Vector Algebra

Brackets do not matter in ordinary multiplication: $(2\times 3)\times 4$ and $2\times (3\times 4)$ both give $24$.

Try the same with the cross product, using the simplest vectors available.

$(\vec{i}\times \vec{i})\times \vec{j}=\vec{0}\times \vec{j}=\vec{0}$

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

One is the zero vector. The other has length $1$. The cross product is not associative, and the brackets are not decoration.

A second habit fails just as quickly. In ordinary algebra $ab=ac$ with $a\ne 0$ gives $b=c$.

$\vec{i}\cdot \vec{j}=0=\vec{i}\cdot \vec{k},\text{and yet }\vec{j}\ne \vec{k}$

Neither failure is an accident, and both come from the same source.

$\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos \theta ,|\vec{a}\times \vec{b}|=|\vec{a}||\vec{b}|\sin \theta$

The dot product measures how much two vectors agree; the cross product measures how much they do not. The dot product is blind to everything perpendicular, which is why cancelling it loses information. The cross product is blind to everything parallel, which is why $\vec{i}\times \vec{i}$ vanishes and the brackets matter.

Put them side by side and nothing is lost at all.

$(\vec{a}\cdot \vec{b}{)}^2+|\vec{a}\times \vec{b}{|}^2=|\vec{a}{|}^2|\vec{b}{|}^2\left({\cos }^2\theta +{\sin }^2\theta \right)=|\vec{a}{|}^2|\vec{b}{|}^2$

That is Lagrange's identity, and it is the chapter in one line: two products, each throwing away half the information, and together throwing away none.

1. Vectors, Scalars and the Basic Operations

Scalar. Magnitude only. Vector. Magnitude and direction.

Addition follows the triangle law, placing vectors nose to tail, or equivalently the parallelogram law, placing them tail to tail. The two are the same statement seen from two corners of the same figure.

2. Components, Position Vectors and the Section Formula

$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k},|\vec{r}|=\sqrt{x^2+y^2+z^2}$

The position vector of a point is the vector from the origin to it, and the vector joining $A$ to $B$ is the difference of their position vectors: $\overrightarrow{AB}=\vec{b}-\vec{a}$.

$\text{dividing }A,B\text{ in }m:n\Rightarrow \vec{r}=\frac{m\vec{b}+n\vec{a}}{m+n}$

Trap. The far point's position vector gets the near ratio. Dividing in $2:3$ from $A$ puts $2$ on $\vec{b}$, not on $\vec{a}$. Check with a midpoint, where $m=n$ makes the question disappear.

Illustration 1

Find the position vector of the point dividing $A(2,-1,3)$ and $B(4,3,-1)$ internally in the ratio $3:1$, and show that $A(-2,3,5)$, $B(1,2,3)$, $C(7,0,-1)$ are collinear.

For the division, the far point's position vector carries the near ratio.

$\vec{r}=\frac{3\vec{b}+1\vec{a}}{4}=\frac{3(4,3,-1)+(2,-1,3)}{4}=\frac{(14,8,0)}{4}=\left(\frac{7}{2},2,0\right)$

Check the ratio directly: from $A$ to $P$ is $\left(\frac{3}{2},3,-3\right)$ and from $P$ to $B$ is $\left(\frac{1}{2},1,-1\right)$, which are in the ratio $3:1$ as required.

For collinearity, build two vectors from a common point and test whether one is a multiple of the other.

$\overrightarrow{AB}=(3,-1,-2),\overrightarrow{AC}=(9,-3,-6)=3\overrightarrow{AB}$

One is exactly three times the other, so the three points lie on a line, with $C$ three times as far from $A$ as $B$ is.

Trap. Collinearity needs the vectors built from a shared point. Comparing $\overrightarrow{AB}$ with $\overrightarrow{BC}$ works too, but comparing $\overrightarrow{AB}$ with $\overrightarrow{CA}$ and forgetting the reversed direction produces a sign error and a wrong verdict on the ordering.

Illustration 2

Explain why $(\vec{a}\times \vec{b})\times \vec{c}$ and $\vec{a}\times (\vec{b}\times \vec{c})$ differ, using the expansion rule.

The rule that settles it is the vector triple product expansion, sometimes remembered as BAC minus CAB.

$\vec{a}\times (\vec{b}\times \vec{c})=(\vec{a}\cdot \vec{c})\vec{b}-(\vec{a}\cdot \vec{b})\vec{c}$

$(\vec{a}\times \vec{b})\times \vec{c}=(\vec{a}\cdot \vec{c})\vec{b}-(\vec{b}\cdot \vec{c})\vec{a}$

The two share their first term and differ in the second, so they agree only when $(\vec{a}\cdot \vec{b})\vec{c}=(\vec{b}\cdot \vec{c})\vec{a}$.

Apply both to $\vec{a}=\vec{b}=\vec{i}$ and $\vec{c}=\vec{j}$.

$\vec{i}\times (\vec{i}\times \vec{j})=(\vec{i}\cdot \vec{j})\vec{i}-(\vec{i}\cdot \vec{i})\vec{j}=\vec{0}-\vec{j}=-\vec{j}$

$(\vec{i}\times \vec{i})\times \vec{j}=(\vec{i}\cdot \vec{j})\vec{i}-(\vec{i}\cdot \vec{j})\vec{i}=\vec{0}$

Both match the direct computation from the opening. Notice what the expansion reveals: the result always lies in the plane of the bracketed pair, so which pair is bracketed decides which plane the answer lives in.

3. The Scalar (Dot) Product

$\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos \theta =a_1{b}_1+a_2{b}_2+a_3{b}_3$

The result is a scalar. That single fact explains most of its behaviour.

The last row is not a failure but a type error: $\vec{a}\cdot \vec{b}$ is a number, and a number cannot be dotted with a vector.

Illustration 3

Find the angle between $\vec{a}=2\hat{i}+2\hat{j}-\hat{k}$ and $\vec{b}=6\hat{i}-3\hat{j}+2\hat{k}$, then verify Lagrange's identity for them.

$\vec{a}\cdot \vec{b}=12-6-2=4,|\vec{a}|=3,|\vec{b}|=7$

$\cos \theta =\frac{4}{21}\Rightarrow \theta \approx 79.0^{\circ }$

Now the cross product, to check the identity.

$\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 2& 2& -1\\ 6& -3& 2\end{array}\right|=(4-3)\hat{i}-(4+6)\hat{j}+(-6-12)\hat{k}=(1,-10,-18)$

$(\vec{a}\cdot \vec{b}{)}^2+|\vec{a}\times \vec{b}{|}^2=16+(1+100+324)=16+425=441$

$|\vec{a}{|}^2|\vec{b}{|}^2=9\times 49=441$

They agree exactly. The identity is also a useful shortcut in reverse: given $|\vec{a}|$, $|\vec{b}|$ and one product, the other follows without finding the angle at all.

Illustration 4

A constant force $\vec{F}=3\hat{i}+2\hat{j}-4\hat{k}$ moves a particle from $(1,2,3)$ to $(5,4,1)$. Find the work done.

Work is the dot product of force with displacement, and it is a scalar precisely because only the component of the force along the motion contributes.

$\vec{d}=(5,4,1)-(1,2,3)=(4,2,-2)$

$W=\vec{F}\cdot \vec{d}=12+4+8=24\text{units}$

The perpendicular part of the force does no work at all, and the dot product discards it automatically. That is the physical reading of the chapter's opening claim: the dot product is blind to everything perpendicular, which here is exactly what should be ignored.

Projections

The scalar projection of $\vec{a}$ on $\vec{b}$ is the length of its shadow; the vector projection points along $\vec{b}$.

$\text{scalar}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|},\text{vector}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}{|}^2}\vec{b}$

Trap. The projection of $\vec{a}$ on $\vec{b}$ divides by $|\vec{b}|$, not $|\vec{a}|$. Read which vector is the screen and which is casting the shadow.

Illustration 5

Find the projection of $\vec{a}=\hat{i}+2\hat{j}+3\hat{k}$ on $\vec{b}=2\hat{i}-\hat{j}+\hat{k}$, as a scalar and as a vector.

$\vec{a}\cdot \vec{b}=2-2+3=3,|\vec{b}|=\sqrt{4+1+1}=\sqrt{6}$

$\text{scalar projection}=\frac{3}{\sqrt{6}}=\frac{\sqrt{6}}{2}\approx 1.22$

For the vector version, multiply the unit vector along $\vec{b}$ by that length, which is the same as dividing by $|\vec{b}{|}^2$.

$\text{vector projection}=\frac{3}{6}(2,-1,1)=\left(1,-\frac{1}{2},\frac{1}{2}\right)$

Check its length: $\sqrt{1+\frac{1}{4}+\frac{1}{4}}=\sqrt{3/2}=\sqrt{6}/2$, matching the scalar answer as it must.

Illustration 6

Find $\lambda$ so that $2\hat{i}+\lambda \hat{j}+\hat{k}$ is perpendicular to $\hat{i}-2\hat{j}+3\hat{k}$.

Perpendicular means the dot product vanishes, which is one linear equation.

$2(1)+\lambda (-2)+1(3)=0\Rightarrow 5-2\lambda =0\Rightarrow \lambda =\frac{5}{2}$

No magnitudes were computed, because the condition lives entirely in the numerator. That is the reason to write the dot product first in any question about perpendicularity: if it is zero, the denominators never matter.

4. The Vector (Cross) Product

$\vec{a}\times \vec{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ a_1& a_2& a_3\\ b_1& b_2& b_3\end{array}\right|,|\vec{a}\times \vec{b}|=|\vec{a}||\vec{b}|\sin \theta$

The result is a vector, perpendicular to both, with direction given by the right-hand rule.

Areas

$\text{parallelogram on }\vec{a},\vec{b}:|\vec{a}\times \vec{b}|;\text{triangle}:\frac{1}{2}|\vec{a}\times \vec{b}|$

For a parallelogram given by its diagonals ${\vec{d}}_1$ and ${\vec{d}}_2$, the area is $\frac{1}{2}|{\vec{d}}_1\times {\vec{d}}_2|$, which is the same formula with an extra halving.

Illustration 7

Find the area of the triangle with vertices $A(1,1,2)$, $B(2,3,5)$ and $C(1,5,5)$, and a unit vector perpendicular to its plane.

Build two edge vectors from the same corner, so the cross product spans the triangle.

$\overrightarrow{AB}=(1,2,3),\overrightarrow{AC}=(0,4,3)$

$\overrightarrow{AB}\times \overrightarrow{AC}=(6-12)\hat{i}-(3-0)\hat{j}+(4-0)\hat{k}=(-6,-3,4)$

$|\overrightarrow{AB}\times \overrightarrow{AC}|=\sqrt{36+9+16}=\sqrt{61}$

$\text{area}=\frac{\sqrt{61}}{2}\approx 3.91$

The same cross product answers the second half for free, since it is already perpendicular to both edges and therefore to the plane.

$\hat{n}=\pm \frac{1}{\sqrt{61}}(-6,-3,4)$

Both signs are correct: a plane has two unit normals, and nothing in the question chooses between them.

Illustration 8

A force $\vec{F}=2\hat{i}+\hat{j}-\hat{k}$ acts at the point $(1,-1,2)$. Find its moment about the origin.

Moment is the cross product of position with force, and it is a vector because a turning effect has an axis as well as a size.

$\vec{M}=\vec{r}\times \vec{F}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& -1& 2\\ 2& 1& -1\end{array}\right|$

$=((-1)(-1)-(2)(1))\hat{i}-((1)(-1)-(2)(2))\hat{j}+((1)(1)-(-1)(2))\hat{k}$

$\vec{M}=-\hat{i}+5\hat{j}+3\hat{k}$

A force acting along the line through the origin would produce no moment, and the cross product returns zero for parallel vectors, which is exactly right. Here the two are not parallel, and the answer points along the axis about which the turning happens.

5. The Scalar Triple Product

$[\vec{a}\vec{b}\vec{c}]=\vec{a}\cdot (\vec{b}\times \vec{c})=\left|\begin{array}{ccc}{a}_1& a_2& a_3\\ b_1& b_2& b_3\\ c_1& c_2& c_3\end{array}\right|$

The result is a scalar, and it is the volume of the parallelepiped the three vectors span, signed by orientation.

$\text{volume of parallelepiped}=|[\vec{a}\vec{b}\vec{c}]|,\text{tetrahedron}=\frac{1}{6}|[\vec{a}\vec{b}\vec{c}]|$

$[\vec{a}\vec{b}\vec{c}]=0⟺\text{the three are coplanar}$

Because it is a determinant, cyclic swaps leave it unchanged and any single swap flips the sign. The dot and the cross may also be exchanged: $\vec{a}\cdot (\vec{b}\times \vec{c})=(\vec{a}\times \vec{b})\cdot \vec{c}$.

Illustration 9

Show that $(1,2,3)$, $(2,3,4)$ and $(3,4,5)$ are coplanar, and find the volume of the tetrahedron on $A(1,1,1)$, $B(2,1,3)$, $C(3,2,2)$, $D(3,3,4)$.

For coplanarity, evaluate the determinant.

$\left|\begin{array}{ccc}1& 2& 3\\ 2& 3& 4\\ 3& 4& 5\end{array}\right|=1(15-16)-2(10-12)+3(8-9)=-1+4-3=0$

Coplanar. The reason is visible in the rows: the third is twice the second minus the first, so one vector is a combination of the other two.

For the tetrahedron, build three edge vectors from one vertex.

$\overrightarrow{AB}=(1,0,2),\overrightarrow{AC}=(2,1,1),\overrightarrow{AD}=(2,2,3)$

$[\overrightarrow{AB}\overrightarrow{AC}\overrightarrow{AD}]=1(3-2)-0(6-2)+2(4-2)=1+4=5$

$V=\frac{1}{6}|5|=\frac{5}{6}$

Trap. A tetrahedron is a sixth of the parallelepiped, not a half or a third. Quoting the triple product alone as the volume is the standard slip.

Illustration 10

Show that $\vec{a}=(1,-2,3)$, $\vec{b}=(-2,3,-4)$ and $\vec{c}=(1,-1,1)$ are coplanar, and express $\vec{c}$ in terms of the other two.

Coplanarity is the determinant vanishing.

$\left|\begin{array}{ccc}1& -2& 3\\ -2& 3& -4\\ 1& -1& 1\end{array}\right|=1(3-4)+2(-2+4)+3(2-3)=-1+4-3=0$

Coplanar. But a zero determinant says more than "they lie in a plane": it says one of them is a combination of the other two, and finding that combination is often what a question really wants.

$\vec{c}=x\vec{a}+y\vec{b}\Rightarrow x-2y=1,-2x+3y=-1,3x-4y=1$

Solve the first two, then use the third as the check that the system is genuinely consistent.

$x=1+2y\Rightarrow -2(1+2y)+3y=-1\Rightarrow -y=1\Rightarrow y=-1,x=-1$

$\text{third equation: }3(-1)-4(-1)=1✓$

$\vec{c}=-\vec{a}-\vec{b}$

The third equation being satisfied is the same fact as the determinant vanishing, seen from the other side: three equations in two unknowns are consistent exactly when the coefficient determinant is zero.

6. The Vector Triple Product

$\vec{a}\times (\vec{b}\times \vec{c})=(\vec{a}\cdot \vec{c})\vec{b}-(\vec{a}\cdot \vec{b})\vec{c}$

Every term on the right is a scalar times a vector, so the answer lies in the plane of $\vec{b}$ and $\vec{c}$ — the bracketed pair. That is the structural reason the brackets cannot be moved.

The mnemonic BAC minus CAB records the order: the middle vector of the bracket comes first, multiplied by the dot product of the outer two.

Illustration 11

Simplify $\hat{i}\times (\hat{j}\times \hat{k})+\hat{j}\times (\hat{k}\times \hat{i})+\hat{k}\times (\hat{i}\times \hat{j})$.

Expand each term by the rule rather than evaluating the inner cross products.

$\hat{i}\times (\hat{j}\times \hat{k})=(\hat{i}\cdot \hat{k})\hat{j}-(\hat{i}\cdot \hat{j})\hat{k}=\vec{0}-\vec{0}=\vec{0}$

Every dot product between two different standard unit vectors is zero, so each of the three terms vanishes for the same reason.

$\text{sum}=\vec{0}$

Checking one directly confirms it: $\hat{j}\times \hat{k}=\hat{i}$, and $\hat{i}\times \hat{i}=\vec{0}$. The expansion was faster because it never required knowing which way round the unit vectors cross.

Summary

The dot product measures how much two vectors agree; the cross product measures how much they do not. Lagrange's identity says that together they lose nothing.

The cross product is not associative and the dot product cannot be cancelled, and both failures come from each product being blind to half the information.

Addition is the triangle law and the parallelogram law, which are the same figure seen from two corners.

The section formula puts the near ratio on the far position vector; check it against the midpoint.

The dot product is a scalar, so $(\vec{a}\cdot \vec{b})\cdot \vec{c}$ is a type error rather than a false statement. It vanishes exactly for perpendicular vectors.

Projection of $\vec{a}$ on $\vec{b}$ divides by $|\vec{b}|$: the second vector is the screen.

The cross product is a vector perpendicular to both, anticommutative, and zero exactly for parallel vectors. Its magnitude is the parallelogram's area, so a triangle is half of it.

A cross product doubles as a normal direction, and a plane always has two unit normals.

A zero triple product says more than coplanar: one vector is a combination of the other two, and finding it is usually what the question wants.

The scalar triple product is a determinant and a volume. Zero means coplanar, cyclic swaps preserve it, and a tetrahedron is one sixth of it.

The vector triple product expands as BAC minus CAB, and always lands in the plane of the bracketed pair, which is why the brackets matter.