Vector Algebra

1. Check this before you revise anything

The Scalar Triple Product is gone. Older editions carried a full subsection on $\vec{a}\cdot (\vec{b}\times \vec{c})$ — the determinant form, its interpretation as the volume of a parallelepiped, and the coplanarity test $[\vec{a}\vec{b}\vec{c}]=0$ used to prove three vectors lie in the same plane. A full-text search of the current 39-page chapter for "triple product," "parallelepiped," or "coplanar" returns zero hits. The section list is:

Nothing beyond 10.6. This matches the syllabus line exactly: "...properties and application of scalar (dot) product of vectors, vector (cross) product of vectors" — full stop, no third product.

The old stub taught the scalar triple product as its own section, with the volume-of-a-parallelepiped interpretation and the coplanarity test as a named technique. Removed entirely from this rebuild.

A genuine error survives in the book's own Miscellaneous Exercise. Question 14 reads: "If $\vec{a},\vec{b},\vec{c}$ are mutually perpendicular vectors of equal magnitude, show that the vector $\vec{c}\cdot \vec{d}=15$ is equally inclined to $\vec{a},\vec{b}$ and $\vec{c}$" — a clause visibly left over from copy-pasting Question 12 (which genuinely does define a vector $\vec{d}$ with $\vec{c}\cdot \vec{d}=15$).

Question 14 has no $\vec{d}$ of its own. The intended, standard question — solved here — is to show that $\vec{a}+\vec{b}+\vec{c}$ is equally inclined to $\vec{a}$, $\vec{b}$, and $\vec{c}$.

Exercise 10.2 also hides two MCQs (Q18–19) past its own visible question list, and Exercise 10.4 hides two more (Q11–12) — both with no "Choose the correct answer" lead-in line, the same pattern found in Chapter 9. All of Exercise 10.1 through the Miscellaneous Exercise were re-verified from 300dpi page renders rather than raw text extraction, since this chapter's heavy use of $\hat{i},\hat{j},\hat{k}$ and vector arrows garbles almost completely under plain PDF text extraction.

2. What a Vector Is, and How It Is Measured (Textbook 10.2)

Some quantities are settled completely by a single number and a unit: mass, length, time, temperature. These are scalars. Others are not. If you are told a body was displaced 5 km, you still do not know where it ended up — you need the direction as well. Quantities that need both a magnitude and a direction are vectors.

The textbook builds every vector from a directed line segment. Given points $A$ and $B$, the segment from $A$ to $B$ with the arrow pointing at $B$ is written $\overrightarrow{AB}$. Here $A$ is the initial point and $B$ the terminal point.

Two pieces of vocabulary follow immediately. The magnitude $|\overrightarrow{AB}|$ is the length of the segment, always a non-negative real number. The direction is the way the arrow points.

Position vectors. Fix the origin $O$. Any point $P(x,y,z)$ in space then has exactly one vector attached to it, $\overrightarrow{OP}$, called its position vector. Writing $\hat{i},\hat{j},\hat{k}$ for unit vectors along the axes:

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

The magnitude formula is not a new rule. It is Pythagoras applied twice — once in the $xy$-plane to reach the foot of the perpendicular, then once vertically to climb to $P$.

Direction cosines. Let $\overrightarrow{OP}$ make angles $\alpha ,\beta ,\gamma$ with the positive $x$, $y$ and $z$ axes. The numbers $l=\cos \alpha$, $m=\cos \beta$, $n=\cos \gamma$ are the direction cosines of the vector.

From the right triangle $OAP$ in the textbook's Fig 10.3, $\cos \alpha =x/r$ where $r=|\overrightarrow{OP}|$, and similarly for the other two. So $l=x/r$, $m=y/r$, $n=z/r$. Squaring and adding gives the relation you will use constantly:

$l^2+m^2+n^2=\frac{x^2+y^2+z^2}{r^2}=\frac{r^2}{r^2}=1$

Any triple $(a,b,c)$ proportional to $(l,m,n)$ is a set of direction ratios. Direction ratios are not unique — every scalar multiple works — which is exactly why they are convenient. Dividing by $\sqrt{a^2+b^2+c^2}$ recovers the direction cosines.

3. Types of Vectors (Textbook 10.3)

This section is pure classification, and Exercise 10.1 tests nothing else. The distinctions matter because later proofs quietly depend on them.

The single most useful idea here is that a vector is free. Because equality ignores the initial point, you may slide any vector anywhere in space without changing it. Every triangle-law diagram in the next section relies on doing exactly that.

A common exam trap lives in this table. Two vectors of equal magnitude are not equal unless their directions agree too, and two collinear vectors need not be equal or even parallel in the same sense — they may point in opposite directions along the line.

4. Addition of Vectors (Textbook 10.4)

Triangle law. Place the initial point of $\vec{b}$ at the terminal point of $\vec{a}$. The vector from the start of $\vec{a}$ to the end of $\vec{b}$ is their sum:

$\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$

An immediate consequence is worth memorising, because it collapses a whole family of exam questions to one line. Since $\overrightarrow{AC}=-\overrightarrow{CA}$, moving everything to one side gives:

$\overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CA}=\vec{0}$

The sides of a triangle taken in order always sum to the zero vector. The same argument extends to any closed polygon.

Parallelogram law. If $\vec{a}$ and $\vec{b}$ are the two adjacent sides of a parallelogram drawn from a common point, their sum is the diagonal from that same point. The triangle and parallelogram laws are the same statement drawn two ways.

Addition obeys two properties the textbook states explicitly. It is commutative, $\vec{a}+\vec{b}=\vec{b}+\vec{a}$, which the parallelogram picture shows at a glance — both routes trace out the same diagonal. It is associative, $(\vec{a}+\vec{b})+\vec{c}=\vec{a}+(\vec{b}+\vec{c})$, so a sum of several vectors needs no brackets.

Finally, $\vec{0}$ is the additive identity, $\vec{a}+\vec{0}=\vec{a}$, and $-\vec{a}$ is the additive inverse, $\vec{a}+(-\vec{a})=\vec{0}$.

5. Scalar Multiplication, Components and the Section Formula (Textbook 10.5)

Multiplying by a scalar. For a real number $\lambda$, the vector $\lambda \vec{a}$ has magnitude $|\lambda ||\vec{a}|$. Its direction is the same as $\vec{a}$ when $\lambda >0$ and opposite when $\lambda <0$.

Two special cases carry the weight. Taking $\lambda =-1$ gives $-\vec{a}$, the negative. Taking $\lambda =1/|\vec{a}|$ gives the unit vector $\hat{a}=\vec{a}/|\vec{a}|$, which is how essentially every "find the unit vector" question is answered.

This also gives the cleanest test for parallelism: $\vec{a}$ and $\vec{b}$ are collinear exactly when $\vec{b}=\lambda \vec{a}$ for some scalar $\lambda$.

Components (10.5.1). Writing $\vec{a}=a_1\hat{i}+a_2\hat{j}+a_3\hat{k}$, the numbers $a_1,a_2,a_3$ are the scalar components and the vectors $a_1\hat{i},a_2\hat{j},a_3\hat{k}$ are the vector components. In this form the algebra becomes arithmetic:

$\vec{a}+\vec{b}=(a_1+b_1)\hat{i}+(a_2+b_2)\hat{j}+(a_3+b_3)\hat{k},\lambda \vec{a}=\lambda a_1\hat{i}+\lambda a_2\hat{j}+\lambda a_3\hat{k}$

Two vectors are equal precisely when all three pairs of components match.

Vector joining two points (10.5.2). For $P_1(x_1,y_1,z_1)$ and $P_2(x_2,y_2,z_2)$, apply the triangle law to $O$, $P_1$, $P_2$: since $\overrightarrow{O{P}_1}+\overrightarrow{P_1{P}_2}=\overrightarrow{O{P}_2}$,

$\overrightarrow{P_1{P}_2}=(x_2-x_1)\hat{i}+(y_2-y_1)\hat{j}+(z_2-z_1)\hat{k}$

Terminal minus initial — in that order. Reversing it is the most frequent slip in this chapter.

Section formula (10.5.3). Let $R$ divide the segment $PQ$ in the ratio $m:n$, with $\vec{p}$ and $\vec{q}$ the position vectors of $P$ and $Q$. For internal division:

$\vec{r}=\frac{m\vec{q}+n\vec{p}}{m+n}$

For external division, the same derivation with the ratio taken as $m:(-n)$ gives:

$\vec{r}=\frac{m\vec{q}-n\vec{p}}{m-n}$

Setting $m=n=1$ in the internal formula recovers the midpoint, $\vec{r}=(\vec{p}+\vec{q})/2$. Note which position vector each coefficient lands on — $m$ multiplies $\vec{q}$, not $\vec{p}$.

6. Scalar (Dot) Product and Projection (Textbook 10.6.1 to 10.6.2)

Definition. For non-zero $\vec{a},\vec{b}$ with angle $\theta$ between them, $0\le \theta \le \pi$:

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

The result is a scalar, not a vector. If either vector is $\vec{0}$, the product is defined to be $0$.

Rearranging gives the formula that answers every "find the angle" question:

$\cos \theta =\frac{\vec{a}\cdot \vec{b}}{|\vec{a}||\vec{b}|}$

Perpendicularity. Since $|\vec{a}|,|\vec{b}|\ne 0$, the product vanishes exactly when $\cos \theta =0$, that is $\theta =\pi /2$. So for non-zero vectors, $\vec{a}\cdot \vec{b}=0$ if and only if they are perpendicular. This is the workhorse of the exercise.

Component form. Applying the definition to the base vectors gives $\hat{i}\cdot \hat{i}=\hat{j}\cdot \hat{j}=\hat{k}\cdot \hat{k}=1$ (angle $0$) and $\hat{i}\cdot \hat{j}=\hat{j}\cdot \hat{k}=\hat{k}\cdot \hat{i}=0$ (angle $\pi /2$). Expanding the product and discarding every cross term:

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

The dot product is commutative, $\vec{a}\cdot \vec{b}=\vec{b}\cdot \vec{a}$, and distributive over addition, $\vec{a}\cdot (\vec{b}+\vec{c})=\vec{a}\cdot \vec{b}+\vec{a}\cdot \vec{c}$. Taking $\vec{b}=\vec{a}$ gives $\vec{a}\cdot \vec{a}=|\vec{a}{|}^2$, which is how identities such as $|\vec{a}+\vec{b}{|}^2=|\vec{a}{|}^2+2\vec{a}\cdot \vec{b}+|\vec{b}{|}^2$ are produced.

Projection (10.6.2). The projection of $\vec{a}$ on a line along $\vec{b}$ is the signed length of the shadow $\vec{a}$ casts on that line:

$\text{projection of }\vec{a}\text{ on }\vec{b}=\frac{\vec{a}\cdot \vec{b}}{|\vec{b}|}=\vec{a}\cdot \hat{b}$

It is a scalar and it can be negative — a negative value means the shadow points against $\vec{b}$. Note the asymmetry: you divide by the magnitude of the vector you are projecting onto. Swapping $\vec{a}$ and $\vec{b}$ generally changes the answer.

7. Vector (Cross) Product (Textbook 10.6.3)

Definition. For non-zero, non-parallel $\vec{a},\vec{b}$ with angle $\theta$ between them:

$\vec{a}\times \vec{b}=|\vec{a}||\vec{b}|\sin \theta \hat{n}$

where $\hat{n}$ is the unit vector perpendicular to both $\vec{a}$ and $\vec{b}$, oriented by the right-hand rule. Unlike the dot product, the result is a vector.

Parallelism. The product vanishes exactly when $\sin \theta =0$, that is $\theta =0$ or $\theta =\pi$. So for non-zero vectors, $\vec{a}\times \vec{b}=\vec{0}$ if and only if they are parallel. Note the clean split from the previous section: the dot product detects perpendicularity, the cross product detects parallelism.

Not commutative. Reversing the order flips the perpendicular direction, so:

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

This is the difference that catches students out most often. Order matters here and did not before.

Determinant form. From $\hat{i}\times \hat{i}=\hat{j}\times \hat{j}=\hat{k}\times \hat{k}=\vec{0}$ and the cyclic rules $\hat{i}\times \hat{j}=\hat{k}$, $\hat{j}\times \hat{k}=\hat{i}$, $\hat{k}\times \hat{i}=\hat{j}$, expanding the product gives:

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

Remember the middle term carries a minus sign when the determinant is expanded along the top row. Dropping it is the single most common computational error in Exercise 10.4.

Areas. The magnitude $|\vec{a}\times \vec{b}|=|\vec{a}||\vec{b}|\sin \theta$ is exactly the area of the parallelogram having $\vec{a}$ and $\vec{b}$ as adjacent sides, since $|\vec{b}|\sin \theta$ is the perpendicular height on base $|\vec{a}|$. The triangle on the same two sides is half of it:

$\text{area of triangle}=\frac{1}{2}|\vec{a}\times \vec{b}|$

For a triangle given by three vertices $A,B,C$, form two edge vectors from a common vertex — say $\overrightarrow{AB}$ and $\overrightarrow{AC}$ — and apply the formula.

Worked, mirroring the textbook's own technique. Show that $A(1,2,7)$, $B(2,6,3)$, $C(3,10,-1)$ are collinear. Compute $\overrightarrow{AB}=(1,4,-4)$ and $\overrightarrow{AC}=(2,8,-8)$. Since $\overrightarrow{AC}=2\overrightarrow{AB}$ exactly, the two vectors are parallel and share the point $A$, so all three points lie on one line.

Equivalently $\overrightarrow{AB}\times \overrightarrow{AC}=\vec{0}$, which is the cross-product test for collinearity — and note it needs no triple product, the technique the current edition no longer carries.

Summary

• A vector has magnitude and direction; a scalar has magnitude alone. Position vector of $P(x,y,z)$ is $x\hat{i}+y\hat{j}+z\hat{k}$ with magnitude $\sqrt{x^2+y^2+z^2}$.
• Direction cosines satisfy $l^2+m^2+n^2=1$; direction ratios are any scalar multiple and must be normalised.
• A vector is free — equality ignores the initial point. The zero vector has undefined direction, which is why the perpendicular and parallel tests carry a "non-zero" caveat.
• Triangle law: $\overrightarrow{AB}+\overrightarrow{BC}=\overrightarrow{AC}$; the sides of any closed polygon taken in order sum to $\vec{0}$.
• $\overrightarrow{P_1{P}_2}$ is terminal minus initial. Section formula: $\frac{m\vec{q}+n\vec{p}}{m+n}$ internally, $\frac{m\vec{q}-n\vec{p}}{m-n}$ externally, midpoint at $m=n=1$.
• $\vec{a}\cdot \vec{b}=|\vec{a}||\vec{b}|\cos \theta =a_1{b}_1+a_2{b}_2+a_3{b}_3$; zero exactly when perpendicular. Projection of $\vec{a}$ on $\vec{b}$ is $(\vec{a}\cdot \vec{b})/|\vec{b}|$, a signed scalar.
• $\vec{a}\times \vec{b}=|\vec{a}||\vec{b}|\sin \theta \hat{n}$; zero exactly when parallel; anti-commutative, $\vec{b}\times \vec{a}=-(\vec{a}\times \vec{b})$.
• $|\vec{a}\times \vec{b}|$ is the parallelogram area; half that is the triangle area.
• The Scalar Triple Product — volume of a parallelepiped, the $[\vec{a}\vec{b}\vec{c}]=0$ coplanarity test — is not part of the current edition.
• The book's own Miscellaneous Exercise Q14 contains a leftover, nonsensical clause from Q12; the standard intended question (solved here) is the $\vec{a}+\vec{b}+\vec{c}$ equal-inclination proof.