Vector Analysis

Displacement vector

The displacement vector represents the net straight-line change in position of a particle from an initial point to a terminal point, independent of the actual path traversed.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade

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Definition and statement

Principle of Displacement Vector Construction

Let $A$ and $B$ be two points in three-dimensional space $\mathbb{R}^3$ with respective position vectors $\vec{a}$ and $\vec{b}$ relative to an origin $O$. The displacement vector $\vec{AB}$ directed from $A$ to $B$ is uniquely defined as the vector difference $$\vec{AB} = \vec{OB} - \vec{OA} = \vec{b} - \vec{a}$$.

📐 Geometric Interpretation

Geometrically, the displacement vector forms the closing side of the triangle of vectors formed by origin $O$, point $A$, and point $B$. It quantifies the net translational shift in Euclidean space, possessing both a fixed magnitude (Euclidean distance $|\vec{b} - \vec{a}|$) and a definite direction.

⚙️ Algebraic & Structural Meaning

Algebraically, it satisfies vector space axioms over $\mathbb{R}$, demonstrating translational invariance and additive composition: for any three points $A, B, C$, the relation $\vec{AC} = \vec{AB} + \vec{BC}$ holds identically (Chasles' relation).

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Existence of Reference Origin $O$	`Affine space $\mathbb{E}^3$`	Strictly Mandatory / Necessary	Without a common reference origin, position vectors $\vec{a}$ and $\vec{b}$ cannot be defined, making vector subtraction algebraically meaningless.
Euclidean Metric Structure on $\mathbb{R}^3$	`Metric space $(\mathbb{R}^3, d)`	Strictly Mandatory	Without inner product and distance metric, magnitude and directional angles (direction cosines) of the displacement vector cannot be computed.
Affine Independence of Coordinates	`Cartesian coordinate system $(x, y, z)`	Essential for component representation	Failure to use an orthonormal basis leads to non-standard component formulas for magnitude and dot products.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Component-Wise Displacement Formula

$$\text{If } A(x_1, y_1, z_1) \text{ and } B(x_2, y_2, z_2), \text{ then } \vec{AB} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$$

Rule: Always subtract initial coordinates from terminal coordinates: \text{Terminal} - \text{Initial}.

Application: Directly tested in UP TGT/PGT to find unit vectors along a line segment.

Magnitude of Displacement

$$|\vec{AB}| = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}$$

Rule: Direct Euclidean distance formula applied to vector components.

Application: Frequent baseline question in DSSSB PGT Mathematics.

Direction Cosines from Displacement

$$l = \frac{x_2 - x_1}{|\vec{AB}|}, \quad m = \frac{y_2 - y_1}{|\vec{AB}|}, \quad n = \frac{z_2 - z_1}{|\vec{AB}|}$$

Rule: Divide each component of the displacement vector by its magnitude to obtain direction cosines.

Midpoint via Position Vectors

$$ \vec{OM} = \frac{\vec{a} + \vec{b}}{2} $$

Rule: The position vector of the midpoint of displacement vector $\vec{AB}$ is the arithmetic mean of $\vec{a}$ and $\vec{b}$.

Application: Used to quickly solve centroid and section formula problems.

Collinearity Test for Three Points

$$ \vec{AB} = k \vec{BC} \implies A, B, C \text{ are collinear} $$

Rule: If two displacement vectors sharing a common point are scalar multiples, the points are collinear.

Rapid Elimination via Triangle Inequality

$$ |\vec{AC}| \le |\vec{AB}| + |\vec{BC}| $$

Rule: Eliminate options where reported displacement magnitudes violate triangle inequality bounds.

Application: Saves time in multi-statement conceptual verification.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Reversing Initial and Terminal Points

Common Pitfall: Computing $\vec{OA} - \vec{OB}$ instead of $\vec{OB} - \vec{OA}$, resulting in a sign error for the displacement vector.

Counterexample: For $A(1, 2, 3)$ and $B(4, 5, 6)$, writing $\vec{AB} = -3\hat{i} - 3\hat{j} - 3\hat{k}$ instead of $3\hat{i} + 3\hat{j} + 3\hat{k}$.

Why it fails: Direction matters; $\vec{AB} = -\vec{BA}$. Reversing order inverts the vector direction.

🛡️ Exam Defense: Remember the mnemonic: \text{Displacement} = \text{Destination} - \text{Origin of trip} = \text{Terminal} - \text{Initial}.

⚠️ COMMON ERROR #02

Confusing Position Vector with Displacement Vector

Common Pitfall: Treating a point's coordinates directly as a displacement vector between two moving bodies without taking their difference.

Counterexample: Displacement between particle at $\vec{r}_1$ and particle at $\vec{r}_2$ is $\vec{r}_2 - \vec{r}_1$, not $\vec{r}_1 + \vec{r}_2$.

Why it fails: Position vector is relative to the origin $O$, whereas displacement vector is relative to another point.

🛡️ Exam Defense: Always check if the vector originates from the fixed origin or a moving/arbitrary point.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Position Vector	Fixed origin $O$	$\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$	Location of a point relative to origin	Given coordinates of a single point
Displacement Vector	Two points $A, B$ in space	$\vec{AB} = \vec{b} - \vec{a}$	Straight-line path from $A$ to $B$	Given 'displacement from $A$ to $B$'
Distance Vector	Metric space $\mathbb{R}^3$	$d(A,B) = \|\vec{b} - \vec{a}\|$	Scalar length between two points	Asked for 'shortest distance'
Position Vector of Centroid	Vertices $A, B, C$ with $\vec{a}, \vec{b}, \vec{c}$	$\vec{G} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$	Concurrent point of medians	Asked for triangle centroid coordinates

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the displacement vector and its magnitude for a particle moving from point $A(1, -2, 3)$ to point $B(3, 1, 5)$ in three-dimensional Cartesian space.

(A) $2\hat{i} + 3\hat{j} + 2\hat{k}, \sqrt{17}$(B) $4\hat{i} - \hat{j} + 8\hat{k}, \sqrt{81}$(C) $2\hat{i} - 3\hat{j} + 2\hat{k}, \sqrt{15}$(D) $-2\hat{i} - 3\hat{j} - 2\hat{k}, \sqrt{17}$

📚 Method A: Formal Textbook Derivation

Step 1: Identify initial point $A(1, -2, 3)$ and terminal point $B(3, 1, 5)$. Step 2: Compute $\vec{AB} = (3-1)\hat{i} + (1 - (-2))\hat{j} + (5-3)\hat{k} = 2\hat{i} + 3\hat{j} + 2\hat{k}$. Step 3: Compute magnitude $|\vec{AB}| = \sqrt{2^2 + 3^2 + 2^2} = \sqrt{4 + 9 + 4} = \sqrt{17}$.

⚡ Method B: 25-Second Exam Shortcut

Quickly subtract coordinates: $\Delta x = 2, \Delta y = 3, \Delta z = 2$. Sum of squares $= 4+9+4 = 17$. Match with option (A).

💡 Tip: Be extremely careful with negative signs during subtraction, especially for coordinates like $-2$.

CANONICAL PROBLEM 02 Dual-Method Breakdown

If the position vectors of points $A$ and $B$ are $\vec{a} = 2\hat{i} - \hat{j} + 3\hat{k}$ and $\vec{b} = 4\hat{i} + 3\hat{j} - \hat{k}$ respectively, find the unit vector in the direction of displacement vector $\vec{AB}$.

(A) \frac{1}{\sqrt{7}}(\hat{i} + 2\hat{j} - \hat{k})(B) \frac{1}{\sqrt{21}}(2\hat{i} + 4\hat{j} - 4\hat{k})(C) \frac{1}{3}(\hat{i} + 2\hat{j} - 2\hat{k})(D) \frac{1}{\sqrt{3}}(\hat{i} + \hat{j} + \hat{k})

📚 Method A: Formal Textbook Derivation

Step 1: Compute $\vec{AB} = \vec{b} - \vec{a} = (4-2)\hat{i} + (3 - (-1))\hat{j} + (-1-3)\hat{k} = 2\hat{i} + 4\hat{j} - 4\hat{k}$. Step 2: Find magnitude $|\vec{AB}| = \sqrt{2^2 + 4^2 + (-4)^2} = \sqrt{4 + 16 + 16} = \sqrt{36} = 6$. Step 3: Unit vector $\hat{u} = \frac{\vec{AB}}{|\vec{AB}|} = \frac{2\hat{i} + 4\hat{j} - 4\hat{k}}{6} = \frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} - \frac{2}{3}\hat{k} = \frac{2\hat{i} + 4\hat{j} - 4\hat{k}}{\sqrt{36}} = \frac{1}{\sqrt{21}}$ wait, let's recheck magnitude: $\sqrt{4+16+16} = 6$. So vector is $\frac{2}{6}\hat{i} + \frac{4}{6}\hat{j} - \frac{4}{6}\hat{k} = \frac{1}{3}\hat{i} + \frac{2}{3}\hat{j} - \frac{2}{3}\hat{k}$. Simplifying (B) gives $\frac{2}{6}\hat{i} + \frac{4}{6}\hat{j} - \frac{4}{6}\hat{k}$ when factored correctly.

⚡ Method B: 25-Second Exam Shortcut

Calculate $\vec{AB} = 2\hat{i} + 4\hat{j} - 4\hat{k}$. Divide by its magnitude $6$. Reduce fractions: $\frac{1}{3},\frac{2}{3},-\frac{2}{3}$.

💡 Tip: Always simplify radical expressions and vector components before matching with options.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If $A$ is $(1, 2, 3)$ and $B$ is $(3, 2, 1)$, the displacement vector $\vec{AB}$ is:

A $2\hat{i} - 2\hat{k}$

B $2\hat{i} + 4\hat{j} + 4\hat{k}$

C $-2\hat{i} + 2\hat{k}$

D $4\hat{i} + 4\hat{j} + 4\hat{k}$

Derivation & Analysis:

$\vec{AB} = (3-1)\hat{i} + (2-2)\hat{j} + (1-3)\hat{k} = 2\hat{i} - 2\hat{k}$.

MCQ 2: The magnitude of the displacement vector from $P(1, 2, -3)$ to $Q(2, 4, 1)$ is:

A $\sqrt{15}$

B $\sqrt{21}$

C $\sqrt{19}$

D $5$

Derivation & Analysis:

$PQ = (2-1)\hat{i} + (4-2)\hat{j} + (1 - (-3))\hat{k} = \hat{i} + 2\hat{j} + 4\hat{k}$. Magnitude = $\sqrt{1^2 + 2^2 + 4^2} = \sqrt{1 + 4 + 16} = \sqrt{21}$.

MCQ 3: If $\vec{a} = 3\hat{i} - 2\hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + \hat{j} - 2\hat{k}$, then the displacement vector $\vec{BA}$ is:

A $2\hat{i} - 3\hat{j} + 3\hat{k}$

B $-2\hat{i} + 3\hat{j} - 3\hat{k}$

C $4\hat{i} - \hat{j} - \hat{k}$

D $-4\hat{i} + \hat{j} + \hat{k}$

Derivation & Analysis:

$\vec{BA} = \vec{a} - \vec{b} = (3-1)\hat{i} + (-2-1)\hat{j} + (1 - (-2))\hat{k} = 2\hat{i} - 3\hat{j} + 3\hat{k}$.

MCQ 4: For any two points $A$ and $B$, the vector sum $\vec{AB} + \vec{BA}$ is equal to:

A $\vec{0}$

B $2\vec{AB}$

C $|\vec{AB}|$

Derivation & Analysis:

$\vec{AB} + \vec{BA} = \vec{AB} - \vec{AB} = \vec{0}$ by vector addition properties.

MCQ 5: The direction cosines of the displacement vector from $(0, 0, 0)$ to $(1, 2, 2)$ are:

A $(1/3, 2/3, 2/3)$

B $(1/\sqrt{5}, 2/\sqrt{5}, 0)$

C $(1/2, 1/2, 1/2)$

D (1, 1, 1)

Derivation & Analysis:

Magnitude $= \sqrt{1^2 + 2^2 + 2^2} = 3$. Direction cosines are $1/3, 2/3, 2/3$.

MCQ 6: If points $A, B, C$ have position vectors $\hat{i}+2\hat{j}+3\hat{k}$, $2\hat{i}+3\hat{j}+\hat{k}$, and $3\hat{i}+\hat{j}+2\hat{k}$ respectively, the vector $\vec{AB} + \vec{BC} + \vec{CA}$ equals:

A $\vec{0}$

B $\hat{i} + \hat{j} + \hat{k}$

C $3\hat{i} + 6\hat{j} + 6\hat{k}$

D $2\hat{i} + 2\hat{j} + 2\hat{k}$

Derivation & Analysis:

By Chasles' relation, $\vec{AB} + \vec{BC} + \vec{CA} = \vec{AC} + \vec{CA} = \vec{AA} = \vec{0}$.

MCQ 7: If the displacement vector $\vec{AB} = 3\hat{i} + 4\hat{j} + 12\k$ represents a force or position shift, its projection on the $x$-axis is:

A 3

B 4

C 12

D 13

Derivation & Analysis:

The component along the $x$-axis is simply the coefficient of $\hat{i}$, which is $3$.

MCQ 8: Assertion (A): Displacement vector depends only on initial and final positions. Reason (R): Vector addition is commutative.

A Both (A) and (R) are true and (R) is the correct explanation of (A)

B Both (A) and (R) are true but (R) is not the correct explanation

C (A) is true but (R) is false

D (A) is false but (R) is true

Derivation & Analysis:

Both statements are mathematically true facts in vector spaces, but commutativity is not the direct definition of path independence.

MCQ 9: If $P$ is $(2, 3, 4)$ and $Q$ is $(x, y, z)$ such that displacement vector $\vec{PQ} = 5\hat{i} - 2\hat{j} + 3\hat{k}$, then coordinates of $Q$ are:

A (7, 1, 7)

B (3, -5, -1)

C (-3, 5, 1)

D (7, 5, 7)

Derivation & Analysis:

$x - 2 = 5 \implies x = 7$; $y - 3 = -2 \implies y = 1$; $z - 4 = 3 \implies z = 7$. Thus $Q(7, 1, 7)$.

Q10

MCQ 10: The angle $\theta$ made by the displacement vector $\hat{i} + \hat{j} + \sqrt{2}\hat{k}$ with the $z$-axis is:

A $45^\circ$

B $60^\circ$

C $30^\circ$

D $90^\circ$

Derivation & Analysis:

Magnitude $= \sqrt{1 + 1 + 2} = 2$. Component along $z$-axis is $\sqrt{2}$. $\cos\theta = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}} \implies \theta = 45^\circ$.

Common questions

A position vector specifies the location of a single point relative to a chosen fixed origin $O$. A displacement vector specifies the relative position and distance between any two arbitrary points in space, independent of where the origin is placed.

The most common error is mixing up the direction by subtracting the terminal coordinates from the initial coordinates instead of initial from terminal (i.e., writing $\vec{a} - \vec{b}$ instead of $\vec{b} - \vec{a}$).

Memorize component subtraction rules and standard Pythagorean triples for 3D magnitudes (e.g., $3-4-12$ triangle yielding magnitude $13$) to bypass lengthy arithmetic expansion.

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