Scalars and vectors

Stand at a crossroads and tell a friend "walk 500 metres" — they still do not know which way to go. Tell them "walk 500 metres north-east" and the instruction is complete. That one missing word, direction, is the entire reason physics splits every quantity you will ever measure into two families: scalars, which are satisfied by a number and a unit, and vectors, which additionally demand a direction and — crucially — a specific rule for how two of them combine. This article is the doorway into the CBSE Class 11 chapter Motion in a Plane. Before you can add two velocities at an angle, resolve a force along an incline, or launch a projectile and predict where it lands, you need working fluency with vectors as objects — how they are drawn, added, subtracted, scaled, split into components, and multiplied by other vectors. Everything downstream — displacement, velocity, acceleration, force, momentum, and eventually the dot and cross products that measure "how much along" and "how much across" — rests on the ideas built here. The payoff is immediate and mechanical. Once you can resolve a vector $\vec A$ making angle $\theta$ with the x-axis into its rectangular components $A_x = A\cos\theta$ and $A_y = A\sin\theta$ , whole classes of two-dimensional problems collapse into two independent one-dimensional problems you already know how to solve. That single trick — split, solve straight-line motion twice, recombine — is the engine behind every projectile-range and circular-motion calculation in this chapter. We build it here, carefully, with worked numbers you can check by hand.

Key takeaways

Scalars need only magnitude and a unit (mass, distance, speed, time, energy); vectors need magnitude and direction, and must obey the triangle/parallelogram law of addition (displacement, velocity, acceleration, force, momentum).
A vector is drawn as an arrow; its length is the magnitude $|\vec A|=A$ and its arrowhead gives the direction. The unit vector $\hat A = \vec A/|\vec A|$ has magnitude 1 and marks direction only.
Two vectors are equal only if they have the same magnitude and the same direction, regardless of where they are drawn; multiplying a vector by a real number $\lambda$ scales its magnitude by $|\lambda|$ and, if $\lambda<0$ , reverses its direction.
Vector addition follows the triangle law (place tail-to-head) or the equivalent parallelogram law (common tail); vector subtraction $\vec A-\vec B$ is addition of $\vec A$ and $(-\vec B)$ .
Resolution into rectangular components: any vector in a plane splits as $A_x=A\cos\theta$ , $A_y=A\sin\theta$ , with $A=\sqrt{A_x^2+A_y^2}$ and $\tan\theta = A_y/A_x$ — this is the working tool for every 2-D motion problem in the chapter.
The scalar (dot) product $\vec A\cdot\vec B=AB\cos\theta$ gives a number (used for work); the vector (cross) product $\vec A\times\vec B=AB\sin\theta\,\hat n$ gives a vector perpendicular to both (used for torque and angular momentum).
Key idea: a vector is fully specified only by magnitude plus direction and a fixed addition rule (triangle/parallelogram law), and resolving it into perpendicular components A cos θ and A sin θ turns any 2-D vector problem into two ordinary 1-D problems.

The core insight

A physical quantity is a vector only if it carries both a magnitude and a direction and obeys the triangle/parallelogram law of addition; resolving such a vector into rectangular components $A_x=A\cos\theta$ , $A_y=A\sin\theta$ is the single tool that converts two-dimensional motion into two independent one-dimensional problems.

Where it fits in the syllabus

This is the opening topic of CBSE Class 11 Physics, chapter Motion in a Plane — everything else in the chapter (position and displacement vectors, addition/subtraction, resolution, uniform-velocity and uniform-acceleration motion in a plane, projectile motion, and uniform circular motion) is built on the vocabulary and tools introduced here. The prescribed scope is scalars versus vectors, equality of vectors, multiplication of a vector by a real number, addition and subtraction by the triangle and parallelogram laws, the unit vector, resolution into rectangular components, and the scalar (dot) and vector (cross) products. Relative velocity in a plane is touched only lightly in this chapter and full river-boat problem chains, non-uniform circular dynamics, and general 3-D vector geometry beyond components are kept out of the Class 11 board syllabus.

The physical picture

A scalar is completely described by a single real number with a unit — no direction attaches to it and it combines by ordinary algebra. Mass, distance, speed, time, temperature, work and energy are scalars: $5\ \text{kg}+3\ \text{kg}=8\ \text{kg}$ , no questions asked. A vector carries both magnitude and direction, and — this is the part students skip — it must combine with other vectors of the same kind by the triangle or parallelogram law, not by simple arithmetic. Displacement, velocity, acceleration, force and momentum are vectors. Graphically, a vector $\vec A$ is a directed line segment: its length (drawn to scale) is the magnitude $|\vec A|=A$ , and the arrowhead marks the direction.

Equality of vectors. Two vectors $\vec A$ and $\vec B$ are equal, written $\vec A=\vec B$ , only when they have the same magnitude and the same direction — where they are drawn on the page does not matter, since a vector can be freely translated (parallel-shifted) without changing its identity.

Multiplication by a real number. If $\lambda$ is a real number, $\lambda\vec A$ is a vector of magnitude $|\lambda|A$ , pointing along $\vec A$ if $\lambda>0$ and opposite to $\vec A$ if $\lambda<0$ . This is exactly how the unit vector is built:

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

where: $\hat A$ is the unit vector along $\vec A$ , a dimensionless vector of magnitude exactly 1 whose sole job is to encode direction; $\vec A$ is the original vector and $|\vec A|=A$ its magnitude. Reading it physically: dividing a vector by its own magnitude strips away "how big" and leaves only "which way." A zero (null) vector $\vec 0$ has magnitude zero and an undefined direction — it is a legitimate vector, not a non-vector, and shows up as the displacement of a body that returns to its starting point.

Addition — triangle and parallelogram laws. To add $\vec A+\vec B$ by the triangle law, draw $\vec A$ , then draw $\vec B$ starting from the tip (head) of $\vec A$ ; the resultant $\vec R$ is the vector from the tail of $\vec A$ to the tip of $\vec B$ . The equivalent parallelogram law draws $\vec A$ and $\vec B$ from a common tail, completes the parallelogram, and takes the diagonal from that common point as $\vec R$ . Both give the same $\vec R$ — they are the same rule seen two ways. Subtraction $\vec A-\vec B$ is simply addition of $\vec A$ and the reversed vector $(-\vec B)$ .

Resolution into rectangular components. Any vector $\vec A$ in a plane, making angle $\theta$ with the x-axis, can be replaced by two perpendicular vectors along the axes whose vector sum reproduces it exactly:

\vec A = A_x\,\hat\imath + A_y\,\hat\jmath,\qquad A_x=A\cos\theta,\quad A_y=A\sin\theta

where: $\hat\imath$ and $\hat\jmath$ are the unit vectors along the x- and y-axes, $A_x$ and $A_y$ are the (scalar) components of $\vec A$ along those axes, and $\theta$ is the angle $\vec A$ makes with the positive x-axis. Physically, $A_x$ and $A_y$ are what a shadow of the arrow would measure if a strong light were shone straight down each axis — the "how much along x" and "how much along y" content of the same single arrow.

Figure

A vector arrow at an angle above the x axis with dashed perpendicular lines dropped to the x and y axes showing its components A cos theta along x and A sin theta along y, next to a triangle law diagram where vector B is placed head to tail after vector A and the resultant R is drawn from the start of A to the tip of B

The dot and cross products. Two vectors combine in two further, distinct ways. The scalar (dot) product returns a number:

\vec A\cdot\vec B = AB\cos\theta

where: $A$ and $B$ are the magnitudes, and $\theta$ is the angle between the two vectors when drawn tail to tail; the result is a plain scalar, largest when the vectors are parallel ( $\theta=0$ ) and zero when they are perpendicular. The vector (cross) product returns a new vector:

\vec A\times\vec B = AB\sin\theta\,\hat n

where: $\hat n$ is the unit vector perpendicular to the plane of $\vec A$ and $\vec B$ , its sense fixed by the right-hand rule; the magnitude $AB\sin\theta$ is largest when the two vectors are perpendicular and zero when they are parallel. These two products are not derived further in this chapter — they resurface later as $W=\vec F\cdot\vec d$ (work) and $\vec\tau=\vec r\times\vec F$ (torque), so it is enough here to know what each produces (a number versus a vector) and when each is maximum or zero.

Derivation AResolving a vector into its rectangular x- and y-components

The goal is to show that any vector $\vec A$ lying in a plane, drawn at angle $\theta$ to the x-axis, can be replaced without any loss of information by two perpendicular vectors along the axes — and to derive the exact scalar magnitudes of those two components, $A_x=A\cos\theta$ and $A_y=A\sin\theta$ , together with the reverse recipe for recovering $A$ and $\theta$ from $A_x$ and $A_y$ . This resolution is the single tool that turns every two-dimensional motion or force problem in the chapter into two independent one-dimensional problems.

Step 1. Place the tail of $\vec A$ at the origin of an x–y coordinate system, with $\vec A$ making angle $\theta$ with the positive x-axis. Drop a perpendicular from the tip of $\vec A$ down to the x-axis, and a perpendicular across to the y-axis. This constructs a right-angled triangle whose hypotenuse is $\vec A$ itself, one leg lying along the x-axis and the other parallel to the y-axis.

Step 2. Read off the two legs using ordinary right-triangle trigonometry. The leg adjacent to $\theta$ has length $A\cos\theta$ and lies along the x-axis; the leg opposite $\theta$ has length $A\sin\theta$ and lies parallel to the y-axis. Call these two legs, as vectors along the axes, $\vec A_x = (A\cos\theta)\hat\imath$ and $\vec A_y=(A\sin\theta)\hat\jmath$ , where $\hat\imath$ and $\hat\jmath$ are the unit vectors along x and y.

Step 3. Check, by the triangle law of addition, that $\vec A_x+\vec A_y$ reproduces $\vec A$ exactly: placing $\vec A_y$ tail-to-head at the tip of $\vec A_x$ walks along the x-leg and then up the y-leg, landing precisely at the tip of the original $\vec A$ . So $\vec A = A\cos\theta\,\hat\imath + A\sin\theta\,\hat\jmath$ , and the pair of numbers $(A_x,A_y)=(A\cos\theta, A\sin\theta)$ carries the same information as the pair $(A,\theta)$ — nothing is lost or gained by resolving.

Step 4. Invert the relation. Given the components $A_x$ and $A_y$ instead of $(A,\theta)$ , apply Pythagoras to the same right triangle to recover the magnitude, $A=\sqrt{A_x^2+A_y^2}$ , and basic trigonometry to recover the direction, $\tan\theta = A_y/A_x$ . This pair of formulas is used constantly in the reverse direction — for instance, once two velocity components are known separately (from independent horizontal and vertical reasoning), the actual speed and direction of motion are recovered from these two lines. Verified: for $A=10$ , $\theta=37^\circ$ , $A_x=10\cos37^\circ=7.99\approx8.0$ and $A_y=10\sin37^\circ=6.02\approx6.0$ ; reversing, $\sqrt{8.0^2+6.0^2}=\sqrt{64+36}=\sqrt{100}=10.0=A$ and $\tan^{-1}(6.0/8.0)=\tan^{-1}(0.75)=36.9^\circ\approx37^\circ$ , matching the starting values.

Worked examples

Board level - classifying quantities, a unit vector, and vector addition at right angles

(a) Classify as scalar or vector, with a one-line reason: (i) distance travelled = 12 m, (ii) displacement = 12 m due east, (iii) mass = 2 kg. (b) A vector $\vec A$ has magnitude 8 units and points due north; find the unit vector $\hat A$ and a vector $\vec B$ of magnitude 3 units in the same direction. (c) A student walks 6 m east then 8 m north. Find the magnitude and direction of the resultant displacement.

Step 1. Classification. Distance (12 m) is a scalar — it records only "how much path," with no direction attached. Displacement (12 m due east) is a vector — it fixes both how far and which way. Mass (2 kg) is a scalar — a single number describing the amount of matter, with no direction possible.

Step 2. Unit vector and scaled vector. Since $\hat A=\vec A/|\vec A|$ , and $|\vec A|=8$ , the unit vector $\hat A$ has magnitude 1 and points due north — "1 unit, due north." A vector $\vec B$ of magnitude 3 in the same direction is then $\vec B = 3\hat A$ , i.e. 3 units due north; this is multiplication of a vector by the positive real number 3, which scales the magnitude and leaves the direction unchanged.

Step 3. Resultant of two perpendicular displacements. The 6 m east and 8 m north legs are two vectors added tail-to-head (triangle law), and because they are perpendicular the resultant magnitude follows directly from Pythagoras: $R=\sqrt{6^2+8^2}=\sqrt{36+64}=\sqrt{100}=10\ \text{m}$ . The direction is $\tan\theta = 8/6 = 1.33$ , so $\theta = \tan^{-1}(1.33)=53.1^\circ$ north of east. Answer: resultant displacement = 10 m, at 53.1° north of east; $\hat A$ = 1 unit due north, $\vec B$ = 3 units due north.

Competitive level - resolving two vectors and combining their components

A block is pulled by two forces in a horizontal plane: $\vec F_1 = 12\ \text{N}$ at $30^\circ$ above the positive x-axis, and $\vec F_2 = 9\ \text{N}$ at $120^\circ$ above the positive x-axis. Find the magnitude and direction of the resultant force $\vec F = \vec F_1+\vec F_2$ by resolving into components.

Step 1. Resolve $\vec F_1$ : $F_{1x}=12\cos30^\circ=12(0.866)=10.39\ \text{N}$ , $F_{1y}=12\sin30^\circ=12(0.5)=6.00\ \text{N}$ .

Step 2. Resolve $\vec F_2$ : $F_{2x}=9\cos120^\circ=9(-0.5)=-4.50\ \text{N}$ , $F_{2y}=9\sin120^\circ=9(0.866)=7.79\ \text{N}$ .

Step 3. Add components separately — this is exactly why resolution is powerful, x and y never mix: $F_x = 10.39+(-4.50)=5.89\ \text{N}$ , $F_y=6.00+7.79=13.79\ \text{N}$ .

Step 4. Recombine into magnitude and direction: $F=\sqrt{F_x^2+F_y^2}=\sqrt{5.89^2+13.79^2}=\sqrt{34.7+190.2}=\sqrt{224.9}=15.0\ \text{N}$ , and $\theta=\tan^{-1}(13.79/5.89)=\tan^{-1}(2.34)=66.8^\circ$ above the positive x-axis. Answer: resultant force $F \approx 15.0\ \text{N}$ at about $66.8^\circ$ above the x-axis. Timing note: about ninety seconds. Resolve each vector into $(\cos,\sin)$ components using its own angle, add the x's together and the y's together — never add magnitudes directly unless the vectors are parallel — then Pythagoras and $\tan^{-1}$ to recombine. The one trap is a sign error on $\cos120^\circ$ (negative, since $120^\circ$ is past the y-axis into the second quadrant); always sketch the angle first to fix the correct sign of each component before plugging into the calculator.

Common traps

Treating any quantity that "has a direction" in everyday speech as automatically a vector — the real test is whether it obeys the triangle/parallelogram addition law; electric current has an associated direction in a circuit but adds algebraically at a junction, not vectorially, so it is a scalar in this course.
Confusing distance (scalar, total path length, always $\ge$ the straight-line separation) with displacement (vector, straight-line separation with direction) — the two are equal in magnitude only for straight-line motion without any reversal.
Adding vector magnitudes directly as if they were scalars, e.g. writing $R=A+B$ for two vectors that are not parallel — the correct route is either the triangle/parallelogram construction or resolving into components and adding component-wise.
Dropping the negative sign on a component when the angle places the vector in the second, third or fourth quadrant — $\cos\theta$ and $\sin\theta$ change sign there, and a missing sign silently flips the direction of the whole answer.
Writing a vector's magnitude alone and calling it the vector, e.g. "the velocity is 5 m/s" with no stated direction — that number is only the speed, the magnitude of the velocity vector, not the vector itself.
Confusing the dot product (a number, maximum when vectors are parallel) with the cross product (a vector, maximum when vectors are perpendicular) — mixing these up misassigns whether a quantity like work or torque should come out as a scalar or a vector.

Common trapA vector is not "any quantity with a direction" — it is defined by obeying the triangle/parallelogram law of addition

The single most common board-exam slip is deciding scalar-versus-vector by ear: "it has a direction in the sentence, so it must be a vector." That test fails for electric current, which is quoted with a direction (through a resistor, into a junction) yet combines by ordinary algebraic addition at a junction, not by the parallelogram rule — so it is classified as a scalar. The reliable test is always the addition rule: does the quantity, when two instances are combined, follow $\vec R = \vec A+\vec B$ constructed geometrically (triangle or parallelogram), giving a resultant that can be smaller than the simple sum of magnitudes when the vectors point in different directions? If yes, it is a vector; if the quantities merely add like ordinary numbers regardless of any stated direction, it is a scalar, however direction-flavoured the sentence sounds.

You should be able to…

Classify a given physical quantity as scalar or vector and justify the classification using the addition-rule test, not just "does it have a direction."
State the equality condition for two vectors (same magnitude and same direction) and describe what multiplying a vector by a positive or negative real number does to it.
Add and subtract two vectors graphically using the triangle law and the parallelogram law, and confirm both give the same resultant.
Resolve a given vector into its rectangular components $A_x=A\cos\theta$ , $A_y=A\sin\theta$ , and recover magnitude and direction from components using $A=\sqrt{A_x^2+A_y^2}$ and $\tan\theta=A_y/A_x$ .
State the formulas for the scalar (dot) and vector (cross) products, and say in one line when each is maximum and when each is zero.
Add two or more vectors given at different angles by resolving each into components, summing components separately, and recombining.

Ask the tutor: Not sure whether to classify a quantity like pressure, electric current, or torque as scalar or vector — or want a walkthrough of resolving a tricky vector at an obtuse angle — ask the AI tutor to work through it with you.

If a vector can be moved anywhere on the page without changing what it is, does that mean position does not matter at all in physics?

For a free vector like a displacement or a velocity, yes — only magnitude and direction define it, and sliding the arrow around the page (without rotating or stretching it) still represents the same vector. But some vectors are tied to a specific point of application, like a force acting at a particular location on a rigid body, where where the force acts changes its physical effect (its torque) even though the force vector itself is unchanged. In this chapter you can treat displacement, velocity and acceleration as free vectors; the point-of-application subtlety becomes important later, in rotational motion.

Why do we bother resolving vectors into components instead of just using the triangle law every time?

The triangle and parallelogram laws are great for adding two vectors, but they get clumsy fast for three, four, or many vectors, or for tracking motion continuously over time. Resolving into perpendicular components turns the problem into ordinary algebra: add all the x-components together, add all the y-components together, and you are done — no protractor, no scale drawing, no accumulated drawing error. It is also exactly what makes projectile motion tractable later in this chapter: horizontal motion (constant velocity) and vertical motion (constant acceleration under gravity) can be solved completely separately, because they are the x- and y-components of the same single motion.

Is the cross product just the dot product with sine instead of cosine, or is something more going on?

The formulas look like siblings, $AB\cos\theta$ versus $AB\sin\theta$ , but the outputs are fundamentally different kinds of object. The dot product $\vec A\cdot\vec B=AB\cos\theta$ produces a plain number — no direction at all — and measures how much of one vector lies along the other. The cross product $\vec A\times\vec B=AB\sin\theta\,\hat n$ produces a brand-new vector, perpendicular to both original vectors, whose direction is fixed by the right-hand rule and whose size measures how "spread apart" (perpendicular) the two original vectors are. You will meet the dot product again as work, $W=\vec F\cdot\vec d$ , and the cross product again as torque, $\vec\tau=\vec r\times\vec F$ , later in the course — for now, just remember: dot gives a number, cross gives a vector.

My teacher says the zero vector has no direction — so how can it still count as a vector at all?

A vector is defined by the pair (magnitude, direction) plus the addition rule it obeys, and the zero vector satisfies both requirements in a limiting way: its magnitude is exactly zero, and because an arrow of zero length has no meaningful orientation, its direction is simply left undefined — that is not the same as "not being a vector." It still obeys vector addition perfectly well: adding $\vec 0$ to any vector $\vec A$ leaves $\vec A$ unchanged, exactly the role zero plays in ordinary number addition. A concrete example: if you walk from your house to the market and back home again, your net displacement for the whole trip is the zero vector — you covered a real, non-zero distance, but your displacement, the vector, is zero.