Products of vectors AI HL

IB Mathematics: Applications and Interpretation HL · Unit 3

Lesson 3.13 · Products of vectors

Use scalar and vector products to calculate angles, perpendicular components, and areas.

Praise for wisdom and understanding

“Great is our Lord, and of great power: his understanding is infinite.” Psalm 147:5 (KJV)

Two vectors can describe directions, but their products reveal more: how closely those directions align and what area they span. As you discover these connections, praise the Lord for the gift of understanding. Ask him to help you use what you learn wisely and for the good of others.

Learning objectives

By the end of this lesson, you should be able to:

calculate and interpret the scalar product of two vectors;
find the angle between vectors and the acute angle between lines;
test whether vectors are perpendicular;
calculate a vector product and determine its direction using the right-hand rule;
find the area of a parallelogram or triangle using a vector product; and
calculate the components of one vector parallel and perpendicular to another.

1. The scalar product

The scalar product, also called the dot product, takes two vectors and produces a number. Multiply corresponding components and add:

$\mathbf{a}\cdot\mathbf{b} =a_1b_1+a_2b_2+a_3b_3$

For vectors in two dimensions, use only the first two terms. The scalar product is also related to the angle $\theta$ between two nonzero vectors:

$\boxed{\mathbf{a}\cdot\mathbf{b} =|\mathbf{a}|\,|\mathbf{b}|\cos\theta}$

Worked example · Calculate a scalar product

Let $\mathbf{a}=\begin{pmatrix}3\\-2\\4\end{pmatrix}$ and $\mathbf{b}=\begin{pmatrix}1\\5\\-1\end{pmatrix}$ . Then:

$\mathbf{a}\cdot\mathbf{b} =3(1)+(-2)(5)+4(-1) =\boxed{-11}$

The answer is a scalar, not a vector. Its negative sign tells us that the angle between these vectors is obtuse.

2. Angles and perpendicular vectors

Rearranging the scalar-product formula gives the angle between two nonzero vectors:

$\boxed{\theta= \cos^{-1}\left(\frac{\mathbf{a}\cdot\mathbf{b}} {|\mathbf{a}|\,|\mathbf{b}|}\right)}$

The angle between vectors lies between $0^\circ$ and $180^\circ$ . A zero scalar product means $\cos\theta=0$ , so the vectors are perpendicular.

$\boxed{\mathbf{a}\cdot\mathbf{b}=0 \quad\Longleftrightarrow\quad \mathbf{a}\perp\mathbf{b}}$

Worked example · Angle between vectors

Find the angle between $\mathbf{u}=\begin{pmatrix}2\\1\\-1\end{pmatrix}$ and $\mathbf{v}=\begin{pmatrix}1\\0\\1\end{pmatrix}$ .

$\mathbf{u}\cdot\mathbf{v}=2(1)+1(0)+(-1)(1)=1$

$|\mathbf{u}|=\sqrt6,\qquad|\mathbf{v}|=\sqrt2$

$\theta=\cos^{-1}\left(\frac1{\sqrt{12}}\right) \approx\boxed{73.2^\circ}$

Angle between lines: A line can be assigned either direction along its length. When an exam asks for the acute angle between two lines, use the absolute value of the dot product:

$\boxed{\alpha= \cos^{-1}\left(\frac{|\mathbf{a}\cdot\mathbf{b}|} {|\mathbf{a}|\,|\mathbf{b}|}\right)}$

Use the lines’ direction vectors for $\mathbf{a}$ and $\mathbf{b}$ . For perpendicular lines, the angle is $90^\circ$ .

3. The vector product

The vector product, also called the cross product, takes two three-dimensional vectors and produces a new vector perpendicular to both. If $\mathbf{a}=\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}$ and $\mathbf{b}=\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix}$ , then:

$\boxed{\mathbf{a}\times\mathbf{b} = \begin{pmatrix} a_2b_3-a_3b_2\\ a_3b_1-a_1b_3\\ a_1b_2-a_2b_1 \end{pmatrix}}$

Its magnitude is related to the angle between the vectors:

$\boxed{|\mathbf{a}\times\mathbf{b}| =|\mathbf{a}|\,|\mathbf{b}|\sin\theta}$

The right-hand rule gives the direction of $\mathbf{a}\times\mathbf{b}$ : point the fingers of your right hand along $\mathbf{a}$ and curl them toward $\mathbf{b}$ through the smaller angle; your thumb points along $\mathbf{a}\times\mathbf{b}$ . Reversing the order reverses the resulting vector.

Worked example · Calculate and check a cross product

Let $\mathbf{a}=\begin{pmatrix}2\\3\\4\end{pmatrix}$ and $\mathbf{b}=\begin{pmatrix}1\\0\\2\end{pmatrix}$ . Then:

$\mathbf{a}\times\mathbf{b} = \begin{pmatrix} 3(2)-4(0)\\ 4(1)-2(2)\\ 2(0)-3(1) \end{pmatrix} = \boxed{\begin{pmatrix}6\\0\\-3\end{pmatrix}}$

Check perpendicularity: $\begin{pmatrix}6\\0\\-3\end{pmatrix}\cdot\begin{pmatrix}2\\3\\4\end{pmatrix}=12-12=0$ , and $\begin{pmatrix}6\\0\\-3\end{pmatrix}\cdot\begin{pmatrix}1\\0\\2\end{pmatrix}=6-6=0$ .

4. Areas from the vector product

Place two nonparallel vectors tail to tail. They form adjacent sides of a parallelogram. The magnitude of their cross product gives its area:

$\boxed{A_{\text{parallelogram}} =|\mathbf{a}\times\mathbf{b}|}$

The triangle formed by the same vectors has half that area:

$\boxed{A_{\text{triangle}} =\frac12|\mathbf{a}\times\mathbf{b}|}$

Worked example · Area of a three-dimensional triangle

Points $A(1,0,2)$ , $B(3,1,2)$ , and $C(2,2,5)$ form a triangle. Find its area.

$\overrightarrow{AB}= \begin{pmatrix}2\\1\\0\end{pmatrix}, \qquad \overrightarrow{AC}= \begin{pmatrix}1\\2\\3\end{pmatrix}$

$\overrightarrow{AB}\times\overrightarrow{AC} = \begin{pmatrix}3\\-6\\3\end{pmatrix}$

$A= \frac12\sqrt{3^2+(-6)^2+3^2} =\boxed{\frac{3\sqrt6}{2}\approx3.67\text{ square units}}$

5. Components parallel and perpendicular to a direction

The signed scalar component of $\mathbf{a}$ in the direction of a nonzero vector $\mathbf{b}$ is:

$\boxed{\operatorname{comp}_{\mathbf{b}}\mathbf{a} =\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|} =|\mathbf{a}|\cos\theta}$

A negative answer means the component points opposite to $\mathbf{b}$ . The magnitude of the component of $\mathbf{a}$ perpendicular to $\mathbf{b}$ , within the plane of the two vectors, is:

$\boxed{\text{Perpendicular magnitude} =\frac{|\mathbf{a}\times\mathbf{b}|}{|\mathbf{b}|} =|\mathbf{a}|\sin\theta}$

To find the parallel vector, multiply the signed scalar component by the unit vector along $\mathbf{b}$ :

$\boxed{\mathbf{a}_{\parallel} =\frac{\mathbf{a}\cdot\mathbf{b}}{|\mathbf{b}|^2}\mathbf{b}, \qquad \mathbf{a}_{\perp}=\mathbf{a}-\mathbf{a}_{\parallel}}$

Worked example · Resolve a force

A force is $\mathbf{F}=\begin{pmatrix}3\\4\\0\end{pmatrix}$ N, and a track runs in the direction $\mathbf{b}=\begin{pmatrix}5\\0\\0\end{pmatrix}$ . Find the signed component along the track and the perpendicular magnitude.

$\operatorname{comp}_{\mathbf{b}}\mathbf{F} =\frac{3(5)}{5} =\boxed{3\text{ N}}$

$\mathbf{F}_{\parallel} =\begin{pmatrix}3\\0\\0\end{pmatrix}\text{ N}, \qquad \mathbf{F}_{\perp} =\begin{pmatrix}0\\4\\0\end{pmatrix}\text{ N}$

Therefore the perpendicular component has magnitude $\boxed{4\text{ N}}$ .

6. Choosing the right product

What you need	Method	Result
Angle or perpendicularity	Scalar product $\mathbf{a}\cdot\mathbf{b}$	A number
Signed component along a direction	$(\mathbf{a}\cdot\mathbf{b})/\|\mathbf{b}\|$	A number
A direction perpendicular to both vectors	Vector product $\mathbf{a}\times\mathbf{b}$	A vector
Area spanned by two vectors	Magnitude $\|\mathbf{a}\times\mathbf{b}\|$	An area
Perpendicular component magnitude	$\|\mathbf{a}\times\mathbf{b}\|/\|\mathbf{b}\|$	A nonnegative number

IB-style Paper 1 · Skill practice

Vectors $\mathbf{a}=\begin{pmatrix}k\\2\\-1\end{pmatrix}$ and $\mathbf{b}=\begin{pmatrix}4\\-1\\3\end{pmatrix}$ are perpendicular. Find $k$ . [3 marks]

Show solution

Set their dot product to zero:

$4k+2(-1)+(-1)(3)=0 \quad\Longrightarrow\quad 4k-5=0 \quad\Longrightarrow\quad \boxed{k=\frac54}$
Find the angle between $\mathbf{c}=\begin{pmatrix}2\\-3\\1\end{pmatrix}$ and $\mathbf{d}=\begin{pmatrix}5\\0\\-2\end{pmatrix}$ , to three significant figures. [4 marks]

Show solution

$\mathbf{c}\cdot\mathbf{d}=10-2=8, \qquad|\mathbf{c}|=\sqrt{14}, \qquad|\mathbf{d}|=\sqrt{29}$

$\theta=\cos^{-1}\left(\frac8{\sqrt{406}}\right) \approx\boxed{66.6^\circ}$
Find $\mathbf{u}\times\mathbf{v}$ for $\mathbf{u}=\begin{pmatrix}1\\-3\\2\end{pmatrix}$ and $\mathbf{v}=\begin{pmatrix}4\\0\\-1\end{pmatrix}$ . [4 marks]

Show solution

$\mathbf{u}\times\mathbf{v} = \begin{pmatrix} (-3)(-1)-2(0)\\ 2(4)-1(-1)\\ 1(0)-(-3)(4) \end{pmatrix} = \boxed{\begin{pmatrix}3\\9\\12\end{pmatrix}}$
The direction vectors of two lines are $\begin{pmatrix}2\\-1\\2\end{pmatrix}$ and $\begin{pmatrix}-1\\3\\0\end{pmatrix}$ . Find the acute angle between the lines, to one decimal place. [4 marks]

Show solution

The dot product is $-5$ , and the magnitudes are $3$ and $\sqrt{10}$ . For the acute angle between lines:

$\alpha= \cos^{-1}\left(\frac5{3\sqrt{10}}\right) \approx\boxed{58.2^\circ}$
Find the area of the parallelogram with adjacent sides $\begin{pmatrix}2\\0\\1\end{pmatrix}$ and $\begin{pmatrix}1\\3\\-1\end{pmatrix}$ . [4 marks]

Show solution

$\begin{pmatrix}2\\0\\1\end{pmatrix} \times\begin{pmatrix}1\\3\\-1\end{pmatrix} =\begin{pmatrix}-3\\3\\6\end{pmatrix}$

$A=\sqrt{(-3)^2+3^2+6^2} =\boxed{3\sqrt6\text{ square units}}$
Find the signed scalar component of $\mathbf{a}=\begin{pmatrix}2\\-1\\3\end{pmatrix}$ in the direction of $\mathbf{b}=\begin{pmatrix}-1\\4\\0\end{pmatrix}$ . [3 marks]

Show solution

$\mathbf{a}\cdot\mathbf{b}=-2-4=-6$ and $|\mathbf{b}|=\sqrt{17}$ , so the signed component is $\boxed{-6/\sqrt{17}\approx-1.46}$ . The negative sign means it acts opposite to $\mathbf{b}$ .

IB-style applications · Linked questions

Each application has connected parts (a), (b), and (c), with a complete solution to reveal after attempting the question.

Application 1 · Installing a triangular solar panel

Three mounting points, measured in metres, are $A(1,0,2)$ , $B(5,1,2)$ , and $C(2,3,4)$ .

Find the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$ . [2 marks]
Find a vector perpendicular to the plane of the panel. [4 marks]
Calculate the area of the triangular panel, to three significant figures. [3 marks]

Show worked solution · Application 1

(a) $\overrightarrow{AB}=\begin{pmatrix}4\\1\\0\end{pmatrix}$ and $\overrightarrow{AC}=\begin{pmatrix}1\\3\\2\end{pmatrix}$ .

(b)

$\overrightarrow{AB}\times\overrightarrow{AC} = \begin{pmatrix} 1(2)-0(3)\\ 0(1)-4(2)\\ 4(3)-1(1) \end{pmatrix} = \boxed{\begin{pmatrix}2\\-8\\11\end{pmatrix}}$

Its negative is also perpendicular to the panel.

(c)

$A=\frac12\sqrt{2^2+(-8)^2+11^2} =\frac{\sqrt{189}}2 \approx\boxed{6.87\text{ m}^2}$

Application 2 · Force along an inclined support

A force $\mathbf{F}=\begin{pmatrix}8\\6\\0\end{pmatrix}$ N acts at a joint. A support points in the direction $\mathbf{b}=\begin{pmatrix}3\\4\\0\end{pmatrix}$ .

Calculate the angle between the force and the support, to one decimal place. [4 marks]
Find the signed scalar component of the force along the support. [3 marks]
Find the magnitude of the component perpendicular to the support. [4 marks]

Show worked solution · Application 2

(a) $\mathbf{F}\cdot\mathbf{b}=8(3)+6(4)=48$ , while $|\mathbf{F}|=10$ and $|\mathbf{b}|=5$ . Hence:

$\theta=\cos^{-1}\left(\frac{48}{50}\right) \approx\boxed{16.3^\circ}$

(b) The signed component is $\boxed{48/5=9.6\text{ N}}$ .

(c) The perpendicular magnitude is:

$\sqrt{|\mathbf{F}|^2-9.6^2} =\sqrt{100-92.16} =\boxed{2.8\text{ N}}$

Equivalently, $|\mathbf{F}\times\mathbf{b}|/|\mathbf{b}|=14/5=2.8$ N.

Application 3 · Angle and area between flight paths

Two straight flight paths have vector equations:

$L_1:\ \mathbf{r} =\begin{pmatrix}2\\-1\\4\end{pmatrix} +s\begin{pmatrix}2\\1\\2\end{pmatrix}$

$L_2:\ \mathbf{r} =\begin{pmatrix}-3\\5\\1\end{pmatrix} +t\begin{pmatrix}1\\-2\\2\end{pmatrix}$

The direction vectors are used to compare the directions of travel. No intersection of the paths is assumed.

Calculate the scalar product of the two direction vectors. [2 marks]
Find the acute angle between the paths. [4 marks]
Find the area of the parallelogram spanned by the two direction vectors. [4 marks]

Show worked solution · Application 3

(a) $2(1)+1(-2)+2(2)=\boxed{4}$ .

(b) Each direction vector has magnitude $3$ . Thus:

$\alpha=\cos^{-1}\left(\frac4{3(3)}\right) \approx\boxed{63.6^\circ}$

(c)

$\begin{pmatrix}2\\1\\2\end{pmatrix} \times\begin{pmatrix}1\\-2\\2\end{pmatrix} = \begin{pmatrix}6\\-2\\-5\end{pmatrix}$

$A=\sqrt{6^2+(-2)^2+(-5)^2} =\boxed{\sqrt{65}\text{ square units}}$

The area describes the parallelogram formed by these vectors when placed tail to tail; it is not an area enclosed by the flight paths themselves.

Application 4 · Torque on a wrench

A wrench extends from a bolt to the point $\mathbf{r}=\begin{pmatrix}0.30\\0\\0\end{pmatrix}$ m. A force $\mathbf{F}=\begin{pmatrix}0\\40\\30\end{pmatrix}$ N acts at that point. The torque vector about the bolt is $\boldsymbol{\tau}=\mathbf{r}\times\mathbf{F}$ .

Find the torque vector. [4 marks]
Find its magnitude. [3 marks]
Calculate the angle between the position and force vectors and explain how it relates to the torque magnitude. [4 marks]

Show worked solution · Application 4

(a)

$\boldsymbol{\tau} = \begin{pmatrix}0.30\\0\\0\end{pmatrix} \times\begin{pmatrix}0\\40\\30\end{pmatrix} = \boxed{\begin{pmatrix}0\\-9\\12\end{pmatrix}\text{ N m}}$

(b) $|\boldsymbol{\tau}|=\sqrt{(-9)^2+12^2}=\boxed{15\text{ N m}}$ .

(c) Since $\mathbf{r}\cdot\mathbf{F}=0$ , the angle is $\boxed{90^\circ}$ . Also $|\mathbf{r}|=0.30$ m and $|\mathbf{F}|=\sqrt{40^2+30^2}=50$ N, so $|\boldsymbol{\tau}|=0.30(50)\sin90^\circ=15$ N m.

Application 5 · Orientation of a sloped panel

A rectangular panel is built along adjacent edge vectors $\mathbf{a}=\begin{pmatrix}2\\0\\1\end{pmatrix}$ m and $\mathbf{b}=\begin{pmatrix}0\\3\\0\end{pmatrix}$ m. A vertical direction is represented by $\mathbf{k}=\begin{pmatrix}0\\0\\1\end{pmatrix}$ .

Show that the two panel edges are perpendicular. [2 marks]
Find a normal vector to the panel and its area. [5 marks]
Find the acute angle between this normal and the vertical, to one decimal place. [4 marks]

Show worked solution · Application 5

(a) $\mathbf{a}\cdot\mathbf{b}=2(0)+0(3)+1(0)=0$ , so the edges are perpendicular.

(b)

$\mathbf{n}=\mathbf{a}\times\mathbf{b} =\begin{pmatrix}-3\\0\\6\end{pmatrix}$

$A=|\mathbf{n}|=\sqrt{(-3)^2+6^2} =\boxed{3\sqrt5\approx6.71\text{ m}^2}$

(c) $\mathbf{n}\cdot\mathbf{k}=6$ , $|\mathbf{n}|=3\sqrt5$ , and $|\mathbf{k}|=1$ . Therefore:

$\theta=\cos^{-1}\left(\frac6{3\sqrt5}\right) =\cos^{-1}\left(\frac2{\sqrt5}\right) \approx\boxed{26.6^\circ}$

Reflection · Learning with purpose

The dot product helps us understand alignment; the cross product helps us understand a direction and the space between two vectors. Psalm 147:5 points us toward the Lord’s understanding. As your own understanding grows, praise God and consider how you can use it to serve someone else.

Lesson 3.13 · Products of vectors

Praise for wisdom and understanding

Learning objectives

1. The scalar product

2. Angles and perpendicular vectors

3. The vector product

4. Areas from the vector product

5. Components parallel and perpendicular to a direction

6. Choosing the right product

IB-style Paper 1 · Skill practice

IB-style applications · Linked questions

Application 1 · Installing a triangular solar panel

Application 2 · Force along an inclined support

Application 3 · Angle and area between flight paths

Application 4 · Torque on a wrench

Application 5 · Orientation of a sloped panel

Reflection · Learning with purpose

Related

Company

Links

Support

Recommend

alphy school publishers

IB Mathematics: Applications and Interpretation HL 6 Sections80 LessonsLifetime

Curriculum

Products of vectors AI HL

Lesson 3.13 · Products of vectors

Praise for wisdom and understanding

Learning objectives

1. The scalar product

2. Angles and perpendicular vectors

3. The vector product

4. Areas from the vector product

5. Components parallel and perpendicular to a direction

6. Choosing the right product

IB-style Paper 1 · Skill practice

IB-style applications · Linked questions

Application 1 · Installing a triangular solar panel

Application 2 · Force along an inclined support

Application 3 · Angle and area between flight paths

Application 4 · Torque on a wrench

Application 5 · Orientation of a sloped panel

Reflection · Learning with purpose

Related

Leave A Reply Cancel reply

Company

Links

Support

Recommend

alphy school publishers

Customise Consent Preferences

Log in with your account

Modal title