Vector Algebra

Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XII, Chapter 10.

Why can 3 km plus 4 km make 5 km?

Walk 3 km north, then 4 km east. You have walked 7 km, yet you stand only 5 km from your starting point, and in a particular direction: about 37° north of east. A plain number like 7 cannot carry that information. A quantity with a direction can. That is a vector, and the whole chapter is the arithmetic that makes "3 plus 4 equals 5" legitimate: you add the arrows, not their lengths.

Once you can add arrows, two multiplications appear naturally. One asks "how much of this arrow lies along that one?" (the dot product, which gives a number). The other asks "what area do these two arrows sweep out, and which way does it face?" (the cross product, which gives a vector). Almost every CUET question in this chapter is one of these two, or a section-formula or direction-cosine problem built on them.

Scalars, vectors and position vectors

A scalar has only magnitude (length, mass, time, speed). A vector has magnitude and direction (displacement, velocity, force). A vector from a point $A$ (initial point) to a point $B$ (terminal point) is written $\vec{AB}$, and its magnitude $\lvert\vec{AB}\rvert$ is the distance between the points. A magnitude is never negative.

The position vector of a point $P(x, y, z)$ is the arrow from the origin $O$ to $P$. Its magnitude comes from Pythagoras applied twice, once in the floor plane and once going up:

$$\vec{OP} = x\hat{i} + y\hat{j} + z\hat{k}, \qquad \lvert\vec{OP}\rvert = \sqrt{x^2 + y^2 + z^2}$$

To get the vector from $P_1(x_1, y_1, z_1)$ to $P_2(x_2, y_2, z_2)$, subtract "head minus tail":

$$\vec{P_1P_2} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$$

Vectors come in several named types. They are easy marks if you keep the table straight.

Direction cosines and direction ratios

Take a vector $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$ with magnitude $r$, and let it make angles $\alpha, \beta, \gamma$ with the positive $x$, $y$, $z$ axes. Drop a perpendicular from the tip to the $x$ axis. In the right triangle that forms, the side next to angle $\alpha$ is $x$ and the hypotenuse is $r$, so $\cos\alpha = x/r$. The other two angles work the same way.

$$l = \cos\alpha = \dfrac{x}{r}, \quad m = \cos\beta = \dfrac{y}{r}, \quad n = \cos\gamma = \dfrac{z}{r}$$

Now add the squares: $l^2 + m^2 + n^2 = \dfrac{x^2 + y^2 + z^2}{r^2} = 1$. So the three direction angles are never free; they must satisfy

$$\cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1$$

The numbers $l, m, n$ are the direction cosines (DCs). Any three numbers $a, b, c$ proportional to them are direction ratios (DRs). For a vector, its components $x, y, z$ are themselves direction ratios, and dividing each by $r$ turns them into direction cosines. A neat way to see this: the DCs are exactly the components of the unit vector $\hat{r} = l\hat{i} + m\hat{j} + n\hat{k}$.

Pause and predict: can a vector make angles of $60°$, $60°$ and $60°$ with the three axes?

Check your answer: No. $\cos^2 60° \times 3 = 3 \times \dfrac{1}{4} = \dfrac{3}{4}$, which is not 1. Angles of $60°, 60°, 45°$ would work, since $\dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{2} = 1$.

Beyond the textbook. For the exam: $\hat{i} + \hat{j} + \hat{k}$ has DCs $\dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}, \dfrac{1}{\sqrt{3}}$, so it is equally inclined to all three axes. In reality: that equal angle is $\cos^{-1}\dfrac{1}{\sqrt{3}} \approx 54.7°$, the angle between any edge of a cube and the diagonal leaving the same corner. The same angle separates the 100 and 111 directions in a cubic crystal.

Adding and stretching vectors

Triangle law. If you walk from $A$ to $B$ and then from $B$ to $C$, your net displacement is the arrow from $A$ to $C$:

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

Parallelogram law. Draw two vectors from the same point as adjacent sides of a parallelogram; their sum is the diagonal from that point. It is the triangle law in disguise, because the opposite side of a parallelogram is an equal vector, so you can slide one arrow to the tip of the other. The same idea extends to a chain of any number of vectors (the polygon law): the sum is the arrow from the first tail to the last head. Subtraction is addition of the negative: $\vec{a} - \vec{b} = \vec{a} + (-\vec{b})$.

In components, everything is done coordinate by coordinate, which is why component form is the exam's favourite way to hold a vector:

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

Multiplying by a scalar $\lambda$ stretches the vector by $\lvert\lambda\rvert$ and flips it if $\lambda < 0$. So $\lambda\vec{a}$ is always collinear with $\vec{a}$. Two consequences matter:

Addition is commutative and associative, $\vec{0}$ is the identity, and scalar multiplication spreads over addition: $\lambda(\vec{a} + \vec{b}) = \lambda\vec{a} + \lambda\vec{b}$.

Worked example: unit vector and a scaled copy. Find the unit vector along $\vec{a} = 2\hat{i} + 3\hat{j} - 6\hat{k}$, and a vector of magnitude 14 in the same direction.

Common wrong step. Adding the components instead of squaring them, so $2 + 3 - 6 = -1$ becomes the "magnitude". A magnitude cannot be negative, and the sign of $-6$ disappears when you square it.

The section formula

Let $P$ and $Q$ have position vectors $\vec{a}$ and $\vec{b}$, and let $R$ divide $PQ$ internally in the ratio $m : n$, meaning $PR : RQ = m : n$. Then $n \cdot \vec{PR} = m \cdot \vec{RQ}$, and both arrows point the same way. Writing $\vec{r}$ for the position vector of $R$:

$$n(\vec{r} - \vec{a}) = m(\vec{b} - \vec{r}) \quad\Longrightarrow\quad \vec{r} = \dfrac{m\vec{b} + n\vec{a}}{m + n}$$

For external division, $R$ lies on the line outside the segment, so $\vec{PR}$ and $\vec{RQ}$ point in opposite directions. The same working with a minus sign gives the second formula. Putting $m = n$ in the internal formula gives the midpoint.

$$\text{Internal: } \dfrac{m\vec{b} + n\vec{a}}{m + n}, \qquad \text{External: } \dfrac{m\vec{b} - n\vec{a}}{m - n}, \qquad \text{Midpoint: } \dfrac{\vec{a} + \vec{b}}{2}$$

A useful check for the internal formula: the point is closer to the end that carries the larger weight. In $\dfrac{m\vec{b} + n\vec{a}}{m + n}$ the weight $m$ sits on $\vec{b}$, so a bigger $m$ drags $R$ towards $Q$. That is exactly what $PR : RQ = m : n$ means.

Worked example: dividing a segment. $P$ has position vector $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $Q$ has $\vec{b} = 4\hat{i} + 5\hat{j} + 6\hat{k}$. Find the point dividing $PQ$ internally in the ratio $2 : 1$, and externally in the ratio $2 : 1$.

Common wrong step. Putting the larger weight on $\vec{a}$, which gives $\dfrac{2\vec{a} + \vec{b}}{3} = 2\hat{i} + 3\hat{j} + 4\hat{k}$. That point is nearer $P$, but $PR : RQ = 2 : 1$ says $R$ is nearer $Q$. The answer $3\hat{i} + 4\hat{j} + 5\hat{k}$ is two thirds of the way along.

Pause and predict: $R$ divides $PQ$ internally in the ratio $3 : 1$. Which end is $R$ closer to?

Check your answer: $Q$. $PR$ is three times $RQ$, so $R$ sits near $Q$. The formula agrees: $\vec{r} = \dfrac{3\vec{b} + \vec{a}}{4}$ puts three quarters of the weight on $\vec{b}$.

Beyond the textbook. For the exam: the formula $\dfrac{m\vec{b} + n\vec{a}}{m + n}$ with the ratio $PR : RQ = m : n$. In reality: the same weighted average is the centre of mass of two bodies, with one subtle twist. Masses $m_1$ at $P$ and $m_2$ at $Q$ balance at $\dfrac{m_1\vec{a} + m_2\vec{b}}{m_1 + m_2}$, so the balance point divides $PQ$ in the ratio $m_2 : m_1$, the inverse of the masses. The heavier body sits closer to the balance point. Do not mix up "weights on the position vectors" with "the ratio of the parts".

The dot product: how much of one arrow lies along another

Let $\theta$ be the angle between $\vec{a}$ and $\vec{b}$, with $0 \le \theta \le \pi$ when both start at the same point. The scalar (dot) product is

$$\vec{a} \cdot \vec{b} = \lvert\vec{a}\rvert\,\lvert\vec{b}\rvert\cos\theta$$

The result is a number. Read it as "the length of $\vec{a}$ times the length of the shadow of $\vec{b}$ on $\vec{a}$". Three cases tell you everything: positive when $\theta$ is acute, zero when $\theta = 90°$, negative when $\theta$ is obtuse. Hence $\vec{a} \cdot \vec{b} = 0$ for non-zero vectors means they are perpendicular. Also $\vec{a} \cdot \vec{a} = \lvert\vec{a}\rvert^2$ and the product is commutative and distributes over addition.

Why the component form works. Expand $(a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \cdot (b_1\hat{i} + b_2\hat{j} + b_3\hat{k})$ using the distributive law. The basis vectors are unit and mutually perpendicular, so $\hat{i}\cdot\hat{i} = \hat{j}\cdot\hat{j} = \hat{k}\cdot\hat{k} = 1$ while every mixed product is 0. Nine terms collapse to three:

$$\vec{a} \cdot \vec{b} = a_1 b_1 + a_2 b_2 + a_3 b_3, \qquad \cos\theta = \dfrac{a_1 b_1 + a_2 b_2 + a_3 b_3}{\lvert\vec{a}\rvert\,\lvert\vec{b}\rvert}$$

The projection of $\vec{a}$ on $\vec{b}$ is the signed length of the shadow of $\vec{a}$ along $\vec{b}$. It is $\lvert\vec{a}\rvert\cos\theta$, and dividing the dot product by $\lvert\vec{b}\rvert$ gives it:

$$\text{Projection of } \vec{a} \text{ on } \vec{b} = \dfrac{\vec{a} \cdot \vec{b}}{\lvert\vec{b}\rvert}$$

Worked example: angle between two vectors. Find the angle between $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.

Common wrong step. Throwing away the minus sign because "an angle between lines is acute". For two vectors the angle runs from $0$ to $\pi$, so a negative cosine is a perfectly good obtuse angle. (Angles between lines are a different convention, taken up in the next chapter.)

Worked example: projection. Find the projection of $\vec{a} = \hat{i} + 3\hat{j} + 7\hat{k}$ on $\vec{b} = 7\hat{i} - \hat{j} + 8\hat{k}$.

Common wrong step. Dividing by $\lvert\vec{a}\rvert = \sqrt{59}$. The projection on $\vec{b}$ always divides by the length of $\vec{b}$. Ask "on what?" and divide by that vector's length.

Pause and predict: if $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c}$ with $\vec{a} \neq \vec{0}$, can you cancel $\vec{a}$ and conclude $\vec{b} = \vec{c}$?

Check your answer: No. The equation says $\vec{a} \cdot (\vec{b} - \vec{c}) = 0$, which only means $\vec{b} - \vec{c}$ is perpendicular to $\vec{a}$. Example: $\vec{a} = \hat{i}$, $\vec{b} = \hat{i} + \hat{j}$, $\vec{c} = \hat{i} + \hat{k}$ give $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 1$ but $\vec{b} \neq \vec{c}$.

Beyond the textbook. For the exam: the dot product is a formula that returns a number and tests for perpendicularity. In reality: it is the physical definition of work. A force $\vec{F}$ moving a body through a displacement $\vec{d}$ does work $W = \vec{F} \cdot \vec{d}$, so only the part of the force along the motion counts. Carrying a bag horizontally with a vertical grip does no work on the bag, because $\theta = 90°$ and $\cos\theta = 0$.

The cross product: area with a direction

The vector (cross) product of $\vec{a}$ and $\vec{b}$ is

$$\vec{a} \times \vec{b} = \lvert\vec{a}\rvert\,\lvert\vec{b}\rvert\sin\theta\ \hat{n}$$

where $\hat{n}$ is the unit vector perpendicular to both, chosen by the right hand rule: curl the fingers of your right hand from $\vec{a}$ towards $\vec{b}$ and your thumb points along $\hat{n}$. The result is a vector.

Why the magnitude is an area. Take $\vec{a}$ as the base of a parallelogram. The height is the part of $\vec{b}$ perpendicular to $\vec{a}$, which is $\lvert\vec{b}\rvert\sin\theta$. So base times height is $\lvert\vec{a}\rvert\lvert\vec{b}\rvert\sin\theta = \lvert\vec{a} \times \vec{b}\rvert$. A triangle with the same two sides has half that area.

$$\text{Parallelogram} = \lvert\vec{a} \times \vec{b}\rvert, \qquad \text{Triangle} = \dfrac{1}{2}\lvert\vec{a} \times \vec{b}\rvert$$

If the vectors are parallel, $\sin\theta = 0$ and the product is $\vec{0}$: the parallelogram has collapsed. Reversing the order reverses the direction, so $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$. Among the basis vectors the pattern is cyclic: $\hat{i}\times\hat{j} = \hat{k}$, $\hat{j}\times\hat{k} = \hat{i}$, $\hat{k}\times\hat{i} = \hat{j}$, and going against the cycle gives a minus sign. Any vector crossed with itself gives zero.

In components, the cross product is a determinant (expand along the first row; the middle term carries a minus sign):

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \cr a_1 & a_2 & a_3 \cr b_1 & b_2 & b_3 \end{vmatrix} = (a_2 b_3 - a_3 b_2)\hat{i} - (a_1 b_3 - a_3 b_1)\hat{j} + (a_1 b_2 - a_2 b_1)\hat{k}$$

The two products are easy to mix up. Keep this table in your head:

Worked example: area of a triangle. Find the area of the triangle with vertices $A(1, 1, 1)$, $B(1, 2, 3)$ and $C(2, 3, 1)$.

Common wrong step. Stopping at $\sqrt{21}$. That is the area of the parallelogram on $\vec{AB}$ and $\vec{AC}$. The triangle is half of it.

Worked example: a unit vector perpendicular to two vectors. Find a unit vector perpendicular to both $\hat{i} + \hat{j}$ and $\hat{j} + \hat{k}$.

Common wrong step. Taking the dot product instead. $\vec{a} \cdot \vec{b}$ is a number and cannot be "a vector perpendicular to both". Also remember that $-\dfrac{1}{\sqrt{3}}(\hat{i} - \hat{j} + \hat{k})$ is equally valid, since a perpendicular line has two directions.

Pause and predict: if $\vec{a} \times \vec{b} = \vec{a} \times \vec{c}$ and $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c}$ with $\vec{a} \neq \vec{0}$, what can you say about $\vec{b}$ and $\vec{c}$?

Check your answer: They are equal. The cross equation makes $\vec{b} - \vec{c}$ parallel to $\vec{a}$, and the dot equation makes it perpendicular to $\vec{a}$. The only vector that is both is $\vec{0}$. Either equation alone is not enough.

Beyond the textbook. For the exam: the cross product uses the right handed system, so $\hat{i}\times\hat{j} = \hat{k}$. In reality: the right hand rule is a convention. If every physicist used a left hand, the direction of a cross product would flip, but its magnitude (an area) would not. Torque $\vec{\tau} = \vec{r} \times \vec{F}$ is the standard example. Its magnitude $rF\sin\theta$ is the same in every convention, and it is largest when the force acts at right angles to the spanner's handle.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Before: Introduction to Three Dimensional Geometry gives the coordinates, distance formula and section formula that this chapter turns into vector form.
• Next: Three Dimensional Geometry uses the dot product for angles between lines and the cross product for shortest distance.
• Sideways: Determinants is the machinery behind the determinant form of the cross product.