Scalars and vectors

Some quantities in physics are settled the moment you give a number and a unit: a mass of 5 kg5\ \text{kg}, a time of 12 s12\ \text{s}, a temperature of 37∘C37^\circ\text{C}. Others refuse to be pinned down that way — saying a car moves at 60 km/h60\ \text{km/h} tells you nothing about whether it heads north or south, and the difference matters enormously for where it ends up. This topic opens the Motion in a Plane portion of Unit 2, Kinematics, by formally splitting every physical quantity into two families: scalars, which need only magnitude, and vectors, which need magnitude and direction together. Getting the split right, along with the closely linked ideas of position, displacement, and equality of vectors, is the foundation on which vector addition, components, and the dot and cross products are built in the rest of the unit.

The core insight

A scalar is fully specified by a magnitude and unit alone, while a vector additionally needs a direction, and this single distinction decides how each quantity is allowed to combine with others of its kind.

The core idea

A scalar quantity has magnitude only, and it obeys ordinary algebra: mass, distance, speed, time, temperature, and work are all scalars. Two scalars of the same kind simply add or subtract as numbers, so 5 kg+3 kg=8 kg5\ \text{kg} + 3\ \text{kg} = 8\ \text{kg} with no further information needed. A vector quantity has both magnitude and direction, and it must be combined using special geometric rules — the triangle and parallelogram laws, covered in the next topic — rather than plain addition. Displacement, velocity, acceleration, force, and momentum are all vectors. A vector is written with an arrow, A⃗\vec A, and drawn as an arrow whose length represents magnitude and whose orientation represents direction; its magnitude alone, a scalar, is written ∣A⃗∣|\vec A| or simply AA. The position vector r⃗\vec r of a point fixes its location relative to a chosen origin. If a particle moves from position r⃗1\vec r_1 to position r⃗2\vec r_2, its displacement vector is

Δr⃗=r⃗2−r⃗1,\Delta \vec r = \vec r_2 - \vec r_1,

where: r⃗1\vec r_1 and r⃗2\vec r_2 are the initial and final position vectors and Δr⃗\Delta \vec r is the net change of position. Displacement depends only on the start and end points, never on the path taken between them — unlike distance travelled, a scalar that does depend on the actual path. Two vectors are equal only when they share both the same magnitude and the same direction; a vector can be slid parallel to itself anywhere on the page without changing its identity, since neither its length nor its orientation changes. Multiplying a vector by −1-1 reverses its direction while leaving its magnitude untouched, so −A⃗-\vec A has the same magnitude as A⃗\vec A but points exactly opposite. Multiplying by a positive scalar kk scales the magnitude without touching direction: ∣kA⃗∣=k∣A⃗∣|k\vec A| = k|\vec A| for k>0k>0. A unit vector has magnitude exactly 11 and encodes pure direction; the unit vector along A⃗\vec A is A^=A⃗/∣A⃗∣\hat A = \vec A/|\vec A|. A null (zero) vector 0⃗\vec 0 has zero magnitude and an arbitrary, undefined direction — it shows up naturally as the displacement of a body that returns exactly to where it started.

Figure
Two arrows drawn from the same origin point, one labelled vector A pointing north east and a second arrow labelled minus A of the same length pointing exactly south west, showing that reversing a vector flips its direction but not its length
Multiplying by minus one reverses direction, not magnitude origin A minus A same length, exactly opposite direction: two equal magnitudes, one reversal
Derivation AMagnitude of a displacement from two position vectors

The goal is to take two position vectors given in component form, r⃗1\vec r_1 and r⃗2\vec r_2, and extract a single number — the magnitude of the displacement between them — using nothing beyond vector subtraction and the Pythagorean theorem, so that a displacement written as components can always be converted into "how far, in a straight line."

Step 1. Write both position vectors in ı^\hat\imath–ȷ^\hat\jmath component form, r⃗1=x1ı^+y1ȷ^\vec r_1 = x_1\hat\imath + y_1\hat\jmath and r⃗2=x2ı^+y2ȷ^\vec r_2 = x_2\hat\imath + y_2\hat\jmath, where xx and yy are the coordinates measured from the chosen origin.

Step 2. Subtract component by component, since vector subtraction acts independently on each axis: Δr⃗=r⃗2−r⃗1=(x2−x1)ı^+(y2−y1)ȷ^\Delta \vec r = \vec r_2 - \vec r_1 = (x_2-x_1)\hat\imath + (y_2-y_1)\hat\jmath. Call these two numbers Δx=x2−x1\Delta x = x_2 - x_1 and Δy=y2−y1\Delta y = y_2 - y_1.

Step 3. Treat Δx\Delta x and Δy\Delta y as the two perpendicular legs of a right triangle, with the displacement vector as the hypotenuse. The Pythagorean theorem then gives the magnitude directly:

∣Δr⃗∣=(Δx)2+(Δy)2.|\Delta \vec r| = \sqrt{(\Delta x)^2 + (\Delta y)^2}.

where: Δx\Delta x and Δy\Delta y are the component-wise changes found in Step 2, and ∣Δr⃗∣|\Delta \vec r| is the straight-line distance between the start and end points, independent of whatever curved path was actually walked.

Step 4. Read off the direction, if needed, from tan⁡θ=Δy/Δx\tan\theta = \Delta y/\Delta x, measured from the positive xx-axis, using the signs of Δx\Delta x and Δy\Delta y to fix which quadrant the angle lies in.

Worked example

A particle's position vector changes from r⃗1=(3ı^+2ȷ^) m\vec r_1 = (3\hat\imath + 2\hat\jmath)\ \text{m} to r⃗2=(7ı^−1ȷ^) m\vec r_2 = (7\hat\imath - 1\hat\jmath)\ \text{m}. Find the displacement vector, its magnitude, and the angle it makes with the xx-axis.

Step 1. Subtract component by component: Δr⃗=r⃗2−r⃗1=(7−3)ı^+(−1−2)ȷ^=4ı^−3ȷ^ m\Delta \vec r = \vec r_2 - \vec r_1 = (7-3)\hat\imath + (-1-2)\hat\jmath = 4\hat\imath - 3\hat\jmath\ \text{m}.

Step 2. Apply the magnitude formula from the derivation above: ∣Δr⃗∣=42+(−3)2=16+9=25=5 m|\Delta \vec r| = \sqrt{4^2 + (-3)^2} = \sqrt{16+9} = \sqrt{25} = 5\ \text{m}.

Step 3. Find the direction using tan⁡θ=Δy/Δx=−3/4\tan\theta = \Delta y/\Delta x = -3/4, so θ=−36.9∘\theta = -36.9^\circ, meaning the displacement points 36.9∘36.9^\circ below the positive xx-axis (since Δx=4>0\Delta x = 4 > 0 and Δy=−3<0\Delta y = -3 < 0 places it in the fourth quadrant).

The particle undergoes a net displacement of magnitude 5 m5\ \text{m}, directed 36.9∘36.9^\circ below the xx-axis along 4ı^−3ȷ^4\hat\imath - 3\hat\jmath — regardless of whatever curved or indirect path it actually followed between the two points.

Competitive stretch

Classify each of the following as scalar or vector, a common opener in board and entrance exams: (a) the distance a hiker walks along a winding trail, (b) the hiker's displacement from start to end, (c) the hiker's average speed, (d) the hiker's average velocity. Distance is total path length, magnitude only: scalar. Displacement is the straight-line change of position with a definite direction: vector. Average speed is total distance divided by total time, both scalars: scalar. Average velocity is displacement divided by time; since displacement is a vector and time is a positive scalar, the quotient is still a vector with a definite direction: vector. The pattern is general — quantities built from path length are scalars, quantities built from net change of position are vectors.

Common traps

  • Treating speed and velocity, or distance and displacement, as interchangeable — in each pair one member is a scalar and the other a vector, and only the vector carries direction.
  • Believing a negative sign in front of a vector, as in −A⃗-\vec A, means "smaller magnitude" — it only reverses direction, since ∣−A⃗∣=∣A⃗∣|-\vec A| = |\vec A|.
  • Adding vectors as if they were ordinary numbers, e.g. 3 m/s+4 m/s=7 m/s3\ \text{m/s} + 4\ \text{m/s} = 7\ \text{m/s}, without accounting for direction — plain addition is valid only when the vectors are collinear.
  • Calling two vectors equal just because they share a magnitude — they must also point in exactly the same direction to be equal.

Ask the tutor: if you are still unsure whether a given quantity, or a given sign in front of a vector, changes its magnitude or only its direction, ask the AI tutor to work through a few more classification and component examples with you.

Why is displacement a vector but distance a scalar, even for the same trip?

Distance adds up every bit of path length you actually cover, so a winding or backtracking route makes it larger, and it never needs a direction to be meaningful — a number in metres is the whole story. Displacement only cares about where you started and where you ended, given by Δr⃗=r⃗2−r⃗1\Delta \vec r = \vec r_2 - \vec r_1, so it needs a direction to say which way that net change points. For a straight-line trip with no doubling back, the two can have the same magnitude, but the moment the path curves or reverses, distance grows while displacement can even shrink back toward zero.

Is a null vector just the number zero?

Not quite. The number zero is a scalar with no direction to speak of, while a null vector 0⃗\vec 0 is a vector whose magnitude happens to be zero, and whose direction is left undefined or arbitrary because there is no meaningful arrow to draw. It appears naturally as the displacement of someone who walks out and returns to exactly the same spot, or as the resultant of two equal and opposite vectors. Treating it as ordinary zero is fine for arithmetic, but remember it still belongs to the vector family, not the scalar one.

Can a vector's magnitude ever be negative?

No. Magnitude is a length, computed for example as ∣Δr⃗∣=(Δx)2+(Δy)2|\Delta \vec r| = \sqrt{(\Delta x)^2 + (\Delta y)^2}, and a square root of a sum of squares is never negative. What can be negative are the individual components, like Δy=−3 m\Delta y = -3\ \text{m} in the worked example above, or the coefficient in front of a vector, like −A⃗-\vec A. A negative component or a negative multiplier changes direction, but once you compute the magnitude itself, the answer is always zero or positive.

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Where this leads