Vectors: magnitude, direction and products

Read vector types and notation

Start by asking what the quantity describes: an amount, a location, or a directed change. Distance is a scalar; displacement needs both a length and a direction. Use the vocabulary below to distinguish these roles before choosing a formula.

The source introduces path integration through ants: model each leg of a zig-zag route by a displacement vector and add the legs to find the resultant. Mathematically, the direct return displacement is the negative of that sum. The ant photograph illustrates this motivation; it does not impose numerical data on a vector problem.

A scalar has magnitude only. The printed examples are mass, area, volume, earthquake strength, pollution index, energy, absolute temperature, distance and speed, represented by a number with suitable units. A vector has magnitude and direction: displacement, velocity, acceleration, force, momentum and torque are the source examples. Force magnitude is measured in newtons; adding forces must account for direction as well as magnitude.

The opening outline covers addition/subtraction/scaling and their geometry; position/displacement/direction; modulus, unit vectors and distance; collinearity and the ratio theorem; scalar/vector products, their properties and angles; and the geometric meanings of $\lvert\mathbf a\cdot\widehat{\mathbf n}\rvert$ and $\lvert\mathbf a\times\widehat{\mathbf n}\rvert$. This reading begins with the vocabulary and basic algebra. The source’s QR opens its O-Level vectors refresher.

Geometrical representation

A directed line segment: its length is the magnitude and its arrow gives direction. The source uses bold letters in print and underlined letters in handwriting.

Modulus

$\lvert\mathbf a\rvert$ is the vector’s magnitude or length, a scalar.

Zero/null vector

$\mathbf0$ has magnitude 0 and can be represented by a point. It has no direction.

Parallel vectors

Nonzero $\mathbf a,\mathbf b$ are parallel iff $\mathbf a=k\mathbf b$ for a nonzero real k. A positive k preserves direction; a negative k reverses it.

Negative vector

$-\mathbf a$ has the same magnitude and opposite direction when $\mathbf a\ne\mathbf0$. For zero, negation leaves zero unchanged.

Unit vector

$\widehat{\mathbf a}=\mathbf a/\lvert\mathbf a\rvert$ for nonzero a; its magnitude is 1.

Free vector

Only magnitude and direction are specified; its representation has no fixed initial or terminal point.

Position vector

$\overrightarrow{OA}$ locates A relative to a chosen origin O.

Displacement vector

$\overrightarrow{AB}$ starts at A and ends at B. It can start at any point; a position vector is the special case that starts at O.

Direction vector

The displacement between two distinct points on a line, or any nonzero scalar multiple of it, supplies a direction for that line.

Keep location separate from displacement

The printed displacement row describes a starting point other than O; the general definition also permits O. Translating a free-vector arrow preserves the vector, but changing the origin changes a point’s position vector.

Scale a unit direction to the required length

Normalisation also gives $\mathbf a=\lvert\mathbf a\rvert\widehat{\mathbf a}$. A vector b parallel to a and in the same direction is $\mathbf b=\lvert\mathbf b\rvert\widehat{\mathbf a}$: the source example of magnitude 6 is $6\widehat{\mathbf a}$. A vector in the opposite direction uses a negative multiple. Zero cannot be normalised.

Add, subtract and scale vectors

Draw the arrows before calculating. Join them head to tail to add, reverse the second arrow to subtract, and check the sign of a scalar before scaling. Moving a free arrow is allowed; rotating it changes the vector.

For nonzero vectors, equality means the same magnitude and direction, regardless of where their free-vector arrows are drawn. In the source grid, a,b,c have equal displacements even though their initial points differ. Zero equals zero too, without needing a direction.

The triangle law places b’s tail at a’s head; a+b joins the first tail to the final head. The source illustrates nonparallel vectors, but addition also works for parallel vectors and zero.

Commutative law:

$$\mathbf a+\mathbf b=\mathbf b+\mathbf a$$

The two boundary routes around the source parallelogram reach the same point and give the same resultant diagonal.

Associative law:

$$\mathbf a+\mathbf b+\mathbf c=(\mathbf a+\mathbf b)+\mathbf c=\mathbf a+(\mathbf b+\mathbf c).$$

The source figure uses the intermediate diagonals a+b and b+c to show two groupings of the same three-leg path.

Subtraction:

$$\mathbf a-\mathbf b=\mathbf a+(-\mathbf b)$$

Reverse b and then apply the triangle law. Its original direction and the reversed direction must remain distinct.

Scalar multiplication: for real $\lambda$,

$$\lvert\lambda\mathbf a\rvert=\lvert\lambda\rvert\lvert\mathbf a\rvert.$$

For nonzero a and nonzero λ, the result is parallel to a. Positive λ keeps the direction and negative λ reverses it; λ=0 produces zero, which has no direction. The source’s parallelism sentence must be read with this zero exception. Its three arrows illustrate a, positive scaling and negative scaling.

Apply the scalar laws

For real scalars m,n and vectors a,b, all four printed properties are:

1. $m\mathbf a=\mathbf a m$ (the scalar may be written on either side; this is scalar multiplication).
2. $(mn)\mathbf a=m(n\mathbf a)=n(m\mathbf a)$ (associativity of scalar multiplication).
3. $(m+n)\mathbf a=m\mathbf a+n\mathbf a$ (distributivity over scalar addition).
4. $m(\mathbf a+\mathbf b)=m\mathbf a+m\mathbf b$ (distributivity over vector addition).

A fixed walking distance multiplies a unit direction, not an arbitrary direction vector. Use the linked Alice question to distinguish $5\widehat{\mathbf p}+3\widehat{\mathbf q}$ from $5\mathbf p+3\mathbf q$. The midpoint proof applies these same path and scaling laws to a triangle.

Work with Cartesian vectors

For a displacement, subtract the starting coordinates from the ending coordinates. For a length, square the components, add and take the square root. Divide by that length only when a unit direction is needed and the vector is nonzero.

In two dimensions, i and j are unit vectors along the positive x- and y-axes. They are perpendicular:

$$\mathbf i=\begin{pmatrix}1\\0\end{pmatrix},\qquad\mathbf j=\begin{pmatrix}0\\1\end{pmatrix}.$$

For the source point A(4,3), the triangle law gives

$$\overrightarrow{OA}=4\mathbf i+3\mathbf j=\begin{pmatrix}4\\3\end{pmatrix},\qquad\lvert\overrightarrow{OA}\rvert=\sqrt{4^2+3^2}=5.$$

Its figure also marks B(2,4), both position vectors and the displacement A→B. In general, $\overrightarrow{OP}=\begin{pmatrix}x\\y\end{pmatrix}$ and $\lvert\overrightarrow{OP}\rvert=\sqrt{x^2+y^2}$. Following the path through O gives the complete displacement chain:

$$\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=-\overrightarrow{OA}+\overrightarrow{OB}=\overrightarrow{OB}-\overrightarrow{OA}.$$

The source translation example starts at P(1,4) with $\mathbf v=\begin{pmatrix}3\\-1\end{pmatrix}$. Then

$$\overrightarrow{OP}+\mathbf v=\begin{pmatrix}1\\4\end{pmatrix}+\begin{pmatrix}3\\-1\end{pmatrix}=\begin{pmatrix}4\\3\end{pmatrix}=\overrightarrow{OQ}.$$

So Q is (4,3): three units in the positive x-direction and one in the negative y-direction. Addition or subtraction of a fixed displacement translates the point.

Extension: coefficient columns

A coordinate column need not describe a spatial point. In the source extension, a real polynomial $a_0+a_1x+\cdots+a_nx^n$ can be represented by the $(n+1)\times1$ coefficient column $\begin{pmatrix}a_0\\a_1\\\vdots\\a_n\end{pmatrix}$, using the fixed ordered basis $1,x,\ldots,x^n$. This represents polynomials of degree at most n, including zero leading coefficients.

Extend to three dimensions

Use three mutually perpendicular unit directions i,j,k along positive x,y,z. Then

$$\overrightarrow{OP}=x\mathbf i+y\mathbf j+z\mathbf k=\begin{pmatrix}x\\y\\z\end{pmatrix},\qquad\lvert\overrightarrow{OP}\rvert=\sqrt{x^2+y^2+z^2}.$$

The source example A(4,3,5) gives $\overrightarrow{OA}=(4,3,5)$ and $\lvert\overrightarrow{OA}\rvert=\sqrt{4^2+3^2+5^2}=5\sqrt2$.

Derive the three-dimensional length

To prove the formulas, the source marks A(x,0,0), B(0,y,0), C(0,0,z), R(x,y,0) and P(x,y,z). In the xy-plane, $\overrightarrow{OR}=x\mathbf i+y\mathbf j$. Since RP is parallel to the z-axis, $\overrightarrow{RP}=z\mathbf k$, so

$$\overrightarrow{OP}=\overrightarrow{OR}+\overrightarrow{RP}=(x\mathbf i+y\mathbf j)+z\mathbf k=x\mathbf i+y\mathbf j+z\mathbf k.$$

Two applications of Pythagoras give

$$\lvert\overrightarrow{OR}\rvert=\sqrt{x^2+y^2},\qquad\lvert\overrightarrow{OP}\rvert=\sqrt{\lvert\overrightarrow{OR}\rvert^2+\lvert\overrightarrow{RP}\rvert^2}=\sqrt{(\sqrt{x^2+y^2})^2+z^2}=\sqrt{x^2+y^2+z^2}.$$

The orthonormal basis columns are $\mathbf i=\begin{pmatrix}1\\0\\0\end{pmatrix}$, $\mathbf j=\begin{pmatrix}0\\1\\0\end{pmatrix}$, $\mathbf k=\begin{pmatrix}0\\0\\1\end{pmatrix}$. For a right-handed system, curl the right-hand fingers from +x towards +y; the thumb points along +z. Viewed from +z towards O, that quarter-turn is anticlockwise. The projected drawing does not show true on-screen lengths or right angles.

Keep the six coordinate results together

For $A(x_1,y_1,z_1)$ and $B(x_2,y_2,z_2)$:

1. Position vector.

$$\overrightarrow{OA}=x_1\mathbf i+y_1\mathbf j+z_1\mathbf k=\begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix}$$

2. Magnitude.

$$\lvert\overrightarrow{OA}\rvert=\sqrt{x_1^2+y_1^2+z_1^2}$$

3. Unit direction, when A≠O.

$$\dfrac{\overrightarrow{OA}}{\lvert\overrightarrow{OA}\rvert}=\dfrac1{\sqrt{x_1^2+y_1^2+z_1^2}}\begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix}$$

4. Scalar multiplication.

$$k\overrightarrow{OA}=\begin{pmatrix}kx_1\\ky_1\\kz_1\end{pmatrix}$$

5. Equality of vectors.

$$\overrightarrow{OA}=\overrightarrow{OB}\iff x_1=x_2,\ y_1=y_2,\ z_1=z_2$$

All three components must agree. 6. Addition and subtraction.

$$\overrightarrow{OA}\pm\overrightarrow{OB}=\begin{pmatrix}x_1\\y_1\\z_1\end{pmatrix}\pm\begin{pmatrix}x_2\\y_2\\z_2\end{pmatrix}=\begin{pmatrix}x_1\pm x_2\\y_1\pm y_2\\z_1\pm z_2\end{pmatrix}$$

Prove collinearity with a common point

Choose two displacements sharing a point, then test whether one is a scalar multiple of the other. The same scalar must work in every coordinate. State the common point when turning the vector equation into a collinearity conclusion.

Three distinct points A,B,C are collinear iff $\overrightarrow{AB}=k\overrightarrow{BC}$ for a nonzero real k. They share B. The picture shows one possible ordering; the scalar can be positive or negative for other orderings.

Justify both directions

For the forward implication, points on the same line give parallel, nonzero AB and BC, hence a scalar multiple. Conversely, the scalar relation gives parallel segments AB and BC; because they share B, their supporting lines coincide, so A,B,C lie on one straight line.

The source gives two equivalent forms: $\overrightarrow{AB}=\lambda\overrightarrow{AC}$ with A common, and $\overrightarrow{AC}=\mu\overrightarrow{BC}$ with C common, for nonzero real λ,μ. The distinct-point condition is essential and also excludes coincident endpoints implicit in some scalar choices.

Use the linked examples to factor a common displacement, compare every coordinate, and justify the common point. To disprove collinearity, show that no scalar satisfies the vector equation. Merely finding one scalar that fails is insufficient. Self-Practice 1 also identifies the nondegenerate parallelogram O,A,B,D from its directed opposite sides.

Divide a segment in a stated ratio

Sketch the point order and mark the two segment lengths first. For an internal point, the endpoint weights are positive and sum to one. For a point on a produced line, identify the middle point before applying the internal formula.

For distinct points $A,B$, write $\overrightarrow{OA}=\mathbf a$, $\overrightarrow{OB}=\mathbf b$. If $P$ divides $AB$ internally with $AP:PB=\lambda:\mu$, where $\lambda,\mu>0$, then the ratio theorem (listed in the source’s MF27) gives

$$\mathbf p=\overrightarrow{OP}=\frac{\mu\mathbf a+\lambda\mathbf b}{\lambda+\mu}.$$

The coefficient of each endpoint is the opposite segment’s share.

Derive the endpoint weights

Begin with the directed fraction of $AB$. The full derivation is:

$$\begin{aligned} \overrightarrow{AP}&=\frac\lambda{\lambda+\mu}\overrightarrow{AB},\\ \overrightarrow{AO}+\overrightarrow{OP}&=\frac\lambda{\lambda+\mu}(\overrightarrow{AO}+\overrightarrow{OB}),\\ -\mathbf a+\mathbf p&=\frac\lambda{\lambda+\mu}(-\mathbf a+\mathbf b),\\ \mathbf p&=\frac\lambda{\lambda+\mu}(-\mathbf a+\mathbf b)+\mathbf a\\ &=\frac\lambda{\lambda+\mu}(-\mathbf a+\mathbf b)+\frac{\lambda+\mu}{\lambda+\mu}\mathbf a\\ &=\frac{-\lambda\mathbf a+\lambda\mathbf b+\lambda\mathbf a+\mu\mathbf a}{\lambda+\mu}\\ &=\frac{\mu\mathbf a+\lambda\mathbf b}{\lambda+\mu}. \end{aligned}$$

All position vectors must start at the same point. That point need not be the origin: replacing $O$ by any common reference $C$ gives

$$\overrightarrow{CP}=\frac{\mu\overrightarrow{CA}+\lambda\overrightarrow{CB}}{\lambda+\mu}.$$

When $\lambda=\mu$, $P$ is the midpoint and $\mathbf p=(\mathbf a+\mathbf b)/2$. The source assumes nonzero endpoint position vectors, but the affine formula also works when an endpoint is the origin. The positive-ratio and distinct-endpoint conditions describe an internal division of a genuine segment.

Read “produced” in order

$AB$ produced continues from $A$ through $B$ and beyond $B$; $BA$ produced continues from $B$ through $A$ and beyond $A$. First sketch the order, then apply the internal theorem to the middle point. Its three cases are $A,P,B$ with $AP:PB=3:7$; $B,A,R$ with $AR:BR=2:7$ (so $BA:AR=5:2$); and $A,B,Q$ with $AQ:BQ=7:3$ (so $AB:BQ=4:3$). Example 6 contrasts an internal $2:3$ point with $AB$ produced at $AP:PB=2:1$, where $B$ is the midpoint of $AP$.

Compare coefficients in two independent directions

Write the unknown point in two ways using the same origin and the same two independent directions. Equate the coefficients of each direction, solve the scalar equations, then check what the parameters say about the point order.

Two nonzero, nonparallel vectors $\mathbf a,\mathbf b$ span their plane: any vector in that plane has form $\lambda\mathbf a+\mu\mathbf b$. Its coefficients are unique:

$$\lambda_1\mathbf a+\mu_1\mathbf b=\lambda_2\mathbf a+\mu_2\mathbf b\quad\Longleftrightarrow\quad\lambda_1=\lambda_2\text{ and }\mu_1=\mu_2.$$

Example 5’s answers show this form: $(7/10)\mathbf a+(3/10)\mathbf b$, $(7/5)\mathbf a-(2/5)\mathbf b$, and $-(3/4)\mathbf a+(7/4)\mathbf b$. They are combinations in the same plane, including negative coefficients for external points. Each pair sums to one because these particular points lie on line $AB$; arbitrary planar combinations need not have that sum.

Why independence matters

The nonparallel condition is essential. Subtract the two representations. If the difference of the $\mathbf b$ coefficients were nonzero, the resulting equation would make $\mathbf b$ a multiple of $\mathbf a$, a contradiction. Both coefficient differences must therefore vanish. Parallel directions could give different coefficient pairs for the same vector.

Use two representations of the intersection

Example 7 uses $P$ midpoint of $AB$, $Q$ midpoint of $OP$ and $R=AQ\cap OB$. Describe $R$ from each line, keeping a common origin:

$$\overrightarrow{OR}=k\mathbf b=\left(1-\frac{3\lambda}4\right)\mathbf a+\frac\lambda4\mathbf b.$$

Compare coefficients only after identifying the independent directions. The source calculation gives $\lambda=4/3$, $k=1/3$; $\lambda>1$ confirms $R$ lies on $AQ$ produced. The full exercise keeps both midpoint steps, the position-to-displacement subtraction, and the final ratio $AQ:QR=3:1$.

Prove the centroid lies on all three medians

To prove that three medians meet, intersect two of them first. Express their intersection in the two independent side directions. Then show that this very point lies on the third median; a diagram alone cannot establish the concurrence.

A median joins a vertex to the midpoint of its opposite side. In a nondegenerate triangle $OAB$, let $M,N,R$ be the midpoints of $OB,OA,AB$. The intersection $G$ of medians $AM$ and $BN$ has two representations:

$$\overrightarrow{OG}=(1-\lambda)\mathbf a+\frac\lambda2\mathbf b=\frac\mu2\mathbf a+(1-\mu)\mathbf b.$$

Check the third median

The complete Self-Reading Example 8 works through both path expansions, compares independent coefficients, and solves $\lambda=\mu=2/3$. Thus $\overrightarrow{OG}=(\mathbf a+\mathbf b)/3$, $AG:GM=2:1$, and $\overrightarrow{OG}=(2/3)\overrightarrow{OR}$. The last vector equality proves that $G$ lies on the third median segment too.

Relabel $O$ as $C$: the three medians of any nondegenerate triangle $ABC$ meet at the centroid, two thirds of the way from each vertex to the opposite midpoint. The source’s final physical remark applies to a uniform triangular lamina in a uniform gravitational field: its centroid is also its centre of gravity. Equal side-midpoint marks and the $2:1$ division refer to different lengths; do not confuse them.

Read the scalar product and its laws

A dot product returns a number. Use it to connect two vector lengths with their angle, or to test perpendicularity. Check that both vectors are nonzero before interpreting a zero product as a right angle.

The scalar product, or dot product, of nonzero vectors is

$$\mathbf a\cdot\mathbf b=|\mathbf a||\mathbf b|\cos\theta,\qquad0\le\theta\le\pi.$$

It produces a scalar, not a vector.

To read $\theta$, translate the free vectors without rotating them so both point away from a common vertex or both point towards it.

Reversing both preserves the dot product; reversing just one changes its sign and usually gives the supplementary angle.

This is the key arrow-choice guidance repeated beside the source definition and Example 13. The source QR for exploring this geometry decodes to the CJC mathematics learning resource.

Use the product laws with their conditions

For nonzero $\mathbf a,\mathbf b$, the six source properties are:

1. Self-product. $\mathbf a\cdot\mathbf a=|\mathbf a|^2$, since $|\mathbf a||\mathbf a|\cos0=|\mathbf a|^2$. In particular, $(2\mathbf a+\mathbf b)\cdot(2\mathbf a+\mathbf b)=|2\mathbf a+\mathbf b|^2$, the additional example supplied in the Teacher note. The self-product identity also holds when its vector is zero.
2. Parallel directions. $|\mathbf a\cdot\mathbf b|=|\mathbf a||\mathbf b|$ if and only if the directions are parallel. Absolute-value bars are essential: $|\cos\theta|=1$ gives $\theta=0$ or $\pi$. Conversely those angles have cosines $1,-1$, giving dot products $|\mathbf a||\mathbf b|$ and $-|\mathbf a||\mathbf b|$, respectively, with equal absolute value.
3. Perpendicular directions. $\mathbf a\cdot\mathbf b=0$ if and only if $\theta=\pi/2$: divide $|\mathbf a||\mathbf b|\cos\theta=0$ by the nonzero lengths, then use $\cos\theta=0$. Conversely $\cos(\pi/2)=0$ makes the product zero. A zero vector also has zero dot product with every vector, but it has no defined angle; do not infer a geometric angle for it.
4. Commutativity. $\mathbf a\cdot\mathbf b=\mathbf b\cdot\mathbf a$: both are the same product $|\mathbf a||\mathbf b|\cos\theta$.
5. Compatibility with scalar multiplication. The source proves the following chain for $k>0$, when scaling preserves the direction:

$$\begin{aligned} \mathbf a\cdot(k\mathbf b)&=|\mathbf a||k\mathbf b|\cos\theta\\ &=k|\mathbf a||\mathbf b|\cos\theta\\ &=k(\mathbf a\cdot\mathbf b)\\ &=|k\mathbf a||\mathbf b|\cos\theta=(k\mathbf a)\cdot\mathbf b. \end{aligned}$$

The algebraic identity holds for every real $k$: for negative $k$, both direction reversal and magnitude scaling must be included; for zero $k$, every term is zero. The source calls this an associative law, but it is not a rule for dotting three vectors. For example, the Teacher’s unit-direction calculation gives

$$\mathbf a\cdot\widehat{\mathbf b}=\mathbf a\cdot\frac{\mathbf b}{|\mathbf b|}=\frac1{|\mathbf b|}(\mathbf a\cdot\mathbf b),\qquad\mathbf b\ne\mathbf0.$$

6. Distributivity. $\mathbf a\cdot(\mathbf b+\mathbf c)=\mathbf a\cdot\mathbf b+\mathbf a\cdot\mathbf c$. The source defers its geometric proof to the length-of-projection section (7.4.3). Use this law with commutativity to expand either factor in a product of sums.

Calculate scalar products from Cartesian components

Multiply matching coordinates and add the results. Keep negative signs with their components. The answer is a scalar; when testing perpendicular directions, compare it with zero and check that neither direction vector is zero.

In an orthonormal basis, write $\mathbf a=(a_1,a_2,a_3)$ and $\mathbf b=(b_1,b_2,b_3)$. Then

$$\mathbf a\cdot\mathbf b=a_1b_1+a_2b_2+a_3b_3=|\mathbf a||\mathbf b|\cos\theta.$$

The basis self-products are $\mathbf i\cdot\mathbf i=\mathbf j\cdot\mathbf j=\mathbf k\cdot\mathbf k=1$. Distinct perpendicular basis directions give $\mathbf i\cdot\mathbf j=\mathbf j\cdot\mathbf k=\mathbf k\cdot\mathbf i=0$, and their reversed products are zero by commutativity. The source basis sketch labels the positive y-direction with another i; use j there, as required by the printed formulas.

Derive the component rule

The complete expansion explains why only matching components survive:

$$\begin{aligned} \mathbf a\cdot\mathbf b&=(a_1\mathbf i+a_2\mathbf j+a_3\mathbf k)\cdot(b_1\mathbf i+b_2\mathbf j+b_3\mathbf k)\\ &=a_1\mathbf i\cdot(b_1\mathbf i+b_2\mathbf j+b_3\mathbf k)\\ &\quad+a_2\mathbf j\cdot(b_1\mathbf i+b_2\mathbf j+b_3\mathbf k)\\ &\quad+a_3\mathbf k\cdot(b_1\mathbf i+b_2\mathbf j+b_3\mathbf k)\\ &=a_1b_1\mathbf i\cdot\mathbf i+a_1b_2\mathbf i\cdot\mathbf j+a_1b_3\mathbf i\cdot\mathbf k\\ &\quad+a_2b_1\mathbf j\cdot\mathbf i+a_2b_2\mathbf j\cdot\mathbf j+a_2b_3\mathbf j\cdot\mathbf k\\ &\quad+a_3b_1\mathbf k\cdot\mathbf i+a_3b_2\mathbf k\cdot\mathbf j+a_3b_3\mathbf k\cdot\mathbf k\\ &=a_1b_1(1)+0+0+0+a_2b_2(1)+0+0+0+a_3b_3(1)\\ &=a_1b_1+a_2b_2+a_3b_3. \end{aligned}$$

Example 9 keeps both methods: form the columns first, or distribute and use self-products. Self-Practice 2 checks signs in two component calculations. Example 10 uses a zero product to find a parameter. Example 11 combines $\cos(\pi/3)=1/2$ with $\mathbf b\perp(\mathbf a-\mathbf b)$ to prove a relation between magnitudes; it does not make the two original vectors parallel.

Find the requested angle without reversing one vector

Name the vertex of the requested angle and choose both arrows from that vertex. Calculate the signed dot product before taking inverse cosine. An absolute value is appropriate for the smaller angle between lines, but can change a requested vector angle.

For two nonzero vectors, rearrange the definition before using inverse cosine:

$$\mathbf a\cdot\mathbf b=|\mathbf a||\mathbf b|\cos\theta \quad\Longrightarrow\quad \cos\theta=\frac{\mathbf a\cdot\mathbf b}{|\mathbf a||\mathbf b|} \quad\Longrightarrow\quad \theta=\cos^{-1}\!\left(\frac{\mathbf a\cdot\mathbf b}{|\mathbf a||\mathbf b|}\right).$$

A negative ratio gives an obtuse angle, with $-1$ giving a straight angle. A positive ratio gives an acute angle, with $1$ giving zero. A zero ratio gives a right angle. #### Distinguish a line angle from a vector angle

The source’s optional “acute angle” convention uses the absolute value:

$$\phi=\cos^{-1}\!\left(\frac{|\mathbf a\cdot\mathbf b|}{|\mathbf a||\mathbf b|}\right).$$

This returns the smaller unoriented angle between the corresponding lines, in $[0,\pi/2]$, including parallel and perpendicular limiting cases.

It does not replace a requested directed-vector angle: Example 12 has negative dot product and an obtuse answer.

Keep the exact inverse cosine until rounding; its printed radian answer 1.63 is corrected to 1.64 (3 s.f.) in the worked solution and retained unchanged in the source evidence.

Self-Practice 3 retains both degree and radian results with their printed precision.

Choose arrows at the named vertex

For $\angle ABC$, choose $\overrightarrow{BA}$ and $\overrightarrow{BC}$, both starting at $B$, or reverse both and use $\overrightarrow{AB},\overrightarrow{CB}$.

Example 13 has $\overrightarrow{BA}=(0,1,2)$, $\overrightarrow{BC}=(1,2,-3)$, dot product $-4$ and angle $118.6^\circ$.

Its source sketch only illustrates arrow choice; the teaching figure shows the calculated obtuse angle.

Self-Practice 4 contrasts both-outgoing, both-incoming and mixed arrows: (2) and (3) are valid identities; (1) is not a general identity, although it happens to agree when the angle is a right angle.

Relate a direction to the positive coordinate axes

For a nonzero vector, compare its direction with each positive coordinate axis. Divide each signed component by the vector length, then take inverse cosine if an angle is requested. A negative component gives an obtuse angle to that positive axis.

The positive x-, y- and z-axes are represented by the unit columns

$$\mathbf i=\begin{pmatrix}1\\0\\0\end{pmatrix},\qquad\mathbf j=\begin{pmatrix}0\\1\\0\end{pmatrix},\qquad\mathbf k=\begin{pmatrix}0\\0\\1\end{pmatrix}.$$

An angle to an axis is therefore another angle between vectors. For a nonzero $\mathbf v=(v_1,v_2,v_3)$ making angles $\theta,\phi,\omega$ with the positive axes,

$$\cos\theta=\frac{\mathbf v\cdot\mathbf i}{|\mathbf v||\mathbf i|}=\frac{v_1}{|\mathbf v|},\qquad \cos\phi=\frac{v_2}{|\mathbf v|},\qquad\cos\omega=\frac{v_3}{|\mathbf v|}.$$

These are directional cosines; their squares sum to one. Example 14 works through all three unit-column substitutions for $\overrightarrow{OA}=(1,4,7)$, giving $1/\sqrt{66},4/\sqrt{66},7/\sqrt{66}$ and the squared sum $(1+16+49)/66=1$. Keep signs when a component is negative.

Check why the squared cosines sum to one

Appendix A gives the general proof. Write $\mathbf a=a_x\mathbf i+a_y\mathbf j+a_z\mathbf k\ne\mathbf0$. For each positive coordinate axis,

$$\begin{aligned} \mathbf a\cdot\mathbf i&=|\mathbf a||\mathbf i|\cos\theta_x,&\cos\theta_x&=\frac{\mathbf a\cdot\mathbf i}{|\mathbf a||\mathbf i|}=\frac{a_x}{|\mathbf a|},\\ \mathbf a\cdot\mathbf j&=|\mathbf a||\mathbf j|\cos\theta_y,&\cos\theta_y&=\frac{\mathbf a\cdot\mathbf j}{|\mathbf a||\mathbf j|}=\frac{a_y}{|\mathbf a|},\\ \mathbf a\cdot\mathbf k&=|\mathbf a||\mathbf k|\cos\theta_z,&\cos\theta_z&=\frac{\mathbf a\cdot\mathbf k}{|\mathbf a||\mathbf k|}=\frac{a_z}{|\mathbf a|}. \end{aligned}$$

Consequently,

$$\begin{aligned} \cos^2\theta_x+\cos^2\theta_y+\cos^2\theta_z &=\frac{a_x^2}{|\mathbf a|^2}+\frac{a_y^2}{|\mathbf a|^2}+\frac{a_z^2}{|\mathbf a|^2}\\ &=\frac{a_x^2+a_y^2+a_z^2}{|\mathbf a|^2}\\ &=\frac{a_x^2+a_y^2+a_z^2}{(\sqrt{a_x^2+a_y^2+a_z^2})^2}=1. \end{aligned}$$

A zero vector has no direction angles, so this division requires $|\mathbf a|>0$. Angles are measured to the positive axes, not to whichever orientation gives the smaller angle.

Resolve parallel and perpendicular components

Start with a nonzero target direction and turn it into a unit vector. Then decide whether the question asks for a signed scalar, a nonnegative length, or a vector component. The same dot product appears in each, but the final objects are different.

Let $\widehat{\mathbf b}=\mathbf b/|\mathbf b|$, with $\mathbf b\ne\mathbf0$. In the source diagrams, $\overrightarrow{PR}=\mathbf a$ and $Q$ is the perpendicular foot from $R$ to the line through $P$ in direction $\mathbf b$. Distinguish three different results:

$$\begin{aligned} PQ&=|\mathbf a\cdot\widehat{\mathbf b}| &&\text{(projection length: scalar)},\\ \overrightarrow{PQ}&=(\mathbf a\cdot\widehat{\mathbf b})\widehat{\mathbf b} &&\text{(parallel component: vector)},\\ \overrightarrow{QR}&=\mathbf a-(\mathbf a\cdot\widehat{\mathbf b})\widehat{\mathbf b} &&\text{(perpendicular component: vector)}. \end{aligned}$$

The signed scalar component $\mathbf a\cdot\widehat{\mathbf b}$ may be negative; a length cannot. The source assumes nonparallel vectors to draw a genuine triangle. The formulas also work for parallel or perpendicular $\mathbf a$ and for $\mathbf a=\mathbf0$; only the target $\mathbf b$ must be nonzero.

Acute case: the projection follows b

Let $\theta$ be the angle between $\mathbf a$ and $\mathbf b$. In the acute diagram, it is also $\angle QPR$. The complete length proof is

$$\frac{PQ}{PR}=\cos\theta,\qquad PQ=PR\cos\theta=|\mathbf a|\cos\theta=|\mathbf a||\widehat{\mathbf b}|\cos\theta=\mathbf a\cdot\widehat{\mathbf b}.$$

Since $\overrightarrow{PQ}$ and $\mathbf b$ have the same unit direction,

$$\frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}=\frac{\mathbf b}{|\mathbf b|} \quad\Longrightarrow\quad \overrightarrow{PQ}=|\overrightarrow{PQ}|\frac{\mathbf b}{|\mathbf b|}=(\mathbf a\cdot\widehat{\mathbf b})\widehat{\mathbf b}.$$

Obtuse case: the projection opposes b

In the source’s second drawing, $\mathbf b$ points left while $\overrightarrow{PR}$ points up-right. The angle between the vectors is obtuse $\theta$, while the triangle’s angle $\angle QPR=\pi-\theta$ is acute. This distinction corrects the source’s unqualified opening “let angle QPR be θ” for the second case. Now

$$\begin{aligned} \frac{PQ}{PR}&=\cos(\pi-\theta),\\ PQ&=-PR\cos\theta\\ &=-|\mathbf a|\cos\theta\\ &=-|\mathbf a||\widehat{\mathbf b}|\cos\theta\\ &=-\mathbf a\cdot\widehat{\mathbf b}. \end{aligned}$$

The unit directions are opposite, so

$$\begin{aligned} \frac{\overrightarrow{PQ}}{|\overrightarrow{PQ}|}&=-\frac{\mathbf b}{|\mathbf b|},\\ \overrightarrow{PQ}&=-|\overrightarrow{PQ}|\frac{\mathbf b}{|\mathbf b|}\\ &=-(-\mathbf a\cdot\widehat{\mathbf b})\frac{\mathbf b}{|\mathbf b|}\\ &=(\mathbf a\cdot\widehat{\mathbf b})\widehat{\mathbf b}. \end{aligned}$$

The Teacher asks why the length formulas have opposite signs: they are the positive and negative cases of the same absolute-value expression. The vector formula is unchanged in both cases. Finally, the source triangle-law proof gives

$$\overrightarrow{PR}=\overrightarrow{PQ}+\overrightarrow{QR},\qquad \mathbf a=(\mathbf a\cdot\widehat{\mathbf b})\widehat{\mathbf b}+\overrightarrow{QR},\qquad \overrightarrow{QR}=\mathbf a-(\mathbf a\cdot\widehat{\mathbf b})\widehat{\mathbf b}.$$

Example 15 applies all three formulas and labels the parallel and perpendicular arrows (ii) and (iii). Its Teacher note stresses that projection of a on b follows b, whereas projection of b on a follows a. Their vectors generally differ even if their lengths happen to agree, as they do in this example. Self-Practice 5 keeps a negative projection coefficient; do not put absolute-value bars into that vector answer.

The earlier scalar-product table refers forward to this projection section for a distributivity proof. The supplied projection pages derive these formulas; a separate proof of distributivity is not printed there.

Reverse the direction and track what changes

Let $\mathbf a=(-3,4,0)$ and $\widehat{\mathbf n}=(1,0,0)$. Find the signed scalar component, projection length, parallel vector component and perpendicular vector component of $\mathbf a$ relative to this direction. Predict which results change when $\widehat{\mathbf n}$ is replaced by $-\widehat{\mathbf n}$, then verify.

After your attempt: compare the reasoning

Initially the signed component is $-3$, the projection length is 3, the parallel component is $(-3,0,0)$ and the perpendicular component is $(0,4,0)$. With the reversed unit direction the signed component becomes 3, but

$[\mathbf a\cdot(-\widehat{\mathbf n})](-\widehat{\mathbf n})=(\mathbf a\cdot\widehat{\mathbf n})\widehat{\mathbf n}.$

Both vector components and the projection length stay unchanged. Also $|\mathbf a\times\widehat{\mathbf n}|=4$ measures the perpendicular length. A signed scalar and a vector component are different objects: reversing the measuring direction changes the scalar’s sign without moving the foot.

Use the cross-product direction and laws

Keep the factor order visible. For nonparallel vectors, curl from the first towards the second with your right hand to choose the normal direction. The cross product is a vector; its magnitude gives a parallelogram area. To find a perpendicular length, cross with a unit direction.

For two nonparallel nonzero vectors in three dimensions, the vector product, or cross product, is

$$\mathbf a\times\mathbf b=(|\mathbf a||\mathbf b|\sin\theta)\widehat{\mathbf n},\qquad0<\theta<\pi.$$

Here $\widehat{\mathbf n}$ is perpendicular to both vectors. Curl the fingers of your right hand from $\mathbf a$ towards $\mathbf b$ through the smaller angle; your thumb selects $\widehat{\mathbf n}$.

The result is a vector. Reversing the order reverses the normal.

The source’s hand illustration supplies this curl/thumb rule; its accompanying physics motivation mentions induced current in a moving wire, generators, dams and windmills and uses the name Fleming’s right-hand rule.

For the mathematics here, apply the stated curl from the first vector to the second convention.

If either vector is zero, the cross product is $\mathbf0$. Parallel nonzero inputs also give zero. In these cases there is no nonzero cross-product direction to normalise; the unit normal in the displayed definition is used only in the nonparallel case.

Apply the cross-product laws

The six source properties and their proof roles are:

1. Magnitude. Taking the norm of $(|\mathbf a||\mathbf b|\sin\theta)\widehat{\mathbf n}$ gives $|\mathbf a\times\mathbf b|=|\mathbf a||\mathbf b|\sin\theta$, since $|\widehat{\mathbf n}|=1$ and $\sin\theta\ge0$ for $\theta\in[0,\pi]$. No absolute value around the sine is needed on this interval.
2. Parallel iff zero product, for nonzero inputs. Parallel directions have $\theta=0$ or $\pi$, so $\sin\theta=0$ and the cross product is zero. Conversely $\mathbf a\times\mathbf b=\mathbf0$ gives $|\mathbf a||\mathbf b|\sin\theta=0$; nonzero lengths force $\sin\theta=0$, hence $\theta=0$ or $\pi$. In particular $\mathbf a\times\mathbf a=\mathbf0$.
3. Perpendicular iff maximal magnitude, for nonzero inputs. If $\theta=\pi/2$, then $|\mathbf a\times\mathbf b|=|\mathbf a||\mathbf b|\sin(\pi/2)=|\mathbf a||\mathbf b|$. Conversely that equality gives $\sin\theta=1$, so $\theta=\pi/2$.
4. Anti-commutativity. The right-hand rule changes $\widehat{\mathbf n}$ to $-\widehat{\mathbf n}$ when the inputs are swapped:

$$\mathbf b\times\mathbf a=|\mathbf b||\mathbf a|\sin\theta(-\widehat{\mathbf n})=-|\mathbf a||\mathbf b|\sin\theta\widehat{\mathbf n}=-\mathbf a\times\mathbf b.$$

Thus $\mathbf a\times\mathbf b=-(\mathbf b\times\mathbf a)$.

5. Compatibility with scalars. $k(\mathbf a\times\mathbf b)=(k\mathbf a)\times\mathbf b=\mathbf a\times(k\mathbf b)$. The source’s magnitude proof reads

$$\begin{aligned} k(\mathbf a\times\mathbf b)&=k|\mathbf a||\mathbf b|\sin\theta\widehat{\mathbf n}\\ &=|k\mathbf a||\mathbf b|\sin\theta\widehat{\mathbf n}=(k\mathbf a)\times\mathbf b\\ &=|\mathbf a||k\mathbf b|\sin\theta\widehat{\mathbf n}=\mathbf a\times(k\mathbf b). \end{aligned}$$

That same-normal magnitude chain requires positive k; for negative k, the angle and normal direction also change. The algebraic identity is still valid for every real k, including zero, as the Cartesian formula confirms. The source calls this an associative law; it concerns a scalar factor, not arbitrary regrouping of three vector cross products.

6. Distributivity, in both orders:

$$\mathbf a\times(\mathbf b+\mathbf c)=\mathbf a\times\mathbf b+\mathbf a\times\mathbf c,\qquad (\mathbf b+\mathbf c)\times\mathbf a=\mathbf b\times\mathbf a+\mathbf c\times\mathbf a.$$

Example 16 combines anti-commutativity and distributivity into a zero cross product. Check that both factors are nonzero before using its geometric parallelism conclusion.

Calculate and check a Cartesian cross product

Calculate the three components in order, taking particular care with the middle sign. Check that a proposed common perpendicular has zero dot product with both inputs. If a direction is required, also check that your result is nonzero.

In a right-handed orthonormal basis, the source coordinate formula and its MF27 version are the same:

$$\begin{pmatrix}a_1\\a_2\\a_3\end{pmatrix}\times\begin{pmatrix}b_1\\b_2\\b_3\end{pmatrix} =\begin{pmatrix}a_2b_3-a_3b_2\\-(a_1b_3-a_3b_1)\\a_1b_2-a_2b_1\end{pmatrix} =\begin{pmatrix}a_2b_3-a_3b_2\\a_3b_1-a_1b_3\\a_1b_2-a_2b_1\end{pmatrix}.$$

This column equals $(|\mathbf a||\mathbf b|\sin\theta)\widehat{\mathbf n}$ in the nonparallel case. The positive cyclic basis products are

$$\mathbf i\times\mathbf j=\mathbf k,\quad\mathbf j\times\mathbf k=\mathbf i,\quad\mathbf k\times\mathbf i=\mathbf j.$$

Reversed order gives

$$\mathbf j\times\mathbf i=-\mathbf k,\quad\mathbf k\times\mathbf j=-\mathbf i,\quad\mathbf i\times\mathbf k=-\mathbf j,$$

and $\mathbf i\times\mathbf i=\mathbf j\times\mathbf j=\mathbf k\times\mathbf k=\mathbf0$. The source QR exploring how the product relates to its inputs decodes to the CJC mathematics learning resource.

Derive the signs from the basis products

To derive the coordinate formula, expand the basis expressions into all nine terms before substituting the cyclic rules:

$$\begin{aligned} \mathbf a\times\mathbf b&=(a_1\mathbf i+a_2\mathbf j+a_3\mathbf k)\times(b_1\mathbf i+b_2\mathbf j+b_3\mathbf k)\\ &=a_1b_1\mathbf i\times\mathbf i+a_1b_2\mathbf i\times\mathbf j+a_1b_3\mathbf i\times\mathbf k\\ &\quad+a_2b_1\mathbf j\times\mathbf i+a_2b_2\mathbf j\times\mathbf j+a_2b_3\mathbf j\times\mathbf k\\ &\quad+a_3b_1\mathbf k\times\mathbf i+a_3b_2\mathbf k\times\mathbf j+a_3b_3\mathbf k\times\mathbf k\\ &=a_1b_1\mathbf0+a_1b_2\mathbf k+a_1b_3(-\mathbf j)\\ &\quad+a_2b_1(-\mathbf k)+a_2b_2\mathbf0+a_2b_3\mathbf i\\ &\quad+a_3b_1\mathbf j+a_3b_2(-\mathbf i)+a_3b_3\mathbf0\\ &=(a_2b_3-a_3b_2)\mathbf i-(a_1b_3-a_3b_1)\mathbf j+(a_1b_2-a_2b_1)\mathbf k\\ &=\begin{pmatrix}a_2b_3-a_3b_2\\-(a_1b_3-a_3b_1)\\a_1b_2-a_2b_1\end{pmatrix}. \end{aligned}$$

Example 17 produces a common perpendicular and checks it using two independent dot products. A zero result would pass those checks too, so verify that the requested perpendicular direction is nonzero. Self-Practice 6 applies the component formula to two pairs.

Use a cross-product magnitude for triangle area

Choose two sides from the same vertex and calculate their cross product. Take its magnitude for a parallelogram area, then halve it for a triangle. Reversing the factor order changes orientation but leaves the area unchanged.

For a nondegenerate triangle $ABC$, choose two side displacements with the same starting vertex. With $\theta=\angle BAC$, the complete source proof is

$$\begin{aligned} \operatorname{Area}(ABC)&=\frac12(AB)(AC)\sin\theta\\ &=\frac12|\overrightarrow{AB}||\overrightarrow{AC}|\sin\theta\\ &=\frac12|\overrightarrow{AB}||\overrightarrow{AC}|\sin\theta|\widehat{\mathbf n}|\\ &=\frac12|\overrightarrow{AB}\times\overrightarrow{AC}|. \end{aligned}$$

Here $\widehat{\mathbf n}$ is a unit normal to the two side directions. The result is a nonnegative scalar area, not the cross-product vector. The source also gives the equivalent choices

$$\operatorname{Area}(ABC)=\frac12|\overrightarrow{BA}\times\overrightarrow{BC}|=\frac12|\overrightarrow{CA}\times\overrightarrow{CB}|.$$

Switching vertices changes the two displacements together; swapping their order changes the normal sign but not its magnitude. The formula also yields zero for a degenerate triangle, without requiring a unit normal there.

Extend from a triangle to a parallelogram

A parallelogram is twice either triangle made by one diagonal. Thus the same two adjacent side vectors give

$$\begin{aligned} \operatorname{Area}(ABCD)&=2\operatorname{Area}(ABD)\\ &=2\left(\frac12|\overrightarrow{AB}\times\overrightarrow{AD}|\right)\\ &=|\overrightarrow{AB}\times\overrightarrow{AD}|. \end{aligned}$$

The length of the cross-product vector is the parallelogram area. If one side is scaled by $t$, the cross-product vector scales by $t$ and its length scales by $|t|$. This explains the Teacher’s suggested comparison between changing vector length and changing area. Example 18 computes both areas from the same cross product; its diagram is schematic.

Form sides from position vectors

For three position vectors, first subtract to form two sides. Example 19 expands $(\mathbf b-\mathbf a)\times(\mathbf c-\mathbf a)$ and obtains

$$\operatorname{Area}(ABC)=\frac12|\mathbf a\times\mathbf b+\mathbf b\times\mathbf c+\mathbf c\times\mathbf a|.$$

Every term inside the norm is a vector: in particular, $\mathbf a\times\mathbf a=\mathbf0$, the zero vector. Reversing the triangle orientation negates the vector sum and leaves its magnitude unchanged. Translating the origin leaves both side differences unchanged, so the area is independent of that choice.

Separate orientation from area

Two sides from the same vertex are $\mathbf u=(2,0,0)$ and $\mathbf v=(1,3,0)$. Find the triangle area. Predict what happens to the cross product and to the area if you swap the sides, then if you replace $\mathbf v$ by $2\mathbf v$. Explain using a base and height as well as a product.

After your attempt: compare the reasoning

Here $\mathbf u\times\mathbf v=(0,0,6)$, so the triangle area is $\tfrac12(6)=3$. The base length is 2 and the perpendicular height is 3, confirming this area. Swapping the sides reverses the cross product to $(0,0,-6)$ but leaves its magnitude and the area unchanged. Doubling the second side gives cross product $(0,0,12)$ and area 6: the height doubles. The cross-product magnitude gives a parallelogram area; a triangle needs the factor $1/2$.

Compare parallel and perpendicular lengths

Use a unit target direction before comparing lengths. The absolute dot product measures the parallel length; the cross-product magnitude measures the perpendicular length. Both answers are scalars and should satisfy the Pythagorean check below.

Let $\mathbf a,\mathbf b$ be nonparallel nonzero vectors, $\widehat{\mathbf b}=\mathbf b/|\mathbf b|$, and $\widehat{\mathbf n}$ their right-hand unit normal. In the source right triangle $PQR$, $\overrightarrow{PR}=\mathbf a$ and $Q$ is the perpendicular foot on the line through $P$ parallel to $\mathbf b$. The acute drawing has $\theta$ at $P$. Its parallel length is $PQ=|\mathbf a\cdot\widehat{\mathbf b}|$, and the complete source height proof is

$$\begin{aligned} \frac{QR}{PR}&=\sin\theta,\\ QR&=|\mathbf a|\sin\theta\\ &=|\mathbf a||\widehat{\mathbf b}|\sin\theta|\widehat{\mathbf n}|\\ &=\big|\big(|\mathbf a||\widehat{\mathbf b}|\sin\theta\big)\widehat{\mathbf n}\big|\\ &=|\mathbf a\times\widehat{\mathbf b}|. \end{aligned}$$

Thus the dot-product magnitude gives the parallel length, and the cross-product magnitude gives the perpendicular length. The cross-product vector itself points normal to their plane; it is not the in-plane perpendicular component. For an obtuse vector angle, the triangle angle is $\pi-\theta$ and $\sin(\pi-\theta)=\sin\theta$, giving the same height. The length formulas extend to parallel directions and $\mathbf a=\mathbf0$ without choosing a unit normal there; $\mathbf b$ must remain nonzero. Pythagoras supplies a useful check:

$$|\mathbf a\cdot\widehat{\mathbf b}|^2+|\mathbf a\times\widehat{\mathbf b}|^2=|\mathbf a|^2.$$

Choose between the two products

The source closes with six comparisons. The geometric “if and only if” statements below assume both vectors nonzero; the algebraic laws also hold for zero vectors. Here $t$ is any real scalar. Here “iff” means “if and only if”.

Self-product

Scalar product:

$$\mathbf a\cdot\mathbf a=\lvert\mathbf a\rvert^2$$

Vector product:

$$\mathbf a\times\mathbf a=\mathbf0$$

Greatest possible magnitude

Scalar product:

$$\lvert\mathbf a\cdot\mathbf b\rvert=\lvert\mathbf a\rvert\lvert\mathbf b\rvert$$

iff parallel

Vector product:

$$\lvert\mathbf a\times\mathbf b\rvert=\lvert\mathbf a\rvert\lvert\mathbf b\rvert$$

iff perpendicular

Zero product

Scalar product:

$$\mathbf a\cdot\mathbf b=0$$

iff perpendicular

Vector product:

$$\mathbf a\times\mathbf b=\mathbf0$$

iff parallel

Reverse order

Scalar product:

$$\mathbf a\cdot\mathbf b=\mathbf b\cdot\mathbf a$$

Vector product:

$$\mathbf a\times\mathbf b=-(\mathbf b\times\mathbf a)$$

Scalar compatibility

Scalar product:

$$\mathbf a\cdot(t\mathbf b)=t(\mathbf a\cdot\mathbf b)=(t\mathbf a)\cdot\mathbf b$$

Vector product:

$$t(\mathbf a\times\mathbf b)=(t\mathbf a)\times\mathbf b=\mathbf a\times(t\mathbf b)$$

Distribute in either position

Scalar product:

$$\mathbf a\cdot(\mathbf b+\mathbf c)=\mathbf a\cdot\mathbf b+\mathbf a\cdot\mathbf c$$ $$(\mathbf b+\mathbf c)\cdot\mathbf a=\mathbf b\cdot\mathbf a+\mathbf c\cdot\mathbf a$$

Vector product:

$$\mathbf a\times(\mathbf b+\mathbf c)=\mathbf a\times\mathbf b+\mathbf a\times\mathbf c$$ $$(\mathbf b+\mathbf c)\times\mathbf a=\mathbf b\times\mathbf a+\mathbf c\times\mathbf a$$

The source calls scalar compatibility an “associative law”; it does not assert that a cross product of three vectors is associative. Use these length relationships later for distances to lines and planes.

Connect vectors with motion and work

Identify what each component measures and keep its units. Velocity and force are vectors, while speed and work are scalars. Use the model assumptions to decide which calculation fits the physical question.

Appendix B uses three physics settings to motivate three-dimensional vectors. Its syllabus references describe the supplied 2025 notes.

Particle motion in 3D. Constant-velocity and constant-acceleration models simplify motion. A boomerang can change velocity and acceleration in multiple directions. The source illustrates a projectile under constant downward acceleration with its linked three-dimensional projectile demonstration.

Collisions. Straight-line particle collisions are a one-dimensional simplification. A baseball meeting a bat and repeated billiard-ball collisions involve more directions. The source links a collision model for two balls of equal volume and mass.

Work done by a force. For a constant force $\mathbf F$ and displacement $\mathbf s$, work is displacement length times the force component along that displacement:

$$W=|\mathbf s|(|\mathbf F|\cos\theta)=\mathbf F\cdot\mathbf s.$$

The sign matters: a force component opposing the displacement does negative work. The two source QR references are vectors and work and the mathematics of work. These are the original external resource addresses; the chapter’s explanation does not depend on opening them.

Distinguish a highest point from coming to rest

A particle has position $\mathbf r(t)=(2t,6t-t^2,0)$ metres for $0\le t\le6$ seconds, with the second coordinate representing height. Find its velocity and acceleration. Locate its highest point and decide whether it is at rest there. State which component controls the height conclusion.

After your attempt: compare the reasoning

Differentiating components gives $\mathbf v(t)=(2,6-2t,0)$ metres per second and $\mathbf a(t)=(0,-2,0)$ metres per second squared. Height is $6t-t^2=9-(t-3)^2$, so the highest point is $(6,9,0)$ at $t=3$. There the vertical velocity is zero, but $\mathbf v(3)=(2,0,0)$ and speed is 2 metres per second. Zero vertical velocity does not mean zero velocity. The downward vertical acceleration also confirms that the stationary height is a maximum.

Review mastery and choose the next practice

Choose one skill you cannot yet explain without looking. Follow its link, then attempt the named example or discussion question before revealing the working. Use the chapter overview to reopen this checklist. Completing a page and being able to use its method are different checkpoints.

Use the same two passes as the source checklist: at the end of the lecture, try these from memory; after the tutorial, revisit the ones you could not explain. The source’s two columns are blank self-assessment spaces, not completed mastery claims. Use each practice question’s existing self-check to record your confidence after attempting it.

Check the foundations from memory

For basic properties in two and three dimensions, check that you can:

1. Read and write vector notation: printed bold $\mathbf a$, handwritten underlined $\underline a$, displacement $\overrightarrow{AB}$, a coordinate column $\begin{pmatrix}x\\y\\z\end{pmatrix}$ and $x\mathbf i+y\mathbf j+z\mathbf k$.
2. Add and subtract vectors and explain their geometry.
3. Multiply a vector by a scalar and explain the change of length and direction.
4. Find the displacement from one given point to another.
5. Calculate a vector’s magnitude or length.
6. Find a unit vector, checking that the original vector is nonzero.
7. Draw a diagram that helps represent a vector problem.
8. Apply the ratio theorem to locate a point dividing a segment in a given ratio.
9. Use collinearity and/or ratio equations to solve vector geometry problems.

Check the product methods from memory

For scalar and vector products, check that you can:

1. Define and calculate a scalar (dot) product.
2. Manipulate scalar-product expressions using its laws.
3. Test perpendicularity with a scalar product and formulate the corresponding equation.
4. Calculate the angle between vectors placed with common initial points.
5. Define and calculate a vector (cross) product in three dimensions.
6. Manipulate vector-product expressions using its laws.
7. Find triangle and parallelogram areas with a vector product.
8. Choose the appropriate scalar or vector product for the length parallel or perpendicular to a given direction.

Choose practice for the skill you need

The source groups practice into eleven question types. Choose a row that matches your uncertainty, follow its lesson link and attempt the named example, then use Discussion and Review for another context. The example numbers here refer to this chapter’s current lecture examples.

Unit vectors

Start with: Example 1(iii)

Check your method: Normalise before scaling; specify orientation.

Magnitude

Start with: Example 1(ii)

Check your method: Take the square root of the sum of squared components.

Equal vectors

Start with: Discussion: Question 2

Check your method: Compare the appropriate components or directed sides.

Ratio theorem

Start with: Example 5

Check your method: Draw internal or produced-line point order before weighting endpoints.

Parallel or collinear vectors

Start with: Example 4

Check your method: Require one consistent scalar multiple and a common point for collinearity.

Dot products

Start with: Example 9

Check your method: Distinguish the scalar result from vector inputs.

Cross products

Start with: Example 17

Check your method: Preserve factor order, then check both dot products.

Geometrical interpretation

Start with: Example 15

Check your method: Distinguish a length from a parallel or residual vector.

Projection length

Start with: Example 15(i)

Check your method: Normalise the target direction and retain absolute value for length.

Angles

Start with: Example 12

Check your method: Use common-tail vectors and the requested rounding.

Triangle or parallelogram area

Start with: Example 18

Check your method: Use adjacent sides, then decide whether a factor one-half is needed.

These method choices use the reviewed current questions. The original source index is retained with its historical subpart references; some refer to earlier Review numbering, so use the chapter’s current links to find questions.