Q.Read each statement below carefully and state with reasons, if it is true or false:

Concept: Scalar vs vector. Precise distinctions between magnitude, components, path length, and displacement resolve each part.

• (a) The magnitude of a vector has no direction, so it is a scalar. True.
• (b) A component of a vector, e.g. $A_x\hat{i}$, still has a direction (along $\hat{i}$) — it is itself a vector, not a scalar. False.
• (c) Path length equals the magnitude of displacement only for straight-line motion without reversal; in general path length $\ge |\text{displacement}|$. False. …

A scalar has magnitude only; a vector has both magnitude and direction. Keeping this distinction precise resolves each statement below.

(a) The magnitude of a vector is always a scalar.

The magnitude of a vector (e.g. $|\vec{v}| = 5\ \text{m/s}$) is a non-negative number with no direction attached — exactly the definition of a scalar.

True.

(b) Each component of a vector is always a scalar.

A vector $\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$ has vector components $A_x\hat{i}$, $A_y\hat{j}$, $A_z\hat{k}$ — each of these still has both a magnitude and a direction (along $\hat{i}$, $\hat{j}$, $\hat{k}$ respectively), so each is itself a vector, not a scalar. (Only the bare coefficients $A_x, A_y, A_z$, called scalar components, are scalars — but "a component of a vector," taken as $A_x\hat{i}$ etc., is a vector.)

False.

(c) The total path length is always equal to the magnitude of the displacement vector.

Path length is the actual distance travelled along the trajectory; displacement is the straight-line vector from start to end. They're equal only for motion in a straight line without reversing — in general (e.g. a curved path, or going out and back) the path length exceeds the magnitude of displacement.

False.

(d) Average speed $\ge$ magnitude of average velocity. …

Concept: Settle Every Claim with an Explicit Concrete Example

Method: Constructive Counterexample / Coordinate-Construction Proof for Each Part

Rather than arguing in words why each statement is true or false, this method either constructs one explicit concrete numeric example that violates a False claim, or gives an explicit coordinate-based proof of a True claim -- a technique that leaves nothing to interpretation.

(a) The magnitude of a vector is always a scalar

For $\vec v = (3,4)$, the magnitude is $|\vec v|=\sqrt{3^2+4^2}=5$ -- just the number $5$, with no direction attached to it (it's meaningless to ask "which way does the magnitude $5$ point?"). This holds for any vector by the very definition $|\vec v|=\sqrt{v_x^2+v_y^2+v_z^2}$, which always outputs a single non-negative real number.

Verdict: True.

(b) Each component of a vector is always a scalar

Take the same $\vec v=(3,4)$. Its component "along $\hat i$" is $3\hat i$ -- this is $3$ units in the $+x$ direction, a genuine vector, not merely the bare number $3$. If instead $\vec v=(-3,4)$, the $x$-component is $-3\hat i$, i.e. $3$ units in the $-x$ direction -- a different vector from $3\hat i$, even though $|-3|=|3|$. This shows the component carries directional information that the scalar coefficient alone does not.

Verdict: False.

(c) Total path length is always equal to the magnitude of the displacement vector

Concrete counterexample: a particle travels a semicircular arc of radius $r=10\ \text{m}$. Path length $=\pi r \approx 31.4\ \text{m}$. But start and end points are diametrically opposite, so displacement magnitude $= 2r = 20\ \text{m}$. Since $31.4 \ne 20$, the statement fails for this explicit case (and in fact $\pi r > 2r$ for any $r>0$, so it fails for every semicircular path, not just this one).

Verdict: False.

(d) Average speed $\ge$ magnitude of average velocity

Reuse the same semicircular example: time taken, say, $\Delta t = 5\ \text{s}$. Average speed $=31.4/5=6.28\ \text{m/s}$; magnitude of average velocity $=20/5=4\ \text{m/s}$. Indeed $6.28 > 4$, consistent with the claim. For a straight-line path (no reversal), e.g. $10\ \text{m}$ east in $5\ \text{s}$: path length $=10\ \text{m}=$ displacement magnitude, so average speed $=2\ \text{m/s}=$ magnitude of average velocity exactly -- equality, not violation. Both the curved and straight explicit cases are consistent with "$\ge$", never "$<$", because path length $\ge$ |displacement| always (dividing by the same positive $\Delta t$ preserves this). …