NCERT Solutions · Class 11 Physics Motion in a Plane

A scalar needs only a magnitude and a unit to be fully specified; a vector needs a magnitude AND a direction — leave out the direction and a vector quantity is not correctly stated.Apply that test to each quantity in the list.Volume — a number in \(\displaystyle \text{m}^3 \) (or litres) says everything there is to say. No direction attaches to "how much space it occupies." Scalar.Mass — a number in kg. Mass is an intrinsic property of a body and has no direction. Scalar. (Aside: mass is not the same as weight — weight is the force \(\displaystyle mg \), and force is a vector. Mass itself never carries a direction.)Speed — the magnitude of velocity, \(\displaystyle |\vec v| \), stated in \(\displaystyle \text{m/s} \). It tells you how fast, not which way. Scalar. (Aside: this is the one to watch — speed and velocity are often confused. Speed is what a speedometer reads; velocity additionally needs "in what direction.")Acceleration — defined as \(\displaystyle \vec a = \dfrac{d\vec v}{dt} \), the rate of change of the velocity vector. Since it is built directly from a vector by differentiation, it has both a magnitude (m/s²) and a direction (the direction in which velocity is changing). Vector.Density — mass divided by volume, \(\displaystyle \rho = m/V \), in \(\displaystyle \text{kg/m}^3 \). Both mass and volume are scalars, and dividing one scalar by another gives a scalar. Scalar.Number of moles — a pure count of \(\displaystyle 6.022\times10^{23} \)-sized groups of particles. A count has no direction. Scalar.Velocity — rate of change of position, \(\displaystyle \vec v = \dfrac{d\vec r}{dt} \), in m/s, with a direction (the direction of motion at that instant). Vector.Angular frequency — defined as \(\displaystyle \omega = \dfrac{2\pi}{T} = 2\pi f \), in rad/s. This \(\displaystyle \omega \) is just a number telling you how rapidly a phase angle is advancing (used for oscillations and SHM) — it is not associated with an axis or a sense of rotation. Scalar.Displacement — the straight-line change in position, \(\displaystyle \vec{\Delta r} = \vec r_2 - \vec r_1 \), in metres, pointing from the initial position to the final position. Vector.Angular velocity — this is the one most often confused with angular frequency. Angular velocity \(\displaystyle \vec\omega \) describes rotation of a rigid body: its magnitude is the rate of change of the angle turned through (rad/s), and its direction is along the axis of rotation, fixed by the right-hand rule (curl the fingers in the sense of rotation, the thumb gives the direction of \(\displaystyle \vec\omega \)). Because it needs an axis/direction to be meaningful, it is a Vector, even though numerically \(\displaystyle |\vec\omega| \) can equal \(\displaystyle \omega \) for a body rotating steadily about a fixed axis.Answer: Scalars — volume, mass, speed, density, number of moles, angular frequency. Vectors — acceleration, velocity, displacement, angular velocity.

A quantity is a scalar only if it has magnitude alone and combines with others by ordinary algebra — not by the vector law of addition (the parallelogram/triangle rule). Having "no direction" is the surface test; the real test is how two of the quantity combine when you add them.Go through the list one by one.Force — has magnitude and direction, and two forces add by the parallelogram law (their resultant depends on the angle between them). Vector.Angular momentum — defined as \(\displaystyle \vec{L} = \vec{r} \times \vec{p} \), a cross product of two vectors. A cross product always produces a vector (it has both a magnitude and a direction, perpendicular to the plane of \(\displaystyle \vec{r} \) and \(\displaystyle \vec{p} \)). Vector.Work — defined as \(\displaystyle W = \vec{F} \cdot \vec{d} \), the dot product of the force and the displacement. A dot product multiplies two vectors down to a single number with no direction attached — the result is just so many joules, however the force and displacement happened to point. Scalar.Current — this is the one that catches people out. Current has a "direction" in the everyday sense (current flows along a wire), so it looks like it should be a vector. But direction alone does not make something a vector — the deciding test is whether it adds by the vector law. Take two wires meeting at a junction, carrying currents \(\displaystyle I_1\) and \(\displaystyle I_2\) into the junction: the current leaving the junction is simply \(\displaystyle I_1+I_2\), added algebraically (Kirchhoff's junction rule), never by a parallelogram construction depending on the angle between the wires. Because current fails the vector-addition test, it is classified as a scalar even though it has an associated sense of flow.Linear momentum — \(\displaystyle \vec{p} = m\vec{v} \), a scalar (\(\displaystyle m\)) times a vector (\(\displaystyle \vec{v}\)); the result keeps the direction of \(\displaystyle \vec{v}\) and adds by the parallelogram law. Vector.Electric field — defined as force per unit charge, \(\displaystyle \vec{E} = \vec{F}/q \); it points in the direction a positive test charge would be pushed, and two fields superpose by vector addition. Vector.Average velocity — \(\displaystyle \vec{v}_{avg} = \Delta\vec{r}/\Delta t \), displacement (a vector) divided by a scalar time interval; it points along the net displacement. Vector.Magnetic moment — \(\displaystyle \vec{m} = I\vec{A} \), current times the vector area of the current loop, direction given by the right-hand rule. Vector.Relative velocity — \(\displaystyle \vec{v}_{AB} = \vec{v}_A - \vec{v}_B \), a difference of two vectors, itself a vector with a definite direction. Vector.So sorting the nine quantities by this test, exactly two survive as scalars: work and current — one because it is built from a dot product, the other because it fails the vector-addition law despite having a "direction" of flow. Every other quantity in the list is a cross product, a scalar multiple of a vector, or a straightforward vector difference/quotient, and stays a vector.Answer: Work and current are the two scalar quantities; the rest — force, angular momentum, linear momentum, electric field, average velocity, magnetic moment, and relative velocity — are all vectors.

Impulse is the only vector in the list — it inherits its direction from the force that produces it.The way to sort this list is to ask, for each quantity, whether it needs a direction to be fully specified, or whether a single number (with a unit) says everything about it.Impulse is defined as \[\vec{J} = \vec{F} \, \Delta t \] where \(\displaystyle \vec{F}\) is the (possibly varying) force acting on a body and \(\displaystyle \Delta t\) is the time for which it acts. Since \(\displaystyle \vec{F}\) is a vector, and multiplying a vector by the scalar \(\displaystyle \Delta t\) only scales its length without touching its direction, \(\displaystyle \vec{J}\) points along the same direction as the force. It is also equal to the change in momentum, \(\displaystyle \vec{J} = \Delta \vec{p} = m\vec{v}_2 - m\vec{v}_1\), which is again a vector difference. So impulse genuinely needs a direction — "an impulse of $\displaystyle 5$ N s" is an incomplete statement until you also say which way it acts.Now check that everything else is a scalar — each is fully described by a magnitude and a unit alone, with no direction attached: Since none of these nine carry direction as part of their definition, while impulse cannot be stated correctly without one, impulse stands out as the vector quantity.Answer: Impulse is the only vector quantity in the list.

Two quantities can be added only if they are the same kind of object and the same physical quantity — matching units is never enough. Multiplication carries no such restriction at all.Everything in this question follows from two rules, so state them first.Rule $\displaystyle 1$ — addition. A sum \(\displaystyle P+Q\) is meaningful only if \(\displaystyle P\) and \(\displaystyle Q\) are both scalars or both vectors, and they represent the same physical quantity. Adding a mass of \(\displaystyle 2.0\ \text{kg}\) to a time of \(\displaystyle 3.0\ \text{s}\) produces nothing; adding \(\displaystyle 2.0\ \text{kg}\) to \(\displaystyle 3.0\ \text{kg}\) produces \(\displaystyle 5.0\ \text{kg}\).Rule $\displaystyle 2$ — multiplication. A product \(\displaystyle PQ\) is always meaningful. The result is simply a new physical quantity whose unit is the product of the two units, e.g. \(\displaystyle (\text{kg})(\text{m s}^{-1}) = \text{kg m s}^{-1}\), which is momentum.Now take the six operations one at a time. Note that (a), (c), (d) and (e) are asked about any two quantities — so a single counterexample is enough to reject the general claim.(a) Adding any two scalars — not meaningful in general. Counterexample: let \(\displaystyle m = 2.0\ \text{kg}\) and \(\displaystyle t = 3.0\ \text{s}\). Both are scalars, but \(\displaystyle m+t\) has no physical meaning, because there is no quantity that is part mass and part time. Matching dimensions does not rescue it either: work and the magnitude of a torque both have dimensions \(\displaystyle \text{M L}^2\text{T}^{-2}\) and both are measured in \(\displaystyle \text{N m}\), yet adding them is meaningless — one is energy transferred, the other is a turning effect. The operation is meaningful in the special case where the two scalars are the same physical quantity (two masses, two time intervals, two energies). As a statement about any two scalars, it fails.(b) Adding a scalar to a vector of the same dimensions — never meaningful. A vector is specified by a magnitude and a direction; a scalar has no direction. There is no operation \(\displaystyle \mathbf{v}+s\) defined, whatever \(\displaystyle s\) is. Concretely, a speed of \(\displaystyle 5\ \text{m s}^{-1}\) and a velocity of \(\displaystyle 5\ \text{m s}^{-1}\) due east cannot be added, even though both are in \(\displaystyle \text{m s}^{-1}\).The trap here is the phrase "of the same dimensions." It is put in deliberately to tempt you. Same dimensions is a necessary condition for addition, never a sufficient one — the two quantities must first be the same kind of object.(c) Multiplying any vector by any scalar — always meaningful. By Rule $\displaystyle 2$, for a scalar \(\displaystyle \lambda\) and a vector \(\displaystyle \mathbf{A}\), the product \(\displaystyle \lambda\mathbf{A}\) is a vector of magnitude \(\displaystyle |\lambda|A\), pointing along \(\displaystyle \mathbf{A}\) if \(\displaystyle \lambda>0\) and opposite to \(\displaystyle \mathbf{A}\) if \(\displaystyle \lambda<0\). Physical example: mass \(\displaystyle \times\) velocity gives momentum, \[\mathbf{p} = m\mathbf{v} = (2.0\ \text{kg})(3.0\ \text{m s}^{-1}\ \text{east}) = 6.0\ \text{kg m s}^{-1}\ \text{east}.\] This is permissible.(d) Multiplying any two scalars — always meaningful. Again by Rule 2. Physical example: speed \(\displaystyle \times\) time gives distance, \[d = v\,t = (3.0\ \text{m s}^{-1})(4.0\ \text{s}) = 12\ \text{m}.\] No restriction on what the two scalars represent. This is permissible.(e) Adding any two vectors — not meaningful in general. Counterexample: a force of \(\displaystyle 10\ \text{N}\) north and a velocity of \(\displaystyle 10\ \text{m s}^{-1}\) north are both vectors, and the parallelogram law would happily give you an arrow — but that arrow stands for nothing. As with (a), the operation is meaningful only when the two vectors are the same physical quantity (two forces, two displacements, two velocities), in which case the triangle or parallelogram law applies. As a statement about any two vectors, it fails.(f) Adding a component of a vector to the same vector — not meaningful. This one turns entirely on what "a component" is. Resolving a vector in the plane gives \[\mathbf{A} = A_x\hat{\imath} + A_y\hat{\jmath},\qquad A_x = A\cos\theta,\quad A_y = A\sin\theta,\] and the quantities called the x- and y-components of \(\displaystyle \mathbf{A}\) are the numbers \(\displaystyle A_x\) and \(\displaystyle A_y\). The number \(\displaystyle A_x\) is not itself a vector; \(\displaystyle A_x\hat{\imath}\) is, and that object has its own name — the component vector. So "a component of \(\displaystyle \mathbf{A}\)" means \(\displaystyle A_x\), a scalar. Then \[\mathbf{A} + A_x\] is a scalar added to a vector, which is exactly case (b) — and so it is not meaningful.This is the step almost everyone gets wrong, so be clear about which object you are holding. If you read "component" as the component vector \(\displaystyle A_x\hat{\imath}\), the sum is fine: \[\mathbf{A} + A_x\hat{\imath} = \left(A_x\hat{\imath} + A_y\hat{\jmath}\right) + A_x\hat{\imath} = 2A_x\hat{\imath} + A_y\hat{\jmath},\] a perfectly good vector of the same physical kind. That is why "yes" is such a common answer to (f). But a component and a component vector are two different objects — the first is a number, the second is that number times a unit vector — and the operation in (f) is written with the number. Adding a number to a vector is not defined.So of the six operations, exactly two are meaningful for any quantities of the stated kind: (c) and (d). Both are multiplications, which is the pattern worth carrying away — multiplication never asks what the quantities mean, addition always does.Answer: Only (c) and (d) are permissible. (c) multiplying any vector by any scalar gives a vector, and (d) multiplying any two scalars gives a scalar, both without restriction. (a) and (e) are meaningful only in the special case that the two scalars, or the two vectors, represent the same physical quantity — not for any two, as asked. (b) is never meaningful, since a scalar has no direction and equal dimensions do not make two quantities addable. (f) is not meaningful either, because a "component" of a vector is the number \(\displaystyle A_x\), not the component vector \(\displaystyle A_x\hat{\imath}\), so it reduces to case (b).

A vector's magnitude, its components, and the distance travelled are three different things — mixing them up is exactly what these five statements are testing.(a) True. The magnitude of a vector \(\displaystyle \vec{A} \), written \(\displaystyle |\vec{A}| \), is a single non‑negative real number — it tells you how big the vector is and carries no information about direction. A quantity that is completely described by one number with a unit, and has no direction, is by definition a scalar. So the magnitude of any vector is always a scalar.(b) False. Resolve a vector along the coordinate axes: \[\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k} \] Each piece on the right, such as \(\displaystyle A_x\hat{i} \), is a vector in its own right — it has a magnitude \(\displaystyle A_x \) and a direction, \(\displaystyle \hat{i} \). That is the actual "component of the vector along the x-axis." Only the number \(\displaystyle A_x \) sitting in front of \(\displaystyle \hat{i} \) (the scalar component) is a scalar; the component itself is not. Since the statement claims every component is always a scalar, and a component is in general a vector, the statement is false.(c) False. Path length is the total length of the actual track a particle follows; displacement is the straight-line vector from the starting point to the ending point. For any real trajectory, the straight line between two points is never longer than the path connecting them, so \[\text{path length} \ge |\vec{\Delta r}| \] The two are equal only in the special case of motion along a fixed straight line without ever reversing direction. As soon as the path curves or doubles back — the extreme case being a particle that goes around a circle and returns to its start — the path length is strictly greater than the displacement magnitude (a full circle has path length \(\displaystyle 2\pi r \) but displacement zero). Because the statement claims equality always holds, it is false.(d) True. Average speed and average velocity are both built by dividing by the same time interval \(\displaystyle t \): \[\text{average speed} = \frac{\text{path length}}{t}, \qquad |\vec{v}_{avg}| = \frac{|\vec{\Delta r}|}{t} \] From part (c), path length \(\displaystyle \ge |\vec{\Delta r}| \) always. Dividing both sides of that inequality by the same positive number \(\displaystyle t \) does not change its direction, so \[\frac{\text{path length}}{t} \ge \frac{|\vec{\Delta r}|}{t} \quad\Rightarrow\quad \text{average speed} \ge |\vec{v}_{avg}| \] Equality is reached only for unidirectional straight-line motion (the case where part (c)'s equality holds). So average speed is always greater than or equal to the magnitude of the average velocity — the statement is true.(e) True. Suppose three vectors \(\displaystyle \vec{A}, \vec{B}, \vec{C} \) — assumed not to lie in one plane — added to a null vector: \[\vec{A} + \vec{B} + \vec{C} = \vec{0} \quad\Rightarrow\quad \vec{C} = -(\vec{A} + \vec{B}) \] The sum \(\displaystyle \vec{A} + \vec{B} \) always lies in the plane containing \(\displaystyle \vec{A} \) and \(\displaystyle \vec{B} \) (that is just the parallelogram law — the resultant of two vectors lies in the plane defined by those two vectors). So \(\displaystyle \vec{C} \), being equal to \(\displaystyle -(\vec{A}+\vec{B}) \), would have to lie in that very same plane as \(\displaystyle \vec{A} \) and \(\displaystyle \vec{B} \) — making all three vectors coplanar after all. That directly contradicts the starting assumption that they do not lie in a plane. So three vectors that are genuinely non-coplanar can never add up to give a null vector; a zero resultant is only possible for vectors that can be arranged as the closed sides of a triangle, which forces them into a single plane.Answer: (a) True (b) False (c) False (d) True (e) True

Every one of these follows from one picture: draw \(\displaystyle \vec a \) and \(\displaystyle \vec b \) as two sides of a triangle, and the third side is forced to be no bigger than their sum and no smaller than their difference.Setting up the triangle. By the triangle law of vector addition, draw \(\displaystyle \overrightarrow{OP} = \vec a \) and then \(\displaystyle \overrightarrow{PQ} = \vec b \) starting from the tip of \(\displaystyle \vec a \). The vector that closes the triangle, \(\displaystyle \overrightarrow{OQ} \), represents \(\displaystyle \vec a + \vec b \). So \(\displaystyle O, P, Q \) are the vertices of a triangle whose three sides have lengths \(\displaystyle |\vec a| \), \(\displaystyle |\vec b| \), and \(\displaystyle |\vec a+\vec b| \) — unless the three points are collinear, in which case the "triangle" flattens onto a line. This single triangle is what proves all four inequalities.(a) \(\displaystyle |\vec a+\vec b| \le |\vec a|+|\vec b| \)In triangle \(\displaystyle OPQ \), the side \(\displaystyle OQ \) cannot exceed the sum of the other two sides: \[OQ \le OP + PQ \quad\Longrightarrow\quad |\vec a+\vec b| \le |\vec a|+|\vec b|. \] This is just the geometric fact that a straight path between two points is never longer than a path via a third point — going straight from \(\displaystyle O \) to \(\displaystyle Q \) is at most as long as going \(\displaystyle O\to P \to Q \).The same result drops out algebraically, which is the cleanest way to pin down when equality holds. Using \(\displaystyle |\vec v|^2 = \vec v \cdot \vec v \) and letting \(\displaystyle \theta \) be the angle between \(\displaystyle \vec a \) and \(\displaystyle \vec b \): \[|\vec a+\vec b|^2 = (\vec a+\vec b)\cdot(\vec a+\vec b) = |\vec a|^2 + |\vec b|^2 + 2\,\vec a\cdot\vec b = |\vec a|^2+|\vec b|^2+2|\vec a||\vec b|\cos\theta . \] Since \(\displaystyle \cos\theta \le 1 \) always, \[|\vec a+\vec b|^2 \le |\vec a|^2+|\vec b|^2+2|\vec a||\vec b| = (|\vec a|+|\vec b|)^2, \] and taking the (non-negative) square root gives \(\displaystyle |\vec a+\vec b| \le |\vec a|+|\vec b| \).Equality needs \(\displaystyle \cos\theta = 1 \), i.e. \(\displaystyle \theta = 0 \): \(\displaystyle \vec a \) and \(\displaystyle \vec b \) point in the same direction. Geometrically the triangle has flattened with \(\displaystyle P \) lying between \(\displaystyle O \) and \(\displaystyle Q \), so the "sum of two sides" literally becomes the third side.(b) \(\displaystyle |\vec a+\vec b| \ge \big||\vec a|-|\vec b|\big| \)Apply the same triangle-side rule the other way around. In triangle \(\displaystyle OPQ \): \[OP \le OQ + QP = OQ + PQ \quad\Longrightarrow\quad |\vec a| - |\vec b| \le |\vec a+\vec b|, \] \[PQ \le PO + OQ \quad\Longrightarrow\quad |\vec b| - |\vec a| \le |\vec a+\vec b|. \] Together these say \(\displaystyle |\vec a+\vec b| \) is at least as large as both \(\displaystyle |\vec a|-|\vec b| \) and \(\displaystyle |\vec b|-|\vec a| \), i.e. at least as large as whichever is positive: \[|\vec a+\vec b| \ge \big||\vec a|-|\vec b|\big|. \] Algebraically, since \(\displaystyle \cos\theta \ge -1 \), \[|\vec a+\vec b|^2 = |\vec a|^2+|\vec b|^2+2|\vec a||\vec b|\cos\theta \ge |\vec a|^2+|\vec b|^2-2|\vec a||\vec b| = (|\vec a|-|\vec b|)^2, \] so \(\displaystyle |\vec a+\vec b| \ge \big||\vec a|-|\vec b|\big| \).Equality needs \(\displaystyle \cos\theta=-1 \), i.e. \(\displaystyle \theta = 180^\circ \): \(\displaystyle \vec a \) and \(\displaystyle \vec b \) point in opposite directions (the triangle again flattens, but now \(\displaystyle \vec a \) and \(\displaystyle \vec b \) overlap back over each other instead of extending each other).(c) \(\displaystyle |\vec a-\vec b| \le |\vec a|+|\vec b| \)Do not re-draw a new triangle — reuse part (a). Let \(\displaystyle \vec c = -\vec b \), so \(\displaystyle |\vec c| = |\vec b| \) (reversing a vector doesn't change its length, only its direction). Then \(\displaystyle \vec a - \vec b = \vec a + \vec c \), and part (a) applied to \(\displaystyle \vec a \) and \(\displaystyle \vec c \) gives \[|\vec a+\vec c| \le |\vec a|+|\vec c| \quad\Longrightarrow\quad |\vec a-\vec b| \le |\vec a|+|\vec b|. \]Equality holds when \(\displaystyle \vec a \) and \(\displaystyle \vec c=-\vec b \) point in the same direction — that means \(\displaystyle \vec a \) and \(\displaystyle \vec b \) themselves point in opposite directions ( \(\displaystyle \theta = 180^\circ \) between \(\displaystyle \vec a \) and \(\displaystyle \vec b \) ). This makes sense: subtracting a vector that already points backward is the same as adding two vectors that point the same way, so their lengths add up fully.(d) \(\displaystyle |\vec a-\vec b| \ge \big||\vec a|-|\vec b|\big| \)Same substitution, now using part (b). With \(\displaystyle \vec c=-\vec b \) and \(\displaystyle |\vec c|=|\vec b|\), part (b) gives \[|\vec a+\vec c| \ge \big||\vec a|-|\vec c|\big| \quad\Longrightarrow\quad |\vec a-\vec b| \ge \big||\vec a|-|\vec b|\big|. \]Equality holds when \(\displaystyle \vec a \) and \(\displaystyle \vec c = -\vec b \) point in opposite directions, which means \(\displaystyle \vec a \) and \(\displaystyle \vec b \) point in the same direction ( \(\displaystyle \theta = 0^\circ \) ).Where people slip. The direction condition flips between the "+" and "−" pairs: for \(\displaystyle |\vec a+\vec b| \), the sum is largest ( \(\displaystyle =|\vec a|+|\vec b|\) ) when the vectors point together, and for \(\displaystyle |\vec a-\vec b| \) it's the opposite pairing that maximizes the length, because subtracting a backward-pointing vector behaves like adding a forward-pointing one. Also, none of these ever become strict "less than only" statements — the problem's inequality signs must be read as \(\displaystyle \le \) and \(\displaystyle \ge \), since equality is genuinely attainable (and is asked about explicitly), including the trivial case where either \(\displaystyle \vec a \) or \(\displaystyle \vec b \) is the zero vector, which satisfies every equality at once.Answer: All four hold as \(\displaystyle \le \) / \(\displaystyle \ge \) inequalities (not strict), proved via the triangle law of vector addition (or equivalently by expanding \(\displaystyle |\vec a\pm\vec b|^2 \) using the dot product). Equality signs apply as follows: (a) \(\displaystyle |\vec a+\vec b|=|\vec a|+|\vec b| \) when \(\displaystyle \vec a \) and \(\displaystyle \vec b \) are parallel, same direction ( \(\displaystyle \theta=0^\circ \) ); (b) \(\displaystyle |\vec a+\vec b|=\big||\vec a|-|\vec b|\big| \) when \(\displaystyle \vec a \) and \(\displaystyle \vec b \) are antiparallel ( \(\displaystyle \theta=180^\circ \) ); (c) \(\displaystyle |\vec a-\vec b|=|\vec a|+|\vec b| \) when \(\displaystyle \vec a \) and \(\displaystyle \vec b \) are antiparallel ( \(\displaystyle \theta=180^\circ \) ); (d) \(\displaystyle |\vec a-\vec b|=\big||\vec a|-|\vec b|\big| \) when \(\displaystyle \vec a \) and \(\displaystyle \vec b \) are parallel, same direction ( \(\displaystyle \theta=0^\circ \) ).

A vector sum being zero says nothing about the individual vectors — it only ties them together as a closed figure. Treat each statement on its own using \(\displaystyle \mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=\mathbf{0} \).(a) False. This equation says the four vectors, placed head to tail, close up into a figure with no gap — it does not say each one has zero length. Four non-zero vectors can easily sum to zero: take \(\displaystyle \mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d} \) as the four sides of a quadrilateral, each traversed the same way around (say, counterclockwise). Walking all four sides brings you back to the start, so their vector sum is \(\displaystyle \mathbf{0} \), yet none of the sides has zero length. So (a) is incorrect.(b) True. Rearranging the given relation, \[\mathbf{a}+\mathbf{c} = -(\mathbf{b}+\mathbf{d}) \] Taking magnitudes of both sides, and using the fact that reversing a vector's direction does not change its length, \(\displaystyle |-\mathbf{x}| = |\mathbf{x}| \), \[|\mathbf{a}+\mathbf{c}| = |-(\mathbf{b}+\mathbf{d})| = |\mathbf{b}+\mathbf{d}| \] So the magnitude of \(\displaystyle (\mathbf{a}+\mathbf{c}) \) does equal the magnitude of \(\displaystyle (\mathbf{b}+\mathbf{d}) \). This statement is correct.(c) True. From the given relation, \[\mathbf{a} = -(\mathbf{b}+\mathbf{c}+\mathbf{d}) \] so \[|\mathbf{a}| = |\mathbf{b}+\mathbf{c}+\mathbf{d}| \] Now use the triangle inequality (repeated for three vectors): the length of a vector sum can never exceed the sum of the individual lengths, because the straight-line path between two points is never longer than a path that goes via detours. That is, \[|\mathbf{b}+\mathbf{c}+\mathbf{d}| \le |\mathbf{b}| + |\mathbf{c}| + |\mathbf{d}| \] Combining the two lines, \[|\mathbf{a}| \le |\mathbf{b}| + |\mathbf{c}| + |\mathbf{d}| \] So \(\displaystyle |\mathbf{a}| \) can never be greater than \(\displaystyle |\mathbf{b}|+|\mathbf{c}|+|\mathbf{d}| \) — equality holds only when \(\displaystyle \mathbf{b}, \mathbf{c}, \mathbf{d} \) all point the same way. This statement is correct.(d) True. From the given relation, \[\mathbf{b}+\mathbf{c} = -(\mathbf{a}+\mathbf{d}) \] The right-hand side is built purely out of \(\displaystyle \mathbf{a} \) and \(\displaystyle \mathbf{d} \) — it is just \(\displaystyle -1 \) times their sum, with no other vector entering. A key idea about combining vectors: any sum (or scalar multiple of a sum) formed only from two given vectors must lie in the geometric space those two vectors span. This statement is correct.The one place students slip is treating \(\displaystyle \mathbf{a}+\mathbf{b}+\mathbf{c}+\mathbf{d}=\mathbf{0} \) as if it meant "each vector is zero" — a sum vanishing only means the vectors close a loop, never that each piece is trivial. That misreading is exactly what makes (a) tempting and wrong.Answer: (a) is incorrect; (b), (c), and (d) are correct.

Displacement depends only on where you start and where you end, never on the road you took to get there — so it is the same vector for all three girls, however oddly each one curves around the rink.P and Q lie on a circle of radius \(\displaystyle r = 200 \text{ m} \), and Q is diametrically opposite P. That means the straight segment joining P and Q passes through the centre of the circle — it is a diameter of the circle.Magnitude of the displacementThe displacement vector \(\displaystyle \vec{d} \) is fixed purely by the position vectors of the start and end points, \[\vec{d} = \vec{r}_Q - \vec{r}_P, \] and this expression does not know or care what curve a skater traced between P and Q — only the two endpoints matter. Since P and Q are diametrically opposite, the straight line joining them is a diameter of the ground: \[d = 2r = 2(200 \text{ m}) = 400 \text{ m}. \]This value, \(\displaystyle 400 \text{ m} \), is the magnitude of the displacement for every one of the three girls, no matter which path each one skated — one may have hugged the rim, another zig-zagged, a third cut straight across. Displacement belongs to the pair of endpoints, not to the trip between them.Direction of the displacementEvery girl's displacement vector also points the same way: straight from P to Q, along the diameter joining them. There is exactly one straight line through two given points, so "from P to Q" already fixes the direction completely — it is independent of which curved route was actually skated.Which girl's path length equals her displacement?For any route between two fixed points, the straight-line segment joining them is the shortest possible path — any bulge, detour, or curve can only add extra length, never remove it. So for each skater, \[\text{path length skated} \;\geq\; |\vec{d}| = 400 \text{ m}, \] with equality holding only when the path itself coincides with the straight segment PQ.Two of the girls follow bent or curved routes (for instance, along an arc of the rim, or some other wandering path), so their path length is strictly greater than $\displaystyle 400$ m — a girl who skated the semicircular rim, say, would cover \(\displaystyle \pi r = \pi(200 \text{ m}) \approx 628 \text{ m} \), well above her $\displaystyle 400$ m displacement.Only the girl who skates in a straight line directly across the ice, along PQ itself, has an actual path length equal to $\displaystyle 400$ m. For her alone, the path length and the displacement magnitude coincide, because her path is the displacement vector — there is no detour whose extra length needs to be discarded.Answer: The displacement magnitude is $\displaystyle 400$ m for all three girls, directed from P to Q along the diameter joining them; this equals the actual path length only for the girl who skates straight across that diameter — the girls on curved paths cover more than $\displaystyle 400$ m.

Net displacement is a straight-line vector from the starting point to the finishing point — and this cyclist ends up right back where he started, at the centre O.The path is: centre \(\displaystyle O \to \) edge \(\displaystyle P \) (along a radius), then \(\displaystyle P \to Q \) along the rim of the park (a quarter of the circle, since \(\displaystyle OP \) and \(\displaystyle OQ \) are perpendicular radii in Fig. $\displaystyle 3.20$), then \(\displaystyle Q \to O \) straight back along the radius.(a) Net displacementDisplacement is defined as the vector from the initial position to the final position, regardless of the path taken in between. Here the initial position is \(\displaystyle O \) and, after the full trip \(\displaystyle O \to P \to Q \to O \), the final position is also \(\displaystyle O \).\[\vec{s} = \vec{r}_{\text{final}} - \vec{r}_{\text{initial}} = \vec{r}_O - \vec{r}_O = 0 \]Net displacement \(\displaystyle = 0 \). This is the step people get wrong: the cyclist has certainly moved — he has covered real distance — but displacement only cares about the net shift in position, and here there is none because the trip is a closed loop.(b) Average velocityAverage velocity is net displacement divided by total time:\[\vec{v}_{avg} = \frac{\vec{s}}{\Delta t} \]Since \(\displaystyle \vec{s} = 0 \) and \(\displaystyle \Delta t = 10\ \text{min} \) (a finite, nonzero time), the average velocity is\[\vec{v}_{avg} = \frac{0}{10\ \text{min}} = 0 \]Average velocity \(\displaystyle = 0 \). Velocity is displacement-based, so a trip that returns to its start always averages to zero velocity, no matter how far or how fast the cyclist actually pedalled.(c) Average speedAverage speed uses the total path length travelled, not displacement:\[v_{avg} = \frac{\text{total distance}}{\text{total time}} \]Total distance = (straight run \(\displaystyle O \to P \)) + (arc \(\displaystyle P \to Q \)) + (straight run \(\displaystyle Q \to O \)), with radius \(\displaystyle r = 1\ \text{km} \). \[PQ = \frac{1}{4}(2\pi r) = \frac{\pi r}{2} = \frac{\pi (1\ \text{km})}{2} = 1.5708\ \text{km} \] Total distance: \[d = OP + PQ + QO = 1\ \text{km} + 1.5708\ \text{km} + 1\ \text{km} = 3.5708\ \text{km} \]Total time, converted to hours so the units match the km distance: \[\Delta t = 10\ \text{min} = \frac{10}{60}\ \text{h} = \frac{1}{6}\ \text{h} = 0.16667\ \text{h} \]Substituting: \[v_{avg} = \frac{3.5708\ \text{km}}{0.16667\ \text{h}} = 21.42\ \text{km/h} \]The radius ($\displaystyle 1$ km) and time ($\displaystyle 10$ min) are each given to about two significant figures once \(\displaystyle \pi \) is treated as exact, so the answer is stated to three significant figures: \(\displaystyle v_{avg} \approx 21.4\ \text{km/h} \). In SI units, that is \[21.4\ \text{km/h} \times \frac{1000\ \text{m}}{3600\ \text{s}} = 5.95\ \text{m/s} \]Answer: (a) Net displacement = $\displaystyle 0$; (b) Average velocity = $\displaystyle 0$; (c) Average speed ≈ $\displaystyle 21.4$ km/h (≈ $\displaystyle 5.95$ m/s).

Displacement is the straight-line vector from the start to wherever the motorist ends up; path length is the actual distance ground he covers along the twisting track. These two numbers can differ enormously once the track loops back on itself — that is exactly what this problem is built to show.Setting up the geometry. The motorist drives $\displaystyle 500$ m, turns $\displaystyle 60$° to his left, drives another $\displaystyle 500$ m, turns $\displaystyle 60$° left again, and so on. A left turn of $\displaystyle 60$° at the end of every straight stretch means the exterior angle of his path is $\displaystyle 60$° at every vertex. A regular polygon whose exterior angle is $\displaystyle 60$° has\[n = \frac{360^{0}}{60^{0}} = 6 \text{ sides,} \]so the track is the perimeter of a regular hexagon of side $\displaystyle 500$ m. After $\displaystyle 6$ turns the motorist is back at his starting point — this is the fact that makes turns $\displaystyle 3$, $\displaystyle 6$ and $\displaystyle 8$ easy to handle without redrawing the whole path each time.Take the direction of the very first $\displaystyle 500$ m stretch as the reference direction (call it \(\displaystyle 0^{0}\)). Because each turn adds $\displaystyle 60$° to the heading, the direction of the \(\displaystyle k\)-th leg is \(\displaystyle (k-1)\times 60^{0}\) measured from that reference. Using this, the position after \(\displaystyle n\) legs (i.e., at the \(\displaystyle n\)-th turn) is the vector sum\[\vec{d}_n=\sum_{k=1}^{n} 500\,\text{m}\,\big(\cos[(k-1)60^{0}],\ \sin[(k-1)60^{0}]\big). \]At the third turn. The three legs point along \(\displaystyle 0^{0},60^{0},120^{0}\):\[x = 500(\cos0^{0}+\cos60^{0}+\cos120^{0}) = 500(1+0.5-0.5)\ \text{m} = 500\ \text{m} \] \[y = 500(\sin0^{0}+\sin60^{0}+\sin120^{0}) = 500(0+0.8660+0.8660)\ \text{m} = 866.0\ \text{m} \]Magnitude (Pythagoras, since \(\displaystyle x\) and \(\displaystyle y\) are perpendicular components):\[|\vec{d}_3| = \sqrt{(500\ \text{m})^{2}+(866.0\ \text{m})^{2}} = \sqrt{1.000\times10^{6}}\ \text{m} = 1000\ \text{m} \]Direction: \(\displaystyle \theta=\tan^{-1}\!\left(\dfrac{866.0}{500}\right)=60.0^{0}\) from the direction of the first $\displaystyle 500$ m stretch.This is not a coincidence of arithmetic — in a regular hexagon of side \(\displaystyle a\), the diagonal joining a vertex to the vertex three steps away (straight through the centre) is exactly \(\displaystyle 2a\). Here \(\displaystyle 2\times500\ \text{m}=1000\ \text{m}\), matching the calculation.Path length after $\displaystyle 3$ legs: \(\displaystyle s_3 = 3\times500\ \text{m} = 1500\ \text{m}\).\[\frac{|\vec{d}_3|}{s_3} = \frac{1000\ \text{m}}{1500\ \text{m}} = 0.667 \]At the sixth turn. Six legs complete the hexagon exactly, so the motorist is back where he began:\[|\vec{d}_6| = 0\ \text{m} \]There is no direction to quote for this one — a zero vector has no direction. Do not read this as "the motorist didn't travel": the odometer still shows the full path length,\[s_6 = 6\times500\ \text{m} = 3000\ \text{m}, \qquad \frac{|\vec{d}_6|}{s_6} = \frac{0}{3000\ \text{m}} = 0. \]This is the case that most sharply separates the scalar (path length, always adding up) from the vector (displacement, which can cancel to zero even after $\displaystyle 3000$ m of driving).At the eighth turn. The first six legs bring the motorist back to the start, so legs $\displaystyle 7$ and $\displaystyle 8$ are just a repeat of legs $\displaystyle 1$ and $\displaystyle 2$, run from the same starting point. That means\[\vec{d}_8 = \vec{d}_2 \]Legs $\displaystyle 1$ and $\displaystyle 2$ point along \(\displaystyle 0^{0}\) and \(\displaystyle 60^{0}\):\[x = 500(\cos0^{0}+\cos60^{0}) = 500(1+0.5)\ \text{m} = 750\ \text{m} \] \[y = 500(\sin0^{0}+\sin60^{0}) = 500(0+0.8660)\ \text{m} = 433.0\ \text{m} \]\[|\vec{d}_8| = \sqrt{(750\ \text{m})^{2}+(433.0\ \text{m})^{2}} = \sqrt{7.4999\times10^{5}}\ \text{m} = 866\ \text{m}\ \ (=500\sqrt3\ \text{m}) \]Direction: \(\displaystyle \theta=\tan^{-1}\!\left(\dfrac{433.0}{750}\right)=30.0^{0}\) from the direction of the first $\displaystyle 500$ m stretch.Path length after $\displaystyle 8$ legs: \(\displaystyle s_8 = 8\times500\ \text{m} = 4000\ \text{m}\).\[\frac{|\vec{d}_8|}{s_8} = \frac{866\ \text{m}}{4000\ \text{m}} = 0.217 \]Answer: At the 3rd turn, displacement = $\displaystyle 1000$ m at \(\displaystyle 60^{0}\) to the initial direction, path length = $\displaystyle 1500$ m, ratio = 0.667. At the 6th turn, displacement = $\displaystyle 0$ m (no direction, since it is the zero vector), path length = $\displaystyle 3000$ m, ratio = 0. At the 8th turn, displacement = $\displaystyle 866$ m (\(\displaystyle =500\sqrt3\) m) at \(\displaystyle 30^{0}\) to the initial direction, path length = $\displaystyle 4000$ m, ratio = 0.217.