Vectors

You push a box with 10 N of force. Your friend pushes the same box with 10 N of force. How much total force acts on the box?

Not 20 N — not necessarily. If you push from opposite sides, the answer is 0. If you push at right angles, it is about 14.1 N. If you both push the same way, then yes — 20 N.

Same magnitudes, three completely different physical outcomes. The difference is direction. And the moment direction matters, ordinary addition breaks. You need a new language.

That language is vectors.

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Scalars and vectors

A scalar is a quantity fully described by a number and a unit. Mass: 5 kg. Temperature: 37°C. Time: 10 s. Speed: 60 km/h. Direction adds nothing to any of these — it makes no physical sense to ask "in which direction is the temperature?"

A vector is a quantity where direction is part of the physical information. Remove direction from force, velocity, acceleration, or displacement, and the physics becomes wrong or meaningless.

A car moving at 60 km/h north is in a fundamentally different physical state from a car moving at 60 km/h south. Same speed. Opposite velocities. They end up in different cities. Everything downstream — position, acceleration needed to stop, collision physics — depends on that direction.

Common scalars: mass, speed, distance, time, temperature, energy, power, pressure, electric charge.

Common vectors: displacement, velocity, acceleration, force, momentum, torque, electric field, magnetic field.

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Representing a vector

A vector is drawn as an arrow. The length represents the magnitude. The direction of the arrow represents the direction of the quantity.

In print: F, v, a (bold). In handwriting: $\vec{F}$, $\vec{v}$, $\vec{a}$ (arrow above). The magnitude of $\vec{F}$ is written $|\vec{F}|$ or simply $F$ (not bold, no arrow).

This distinction matters constantly. $F = 10$ N means the magnitude is 10 N. $\vec{F} = 10 \text{ N north}$ means the full vector.

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Types of vectors

Zero vector (null vector) — magnitude zero, direction undefined. Written $\vec{0}$. It arises when two equal and opposite vectors add. It is not "nothing" — it is a vector quantity with the value zero. A particle at rest has zero velocity, not no velocity.

Unit vector — magnitude exactly 1. Its only job is to carry direction, no magnitude information. Any vector $\vec{A}$ has a unit vector in its direction:

$\hat{A} = \frac{\vec{A}}{|\vec{A}|}$

Divide by the magnitude. Direction survives, magnitude becomes 1.

Equal vectors — same magnitude, same direction. Position in space is irrelevant. Two arrows of the same length pointing the same way are equal vectors even if drawn in different places.

Negative vector — same magnitude, opposite direction. $-\vec{A}$ is the vector that when added to $\vec{A}$ gives $\vec{0}$.

Position vector — the displacement of a point from a fixed origin. The position vector of a point $P$ is written $\vec{r}$ or $\vec{OP}$. It specifies where $P$ is, not just how far.

Collinear vectors — vectors along the same line (or parallel lines), regardless of magnitude or sense. They can be parallel (same direction) or antiparallel (opposite direction).

Coplanar vectors — vectors lying in the same plane.

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Vector addition

The triangle law

Place the second vector tail-to-tip of the first. The resultant $\vec{R}$ is the vector from the tail of the first to the tip of the second — it closes the triangle.

For three or more vectors: chain them tip-to-tail in any order. The resultant closes the chain from the very first tail to the very last tip. This is the polygon law. The order of chaining does not affect the result — vector addition is commutative and associative:

$\vec{A} + \vec{B} = \vec{B} + \vec{A}$ $(\vec{A} + \vec{B}) + \vec{C} = \vec{A} + (\vec{B} + \vec{C})$

{/* DIAGRAM: Triangle law — A tip-to-tail with B, R closes the triangle */}

The parallelogram law

Place both vectors tail-to-tail from the same point. Complete the parallelogram. The diagonal from the common tail is the resultant.

This is the same operation as triangle law — just a different way of drawing it. The parallelogram law is the standard exam statement: "If two vectors acting simultaneously on a particle are represented in magnitude and direction by two adjacent sides of a parallelogram, their resultant is represented by the diagonal drawn from the same point."

{/* DIAGRAM: Parallelogram law — A and B tail-to-tail, diagonal R */}

Magnitude and direction of the resultant

For two vectors $\vec{A}$ and $\vec{B}$ at angle $\theta$ to each other:

$R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$

The resultant makes angle $\alpha$ with $\vec{A}$:

$\tan\alpha = \frac{B\sin\theta}{A + B\cos\theta}$

Where this comes from: the component of $\vec{B}$ along $\vec{A}$ is $B\cos\theta$, and perpendicular to $\vec{A}$ is $B\sin\theta$. The total component along $\vec{A}$ is $A + B\cos\theta$. The angle $\alpha$ is simply the arctangent of perpendicular over parallel. You can re-derive this in thirty seconds if you forget it — which is better than memorising it.

Check the special cases:

• $\theta = 0°$: $R = A + B$. Vectors aligned — maximum resultant.
• $\theta = 180°$: $R = |A - B|$. Vectors opposed — minimum resultant.
• $\theta = 90°$: $R = \sqrt{A^2 + B^2}$. Pythagorean theorem — they were perpendicular.

These are not three formulas. They are one formula at three angles.

Illustration. A boat can travel at 4 m/s in still water. A river flows at 3 m/s perpendicular to the boat's intended direction. What is the actual velocity of the boat?

The two velocities are perpendicular, so $\theta = 90°$: $v = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = 5 \text{ m/s}$

Direction: $\tan\alpha = 3/4$, so $\alpha \approx 37°$ downstream from the intended direction.

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Vector subtraction

$\vec{A} - \vec{B}$ is defined as $\vec{A} + (-\vec{B})$. Reverse $\vec{B}$ to get $-\vec{B}$, then add it to $\vec{A}$ using the triangle law.

The magnitude of $\vec{A} - \vec{B}$:

$|\vec{A} - \vec{B}| = \sqrt{A^2 + B^2 - 2AB\cos\theta}$

Note the minus sign — it is the resultant formula with $\theta$ replaced by $180° - \theta$, since reversing $\vec{B}$ changes the angle between them.

Relative velocity is the canonical application. If $\vec{v}_A$ is the velocity of object A and $\vec{v}_B$ is the velocity of object B, then the velocity of A as seen from B is $\vec{v}_{AB} = \vec{v}_A - \vec{v}_B$.

Illustration. Two cars travel at 60 km/h, one north and one east. What is the velocity of the first car relative to the second?

$\vec{v}_1 = 60\hat{j}$ km/h (north), $\vec{v}_2 = 60\hat{i}$ km/h (east).

$\vec{v}_{12} = \vec{v}_1 - \vec{v}_2 = -60\hat{i} + 60\hat{j}$ km/h

Magnitude: $\sqrt{60^2 + 60^2} = 60\sqrt{2} \approx 84.9$ km/h, directed northwest.

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Scalar multiplication

Multiplying vector $\vec{A}$ by a scalar $k$:

• If $k > 0$: same direction as $\vec{A}$, magnitude $kA$.
• If $k < 0$: opposite direction, magnitude $|k|A$.
• If $k = 0$: the zero vector.

Scalar multiplication is distributive over both vector addition and scalar addition:

$k(\vec{A} + \vec{B}) = k\vec{A} + k\vec{B}$ $(k_1 + k_2)\vec{A} = k_1\vec{A} + k_2\vec{A}$

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Components and resolution

The triangle law is clean geometrically, but for more than two vectors — or for 3D — it becomes unwieldy. Components solve this.

Every vector in a plane can be split into two perpendicular pieces. Choose x and y axes. A vector $\vec{A}$ of magnitude $A$ making angle $\theta$ with the positive x-axis has:

$A_x = A\cos\theta \qquad A_y = A\sin\theta$

These are scalars — ordinary numbers with sign. Positive means along the positive axis. Negative means along the negative axis.

{/* DIAGRAM: Vector A at angle θ with x-axis, showing Ax and Ay */}

To add vectors using components:

$R_x = \sum A_x \qquad R_y = \sum A_y$

$R = \sqrt{R_x^2 + R_y^2} \qquad \theta_R = \tan^{-1}\left(\frac{R_y}{R_x}\right)$

This converts vector addition into scalar addition — done separately for each axis. The geometry is handled twice: once when resolving, once when reconstructing. Everything between is arithmetic.

The quadrant trap. $\tan^{-1}$ always returns an angle between $-90°$ and $+90°$. Check the signs of $R_x$ and $R_y$ to identify the correct quadrant. If $R_x < 0$ and $R_y > 0$, you are in the second quadrant — add $180°$ to what $\tan^{-1}$ gives.

In 3D, the same idea extends with a third axis:

$A_x = A\cos\alpha, \quad A_y = A\cos\beta, \quad A_z = A\cos\gamma$

where $\alpha$, $\beta$, $\gamma$ are the angles $\vec{A}$ makes with the x, y, z axes respectively. These are called direction cosines. They satisfy:

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

Along non-orthogonal axes. Sometimes a problem is easier in axes that are not perpendicular — for instance, along and perpendicular to an inclined plane. Resolution still works: draw the component parallelogram along the chosen axes using the sine rule. The components are no longer $A\cos\theta$ — they depend on the geometry of the particular axis system. This is rare in JEE but worth knowing exists.

Illustration. Three forces act on a point: $\vec{F_1} = 10$ N along positive x, $\vec{F_2} = 6$ N at $60°$ above x, $\vec{F_3} = 8$ N along negative y. Find the resultant.

Resolve: $F_{1x} = 10, \quad F_{1y} = 0$ $F_{2x} = 6\cos60° = 3, \quad F_{2y} = 6\sin60° = 5.196$ $F_{3x} = 0, \quad F_{3y} = -8$

Add: $R_x = 10 + 3 + 0 = 13 \text{ N}, \quad R_y = 0 + 5.196 - 8 = -2.804 \text{ N}$

Reconstruct: $R = \sqrt{13^2 + 2.804^2} \approx 13.3 \text{ N}$ $\theta = \tan^{-1}\left(\frac{-2.804}{13}\right) \approx -12.2°$

$R_x > 0$, $R_y < 0$ — fourth quadrant. Resultant is $13.3$ N at $12.2°$ below the positive x-axis.

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Standard unit vectors

The three standard unit vectors along the coordinate axes are:

$\hat{i} \text{ — along x} \qquad \hat{j} \text{ — along y} \qquad \hat{k} \text{ — along z}$

Any vector can be written:

$\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$

This notation makes all operations clean. Addition:

$\vec{A} + \vec{B} = (A_x + B_x)\hat{i} + (A_y + B_y)\hat{j} + (A_z + B_z)\hat{k}$

Magnitude:

$|\vec{A}| = \sqrt{A_x^2 + A_y^2 + A_z^2}$

Unit vector in the direction of $\vec{A}$:

$\hat{A} = \frac{A_x\hat{i} + A_y\hat{j} + A_z\hat{k}}{\sqrt{A_x^2 + A_y^2 + A_z^2}}$

Illustration. Find the unit vector in the direction of $\vec{A} = 3\hat{i} - 4\hat{j}$.

$|\vec{A}| = \sqrt{9 + 16} = 5$ $\hat{A} = \frac{3\hat{i} - 4\hat{j}}{5} = 0.6\hat{i} - 0.8\hat{j}$

Check: magnitude of $\hat{A}$ is $\sqrt{0.36 + 0.64} = 1$. ✓

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Lami's theorem

If three concurrent forces keep a particle in equilibrium, and the angles between them (measured going around) are $\alpha$, $\beta$, $\gamma$ (where $\alpha + \beta + \gamma = 360°$), then:

$\frac{F_1}{\sin\alpha} = \frac{F_2}{\sin\beta} = \frac{F_3}{\sin\gamma}$

Where this comes from: equilibrium means the three vectors form a closed triangle (they add to zero). Applying the sine rule to that triangle gives Lami's theorem. It is not a separate law — it is the sine rule in disguise.

When to use it. Three forces, equilibrium, angles given between forces. Lami's theorem gives the magnitudes directly without resolving components. Faster than the component method for three-force problems.

Illustration. A traffic signal of weight 200 N is hung from two cables making $30°$ and $45°$ with the vertical. Find the tension in each cable.

The three forces (two tensions + weight) are in equilibrium. The angle between $T_1$ (at $30°$ from vertical) and the weight (downward) is $180° - 30° = 150°$. The angle between $T_2$ (at $45°$ from vertical) and the weight is $180° - 45° = 135°$. The angle between $T_1$ and $T_2$ is $360° - 150° - 135° = 75°$.

By Lami's theorem: $\frac{200}{\sin75°} = \frac{T_1}{\sin135°} = \frac{T_2}{\sin150°}$

$T_1 = \frac{200\sin135°}{\sin75°} = \frac{200 \times 0.707}{0.966} \approx 146.4 \text{ N}$

$T_2 = \frac{200\sin150°}{\sin75°} = \frac{200 \times 0.5}{0.966} \approx 103.5 \text{ N}$

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The dot product

The question it answers

A force pushes a box. The box moves. How much work did the force do?

Not all of the force did useful work. Only the component along the direction of motion contributed. The perpendicular component pushed the box into the floor — the floor pushed back, and the box didn't move sideways. Physics needed an operation that automatically extracts "how much of this vector lies along that one." That operation is the dot product.

Definition

$\vec{A} \cdot \vec{B} = AB\cos\theta$

where $\theta$ is the angle between $\vec{A}$ and $\vec{B}$ ($0 \le \theta \le 180°$), and $A$, $B$ are magnitudes.

The result is a scalar — a plain number with no direction.

Geometrically: $AB\cos\theta$ is the magnitude of $\vec{A}$ multiplied by the projection of $\vec{B}$ onto $\vec{A}$. Or equivalently, the magnitude of $\vec{B}$ multiplied by the projection of $\vec{A}$ onto $\vec{B}$. Either reading gives the same number — which is why the dot product is commutative:

$\vec{A} \cdot \vec{B} = \vec{B} \cdot \vec{A}$

Special angles:

• $\theta = 0°$: $\vec{A} \cdot \vec{B} = AB$. Fully aligned — maximum.
• $\theta = 90°$: $\vec{A} \cdot \vec{B} = 0$. Perpendicular — nothing in common.
• $\theta = 180°$: $\vec{A} \cdot \vec{B} = -AB$. Fully opposed — minimum.

Why work is a dot product

$W = \vec{F} \cdot \vec{d} = Fd\cos\theta$

The dot product is not a formula someone chose for convenience. Work is a dot product — because the physics demands exactly that "extract the parallel component" operation.

Push horizontally, box moves horizontally: $\theta = 0$, $W = Fd$. Full work done. Hold a bag and walk horizontally: force is vertical, displacement is horizontal, $\theta = 90°$, $W = 0$. No work, however tired you feel. Friction opposing motion: $\theta = 180°$, $W = -Fd$. Negative work — energy removed from the box.

Component form

The dot products of unit vectors:

$\hat{i}\cdot\hat{i} = \hat{j}\cdot\hat{j} = \hat{k}\cdot\hat{k} = 1 \quad \text{(parallel, angle 0°)}$ $\hat{i}\cdot\hat{j} = \hat{j}\cdot\hat{k} = \hat{k}\cdot\hat{i} = 0 \quad \text{(perpendicular, angle 90°)}$

Expanding $\vec{A} \cdot \vec{B}$ using these:

$\vec{A} \cdot \vec{B} = (A_x\hat{i} + A_y\hat{j} + A_z\hat{k})\cdot(B_x\hat{i} + B_y\hat{j} + B_z\hat{k})$

$\vec{A} \cdot \vec{B} = A_xB_x + A_yB_y + A_zB_z$

All cross terms vanish because $\hat{i}\cdot\hat{j} = 0$ etc. Only matching components survive.

Finding the angle between two vectors

$\cos\theta = \frac{\vec{A}\cdot\vec{B}}{AB} = \frac{A_xB_x + A_yB_y + A_zB_z}{\sqrt{A_x^2+A_y^2+A_z^2}\cdot\sqrt{B_x^2+B_y^2+B_z^2}}$

This is the most practical application of the dot product when vectors are given in component form.

Properties

• Commutative: $\vec{A}\cdot\vec{B} = \vec{B}\cdot\vec{A}$
• Distributive: $\vec{A}\cdot(\vec{B}+\vec{C}) = \vec{A}\cdot\vec{B} + \vec{A}\cdot\vec{C}$
• $\vec{A}\cdot\vec{A} = A^2$ (angle between a vector and itself is zero)
• If $\vec{A}\cdot\vec{B} = 0$ and neither is a zero vector, then $\vec{A} \perp \vec{B}$

Illustration. $\vec{A} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{B} = \hat{i} - 2\hat{j} + 2\hat{k}$. Find the angle between them.

$\vec{A}\cdot\vec{B} = (2)(1) + (3)(-2) + (-1)(2) = 2 - 6 - 2 = -6$

$A = \sqrt{4+9+1} = \sqrt{14}, \quad B = \sqrt{1+4+4} = 3$

$\cos\theta = \frac{-6}{3\sqrt{14}} = \frac{-2}{\sqrt{14}} \approx -0.535$

$\theta \approx 122.3°$

The negative dot product confirms the angle is obtuse — the vectors lean away from each other.

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The cross product

The question it answers

A force acts on a door. The door rotates about its hinges. What determines how much rotation the force causes?

Not just the magnitude of the force. Not just where it is applied. The rotation depends on both — specifically, on the component of force perpendicular to the position vector from the hinge. And the rotation happens about an axis — a direction in space, not just a number.

Physics needed an operation that takes two vectors and returns a third vector representing: "the axis and extent of the rotation they together define." That operation is the cross product.

Why the result is perpendicular to both — the real reason

The rotation caused by a force at position $\vec{r}$ happens about an axis. That axis is not along $\vec{r}$. It is not along $\vec{F}$. It is perpendicular to the plane containing both — because that is the axis about which things actually spin.

More fundamentally: $\vec{r}$ and $\vec{F}$ together define a plane. The most natural vector we can construct from two vectors that characterises their relationship as a plane — not their individual directions, not their alignment — is the normal to that plane. The normal is the one direction neither vector individually contains.

Ask it as a requirement: we want an operation $\vec{A} \otimes \vec{B}$ that (i) gives zero when $\vec{A}$ and $\vec{B}$ are parallel (they define no plane), (ii) reverses sign when $\vec{A}$ and $\vec{B}$ are swapped (reversing the rotation), (iii) is linear in each argument. In three dimensions, these three requirements together force the result to be perpendicular to both. Perpendicularity is not imposed on the cross product — it emerges from what the cross product is trying to do.

Why AB sinθ — the area connection

Draw $\vec{A}$ and $\vec{B}$ from the same point. They form a parallelogram.

Area of parallelogram = base × height = $A \times (B\sin\theta)$ = $AB\sin\theta$.

That is the magnitude of the cross product — exactly.

This is not a coincidence. Every physical situation where a cross product appears is secretly asking about area:

• Torque: the moment arm is the perpendicular distance from the axis to the line of force — it is the height of the parallelogram with sides $\vec{r}$ and $\vec{F}$.
• Angular momentum: $|\vec{r} \times \vec{p}|$ is the area swept per unit time, times mass. Kepler's second law (equal areas in equal times) is literally $|\vec{r} \times \vec{v}| =$ constant.
• Magnetic force: $\vec{F} = q\vec{v}\times\vec{B}$ — only the component of velocity perpendicular to $\vec{B}$ creates force. The cross product's $\sin\theta$ extracts it.

The cross product is the natural tool whenever physics asks "how much of this plane is exposed to that influence."

Definition

$\vec{A}\times\vec{B} = AB\sin\theta\ \hat{n}$

where $\theta$ is the angle between $\vec{A}$ and $\vec{B}$ ($0 \le \theta \le 180°$), and $\hat{n}$ is the unit vector perpendicular to both, with direction given by the right-hand rule.

Right-hand rule: Point the fingers of your right hand along $\vec{A}$. Curl them toward $\vec{B}$ through the smaller angle. Your thumb points in the direction of $\vec{A}\times\vec{B}$.

{/* DIAGRAM: Right-hand rule — fingers from A curling toward B, thumb pointing up as n̂ */}

Special angles:

• $\theta = 0°$ or $180°$: $|\vec{A}\times\vec{B}| = 0$. Parallel or antiparallel vectors define no plane, no unique normal.
• $\theta = 90°$: $|\vec{A}\times\vec{B}| = AB$. Maximum — they span the plane most fully.

Why this particular perpendicular direction — the convention

There are two normals to any plane: one up, one down. The right-hand rule is a convention that makes the cross product consistent with our right-handed coordinate system (where $\hat{i}\times\hat{j} = \hat{k}$). A physicist in a mirror universe would use a left-hand rule and get identical physics — every cross product would point the other way, every rotation would be described in the opposite sense, but the physical predictions would be unchanged.

This is actually a deep point. The cross product — torque, angular momentum, magnetic field — changes sign in a mirror. These are called pseudovectors or axial vectors, as distinct from true vectors like force and velocity which have absolute direction. For JEE, you do not need to use the word pseudovector — but knowing that the right-hand rule is a convention, not a physical law, prevents one category of conceptual confusion.

The cross product is not commutative

$\vec{A}\times\vec{B} = -(\vec{B}\times\vec{A})$

Swap the order, reverse the direction. Curl from $\vec{B}$ to $\vec{A}$ instead of $\vec{A}$ to $\vec{B}$ — the thumb points the other way. The magnitude is unchanged; the sign flips.

This is anti-commutativity, and it is not a quirk — it reflects the physical reality that rotating about an axis clockwise is the opposite of rotating anticlockwise.

Consequence: $\vec{A}\times\vec{A} = \vec{0}$ always (angle is zero, $\sin 0 = 0$).

Component form — derived, not stated

The cross products of unit vectors, from the definition and right-hand rule:

$\hat{i}\times\hat{i} = \hat{j}\times\hat{j} = \hat{k}\times\hat{k} = \vec{0} \quad (\theta = 0°)$

For the cross terms — apply right-hand rule to each:

$\hat{i}\times\hat{j} = \hat{k} \qquad \hat{j}\times\hat{k} = \hat{i} \qquad \hat{k}\times\hat{i} = \hat{j}$

And reversing (anti-commutativity):

$\hat{j}\times\hat{i} = -\hat{k} \qquad \hat{k}\times\hat{j} = -\hat{i} \qquad \hat{i}\times\hat{k} = -\hat{j}$

Memory device: going $\hat{i}\to\hat{j}\to\hat{k}\to\hat{i}$ in a cycle gives $+$. Going backward gives $-$.

Now expand $\vec{A}\times\vec{B}$ using distributivity:

$\vec{A}\times\vec{B} = (A_x\hat{i} + A_y\hat{j} + A_z\hat{k})\times(B_x\hat{i} + B_y\hat{j} + B_z\hat{k})$

Expanding (cross terms with same unit vector vanish):

$= A_xB_y(\hat{i}\times\hat{j}) + A_xB_z(\hat{i}\times\hat{k}) + A_yB_x(\hat{j}\times\hat{i}) + A_yB_z(\hat{j}\times\hat{k}) + A_zB_x(\hat{k}\times\hat{i}) + A_zB_y(\hat{k}\times\hat{j})$

$= A_xB_y\hat{k} - A_xB_z\hat{j} - A_yB_x\hat{k} + A_yB_z\hat{i} + A_zB_x\hat{j} - A_zB_y\hat{i}$

Collecting by component:

$\vec{A}\times\vec{B} = (A_yB_z - A_zB_y)\hat{i} - (A_xB_z - A_zB_x)\hat{j} + (A_xB_y - A_yB_x)\hat{k}$

Once you learn matrices, you will recognise this as a $3\times3$ determinant with $\hat{i}$, $\hat{j}$, $\hat{k}$ in the first row. That is a cleaner way to write the same thing — but you do not need matrices to derive or use this result.

Properties

• Anti-commutative: $\vec{A}\times\vec{B} = -\vec{B}\times\vec{A}$
• Distributive: $\vec{A}\times(\vec{B}+\vec{C}) = \vec{A}\times\vec{B} + \vec{A}\times\vec{C}$
• Not associative: $\vec{A}\times(\vec{B}\times\vec{C}) \ne (\vec{A}\times\vec{B})\times\vec{C}$ in general
• $\vec{A}\times\vec{A} = \vec{0}$
• If $\vec{A}\times\vec{B} = \vec{0}$ and neither is a zero vector, then $\vec{A}\parallel\vec{B}$

Illustration. $\vec{A} = 3\hat{i} + 4\hat{j}$ and $\vec{B} = 2\hat{i} - 3\hat{j}$. Find $\vec{A}\times\vec{B}$.

Both vectors are in the xy-plane, so $A_z = B_z = 0$.

$\vec{A}\times\vec{B} = (A_yB_z - A_zB_y)\hat{i} - (A_xB_z - A_zB_x)\hat{j} + (A_xB_y - A_yB_x)\hat{k}$ $= (0 - 0)\hat{i} - (0 - 0)\hat{j} + ((3)(-3) - (4)(2))\hat{k}$ $= (-9 - 8)\hat{k} = -17\hat{k}$

The result is in the $-z$ direction — into the page. Magnitude 17, which is the area of the parallelogram formed by $\vec{A}$ and $\vec{B}$.

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Dot and cross product — side by side

	Dot product	Cross product
Symbol	$\vec{A}\cdot\vec{B}$	$\vec{A}\times\vec{B}$
Result	Scalar	Vector
Formula	$AB\cos\theta$	$AB\sin\theta\ \hat{n}$
Maximum when	Parallel ($\theta = 0°$)	Perpendicular ($\theta = 90°$)
Zero when	Perpendicular	Parallel
Commutative?	Yes	No — reverses sign
Geometric meaning	Projection × magnitude	Area of parallelogram
Physics: scalar result	Work, power, potential energy	—
Physics: vector result	—	Torque, angular momentum, magnetic force

The choice between them is never arbitrary. When the physics gives a scalar that depends on alignment — dot product. When the physics gives a vector along an axis of rotation or a field normal — cross product.

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Scalar triple product

The scalar triple product of three vectors $\vec{A}$, $\vec{B}$, $\vec{C}$ is:

$\vec{A}\cdot(\vec{B}\times\vec{C})$

The result is a scalar.

Geometric meaning: It is the volume of the parallelepiped (the 3D parallelogram) formed by $\vec{A}$, $\vec{B}$, $\vec{C}$. The cross product $\vec{B}\times\vec{C}$ gives the area of the base (parallelogram of $\vec{B}$ and $\vec{C}$) times the normal direction. Dotting with $\vec{A}$ then gives the height (projection of $\vec{A}$ along the normal) times the base area — which is the volume.

$V = |\vec{A}\cdot(\vec{B}\times\vec{C})|$

Key property:

$\vec{A}\cdot(\vec{B}\times\vec{C}) = \vec{B}\cdot(\vec{C}\times\vec{A}) = \vec{C}\cdot(\vec{A}\times\vec{B})$

Cyclic permutation of the three vectors leaves the scalar triple product unchanged. Swapping any two vectors reverses the sign.

If the scalar triple product is zero, the three vectors are coplanar — they lie in the same plane and form no volume.

In component form:

$\vec{A}\cdot(\vec{B}\times\vec{C}) = A_x(B_yC_z - B_zC_y) - A_y(B_xC_z - B_zC_x) + A_z(B_xC_y - B_yC_x)$

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Vector triple product

The vector triple product is:

$\vec{A}\times(\vec{B}\times\vec{C})$

The result is a vector lying in the plane of $\vec{B}$ and $\vec{C}$ (not along $\vec{A}$).

The identity — called the BAC-CAB rule — is:

$\vec{A}\times(\vec{B}\times\vec{C}) = \vec{B}(\vec{A}\cdot\vec{C}) - \vec{C}(\vec{A}\cdot\vec{B})$

How to remember it: BAC minus CAB. The first letter of each term (B, C) matches the vectors not being "crossed" in the original, and the dot products are with $\vec{A}$.

Note that the cross product is not associative: $\vec{A}\times(\vec{B}\times\vec{C}) \ne (\vec{A}\times\vec{B})\times\vec{C}$ in general. The BAC-CAB rule only applies when parentheses are on the right.

For $(\vec{A}\times\vec{B})\times\vec{C}$: swap $\vec{A}$ and $\vec{C}$ and negate — $(\vec{A}\times\vec{B})\times\vec{C} = -\vec{C}\times(\vec{A}\times\vec{B}) = -\vec{A}(\vec{C}\cdot\vec{B}) + \vec{B}(\vec{C}\cdot\vec{A})$.

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Common mistakes

Adding magnitudes when directions differ. Two 5 N forces do not always give 10 N. Always check the angle.

Wrong angle in $A\cos\theta$. The angle is always between the vector and the axis you are resolving along. If the problem gives an angle from the y-axis and you need the x-component, it is $A\sin\theta$, not $A\cos\theta$.

Dropping signs in components. A vector pointing left has negative x-component. Dropping the sign gives the wrong resultant direction.

Wrong quadrant from $\tan^{-1}$. $\tan^{-1}$ returns values between $-90°$ and $+90°$. Check the signs of both $R_x$ and $R_y$ before stating the direction.

Mixing sinθ and cosθ for dot and cross. Dot product uses $\cos\theta$ — parallel alignment. Cross product uses $\sin\theta$ — perpendicular extent. Swapping them makes work zero when force and displacement are parallel, which is exactly backwards.

Forgetting direction for cross product. The cross product is a vector. Giving only the magnitude is an incomplete answer. Always state the direction — from the right-hand rule or from the component calculation.

Assuming $\vec{A}\times\vec{B} = \vec{B}\times\vec{A}$. It does not. The cross product is anti-commutative. Order always matters.

Associativity of cross product. $\vec{A}\times(\vec{B}\times\vec{C}) \ne (\vec{A}\times\vec{B})\times\vec{C}$. Parentheses cannot be moved freely.

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What to remember

Direction is part of the quantity — not a label attached to it. Remove direction from force or velocity and the physics collapses.

Vectors add geometrically. The resultant depends on both magnitude and angle. $R = \sqrt{A^2 + B^2 + 2AB\cos\theta}$ — one formula, not three.

Components convert vector addition into scalar addition. Resolve, add numbers by axis, reconstruct. Check the quadrant.

The dot product extracts parallel alignment. $\vec{A}\cdot\vec{B} = AB\cos\theta$. Result is a scalar. Maximum when parallel, zero when perpendicular. Use it for work, power, and finding angles.

The cross product captures perpendicular extent. $\vec{A}\times\vec{B} = AB\sin\theta\ \hat{n}$. Result is a vector perpendicular to both. Magnitude is the area of the parallelogram. Maximum when perpendicular, zero when parallel. Use it for torque, angular momentum, and magnetic force.

The scalar triple product gives the volume of a parallelepiped. If it is zero, the three vectors are coplanar.

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For practice problems on all topics above — from direct computation to JEE-level applications — see Vectors: Problem Bank.