B)$ = $44$

The following code will draw the vectors in Figure 1.3.2(a). The additional text was added to the figure interactively via the Drawing toolbar.

$use Student :- VectorCalculus, plots in module () local A, B, V, p, q, r &semi; A, B ≔ ⟨3, 2, 7⟩, ⟨4, - 5, 6⟩ &semi; V ≔ Student :- MultivariateCalculus :- Projection (B, A) &semi; p ≔ spacecurve ([convert (B, list), convert (V, list)], numpoints = 2, linestyle = dot, color = blue) &semi; q ≔ PlotVector ([A, B, V, B - V], color = [red, green, black, blue], transparency = .5) &semi; r ≔ display (p, q, scaling = constrained, axes = frame, labels = [x, y, z], orientation = [- 20, 75, 0], tickmarks = [5, 4, 7], lightmodel = none, glossiness = 0) &semi; print (r) &semi; end module &colon; end use &colon;$

Part (b)

Recall that the Student MultivariateCalculus package was installed in Part (a).

Compute $θ$ via the Angle command

$Angle (A, B)$ = $\arccos (\frac{2}{217} \sqrt{62} \sqrt{77})$

Compute $θ$ from first principles

$\arccos (\frac{DotProduct (A, B)}{Norm (A) Norm (B)})$ = $\arccos (\frac{2}{217} \sqrt{62} \sqrt{77})$

Part (c)

Recall that the Student MultivariateCalculus package was installed in Part (a).

$Norm {(A)}^{2}$ = $62$

$DotProduct (A, A)$ = $62$

Part (d)

Recall that the Student MultivariateCalculus package was installed in Part (a).

$\frac{DotProduct (A, B)}{Norm (A)}$ = $\frac{22}{31} \sqrt{62}$

Part (e)

Recall that the Student MultivariateCalculus package was installed in Part (a).

Compute $B_{A}$ via the Projection command in the Student MultivariateCalculus package

$Projection (B, A)$ =

Compute $B_{A}$ from first principles

$\frac{DotProduct (B, A)}{DotProduct (A, A)} A$ =

Part (f)

Recall that the Student MultivariateCalculus package was installed in Part (a).

Compute $B_{⊥ A}$ via the Projection command in the Student MultivariateCalculus package

$B - Projection (B, A)$ =

Compute $B_{⊥ A}$ from first principles

$B - \frac{DotProduct (B, A)}{DotProduct (A, A)} A$ =

Part (g)

Recall that the Student MultivariateCalculus package was installed in Part (a).

Use the Normalize command to obtain the vector $T = A / ∥A∥$ .	$T ≔ Normalize (A) &colon;$
Direction Cosines	Direction Angles
$\cos (α) = T_{1}$ $\cos (α) = \frac{3}{62} \sqrt{62}$	$\cos^{- 1} (T_{1})$ = $\arccos (\frac{3}{62} \sqrt{62})$ $\overset{at 5 digits}{\to}$ $1.1799$
$\cos (β) = T_{2}$ $\cos (β) = \frac{1}{31} \sqrt{62}$	$\cos^{- 1} (T_{2})$ = $\arccos (\frac{1}{31} \sqrt{62})$ $\overset{at 5 digits}{\to}$ $1.3140$
$\cos (γ) = T_{3}$ $\cos (γ) = \frac{7}{62} \sqrt{62}$	$\cos^{- 1} (T_{3})$ = $\arccos (\frac{7}{62} \sqrt{62})$ $\overset{at 5 digits}{\to}$ $0.47571$

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