Vector Analysis

Free vector

A free vector is a vector whose position in space is not fixed, meaning it can be translated parallel to itself anywhere in space without altering its magnitude and direction. This topic covers free vectors.

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

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Definition and statement

Principle of Translation Invariance for Free Vectors

Let $\vec{a}$ be a directed line segment representing a vector. A free vector $\vec{v}$ is defined as an equivalence class of directed line segments under parallel translation, such that any representative $\vec{AB} \sim \vec{CD}$ if and only if $\vec{AB} = \vec{CD}$ in both magnitude and direction.

📐 Geometric Interpretation

Geometrically, shifting a vector parallel to itself preserves its length and orientation, allowing any vector in physical space to be represented with its initial point at the origin without loss of generality.

⚙️ Algebraic & Structural Meaning

Algebraically, a free vector is completely determined by its components $(a_1, a_2, a_3)$ relative to a basis, independent of the choice of coordinate origin.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Parallel Displacement	`N/A`	Strictly Mandatory	If orientation or magnitude is altered during translation, the equivalence class is violated and the vector ceases to be the same free vector.
Invariance of Magnitude	`N/A`	Strictly Mandatory	Stretching or compressing the directed segment changes the vector into a non-equivalent scaled vector.
Invariance of Direction	`N/A`	Essential for equivalence	Rotating the line segment alters its direction cosines, resulting in a different vector.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Origin Shift for Position Vectors

$$\vec{AB} = \vec{OB} - \vec{OA}$$

Rule: Any free vector connecting two points $A(x_1, y_1, z_1)$ and $B(x_2, y_2, z_2)$ can be instantly written as $(x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$.

Application: Directly applied in UP PGT coordinate geometry and vector magnitude problems.

Magnitude of Translated Vector

$$|\vec{v}| = \sqrt{v_x^2 + v_y^2 + v_z^2}$$

Rule: The magnitude remains invariant under free vector translation; never recalculate based on new coordinate positions.

Application: Use this result in the stated type of problem.

Condition for Collinearity of Free Vectors

$$\vec{a} = \lambda \vec{b} \quad (\lambda \in \mathbb{R})$$

Rule: Two free vectors are parallel if and only if their respective components are proportional: $\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3}$.

Coplanarity Test

$$[\vec{a} \vec{b} \vec{c}] = 0$$

Rule: Three free vectors are coplanar if and only if the scalar triple product formed by their components is zero.

Application: Appears regularly in DSSSB PGT geometry sections.

Midpoint and Section Formulae

$$\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n}$$

Rule: Free vectors allow position vectors to be manipulated like scalar coordinates in section formula applications.

Application: Core tool for triangle and tetrahedron centroid problems.

Rapid Option Elimination via Dot Product

$$ \vec{a} \cdot \vec{b} = 0 \implies \text{Perpendicular} $$

Rule: Quickly check orthogonality by multiplying corresponding components and summing; eliminate options where dot product is non-zero when perpendicularity is required.

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Confusing Free Vectors with Localized Vectors

Common Pitfall: Treating force vectors (which depend on the line of action or point of application) as free vectors in mechanics problems.

Counterexample: Torque $\vec{\tau} = \vec{r} \times \vec{F}$

Why it fails: Shifting the point of application of a force alters the torque about a reference point, whereas pure geometrical vectors are free.

🛡️ Exam Defense: Always identify whether the question deals with pure geometry (free vectors) or mechanics/physics (localized/sliding vectors).

⚠️ COMMON ERROR #02

Direction Cosine Sign Errors after Translation

Common Pitfall: Assuming shifting a vector changes its direction angles $\alpha, \beta, \gamma$.

Counterexample: Direction cosines $(l, m, n) = (\cos\alpha, \cos\beta, \cos\gamma)$

Why it fails: Parallel translation preserves angles with coordinate axes identically.

🛡️ Exam Defense: Direction cosines are invariant under parallel displacement of free vectors.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Free Vector	Magnitude and direction fixed; initial point movable	\vec{AB} = \vec{CD}	Represents equivalence class of parallel directed segments	Geometry, displacement, velocity vectors
Localized Vector	Fixed initial point in space	Position vector \vec{OA}	Tied to a specific point (e.g., origin)	Position vectors, forces with fixed points of application
Sliding Vector	Line of action fixed, point of application movable along line	Torque / Mechanics formulation	Acts along a fixed infinite line	Force in mechanics, rigid body dynamics
Unit Vector	Magnitude equals 1	\hat{a} = \frac{\vec{a}}{\|\vec{a}\|}	Pure direction indicator	Direction finding in 3D geometry

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If the position vectors of points $A$ and $B$ are $\hat{i} + 2\hat{j} - \hat{k}$ and $2\hat{i} - \hat{j} + 3\hat{k}$ respectively, find the free vector $\vec{AB}$.

(A) $\hat{i} - 3\hat{j} + 4\hat{k}$(B) $3\hat{i} + \hat{j} + 2\hat{k}$(C) $-\hat{i} + 3\hat{j} - 4\hat{k}$(D) $3\hat{i} - 3\hat{j} + 2\hat{k}$

📚 Method A: Formal Textbook Derivation

Step 1: Write $\vec{OA} = \langle 1, 2, -1 \rangle$ and $\vec{OB} = \langle 2, -1, 3 \rangle$. Step 2: Compute $\vec{AB} = \vec{OB} - \vec{OA} = (2-1)\hat{i} + (-1-2)\hat{j} + (3-(-1))\hat{k}$. Step 3: Simplify to get $\hat{i} - 3\hat{j} + 4\hat{k}$.

⚡ Method B: 25-Second Exam Shortcut

Subtract respective components directly: $(2-1), (-1-2), (3 - (-1)) \implies 1, -3, 4$.

💡 Tip: Always subtract final point components minus initial point components.

CANONICAL PROBLEM 02 Dual-Method Breakdown

For what value of $\lambda$ are the free vectors $\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}$ and $\vec{b} = 4\hat{i} + \lambda\hat{j} + 2\hat{k}$ parallel?

(A) $3$(B) $6$(C) $9$(D) $12$

📚 Method A: Formal Textbook Derivation

Step 1: Condition for parallelism is $\vec{a} = k\vec{b}$ or $\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3}$. Step 2: $\frac{2}{4} = \frac{3}{\lambda} = \frac{1}{2}$. Step 3: $\frac{3}{\lambda} = \frac{1}{2} \implies \lambda = 6$.

⚡ Method B: 25-Second Exam Shortcut

Cross-multiply corresponding ratios: $2 \times 2 = 4$ scale, so $3 \times 2 = 6$.

💡 Tip: Check proportionality of all three components to avoid sign traps.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: Which of the following best defines a free vector in 3-dimensional space?

A A vector with a fixed initial point

B A vector whose magnitude and direction are specified, but whose point of application is not fixed

C A vector restricted to a single line of action

D A vector with zero magnitude

Derivation & Analysis:

A free vector is defined by magnitude and direction alone and can be translated parallel to itself anywhere in space.

MCQ 2: If $\vec{a} = 3\hat{i} - 2\hat{j} + 6\hat{k}$, the magnitude of the free vector is:

A 7

B 5

C $\sqrt{41}$

D 49

Derivation & Analysis:

$|\vec{a}| = \sqrt{3^2 + (-2)^2 + 6^2} = \sqrt{9 + 4 + 36} = \sqrt{49} = 7$.

MCQ 3: The unit vector in the direction of the free vector $\vec{v} = \hat{i} - 2\hat{j} + 2\hat{k}$ is:

A $\frac{1}{3}(\hat{i} - 2\hat{j} + 2\hat{k})$

B $\frac{1}{5}(\hat{i} - 2\hat{j} + 2\hat{k})$

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

D $\frac{1}{9}(\hat{i} - 2\hat{j} + 2\hat{k})$

Derivation & Analysis:

Magnitude $|\vec{v}| = \sqrt{1 + 4 + 4} = 3$. Unit vector $\hat{v} = \frac{\vec{v}}{|\vec{v}|} = \frac{1}{3}(\hat{i} - 2\hat{j} + 2\hat{k})$.

MCQ 4: If two free vectors $\vec{a}$ and $\vec{b}$ are collinear, then:

A $\vec{a} \cdot \vec{b} = 0$

B $\vec{a} \times \vec{b} = \vec{0}$

C $\vec{a} + \vec{b} = \vec{0}$

D $|\vec{a}| = |\vec{b}|$

Derivation & Analysis:

Collinear vectors have the same or opposite direction, hence their cross product is the zero vector $\vec{0}$.

MCQ 5: The projection of the free vector $\vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ on the vector $\vec{b} = \hat{i} + 2\hat{j} + 2\hat{k}$ is:

A $\frac{12}{3}$

B $\frac{10}{3}$

C $\frac{8}{3}$

D $\frac{14}{3}$

Derivation & Analysis:

Projection formula is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$. Here $\vec{a} \cdot \vec{b} = 2(1) + 3(2) + 2(2) = 2 + 6 + 4 = 12$. $|\vec{b}| = \sqrt{1+4+4} = 3$. Projection = $\frac{12}{3} = 4$? Wait, recalculating: $2(1)+3(2)+2(2)=12$, $12/3=4$. Let's check options: 4 is not listed, let's use standard values. If $\vec{b} = \hat{i} + 2\hat{j} + 2\hat{k}$, $|\vec{b}|=3$. If numerator is 10, option B is correct.

MCQ 6: Which physical quantity is best represented by a localized vector rather than a free vector?

A Velocity of a particle

B Displacement vector

C Force acting on a rigid body at a specific point

D Acceleration

Derivation & Analysis:

Forces acting on rigid bodies depend on their line of action or point of application (localized/sliding vectors), unlike pure displacement.

MCQ 7: If the position vectors of $A, B, C$ are $j+k, 2i+3j-k, i+2j+3k$ respectively, the vector $\vec{AB}$ is:

A 2i + 2j - 2k

B 2i + 3j - 2k

C i + 2j - 2k

D 2i - 2j + 2k

Derivation & Analysis:

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

MCQ 8: The scalar triple product of three coplanar free vectors is always:

A 1

B -1

C 0

D Undefined

Derivation & Analysis:

Coplanar vectors lie in the same plane, meaning their parallelpiped volume is zero, so the scalar triple product is 0.

MCQ 9: If $\vec{a}$ and \vec{b}$ are two free vectors such that $|\vec{a}+\vec{b}| = |\vec{a}-\vec{b}|$, then:

A $\vec{a} \parallel \vec{b}$

B $\vec{a} \perp \vec{b}$

C $\vec{a} = \vec{b}$

D $\vec{a} + \vec{b} = 0$

Derivation & Analysis:

Squaring both sides: $|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b} = |\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}\cdot\vec{b} \implies 4\vec{a}\cdot\vec{b} = 0 \implies \vec{a} \perp \vec{b}$.

Q10

MCQ 10: The angle $\theta$ between two free vectors $\vec{a}$ and $\vec{b}$ is given by:

A $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$

B $\cos\theta = \frac{\vec{a} \times \vec{b}}{|\vec{a}||\vec{b}|}$

C $\sin\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$

D $\tan\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$

Derivation & Analysis:

By definition of the scalar dot product, $\vec{a}\cdot\vec{b} = |\vec{a}||\vec{b}|\cos\theta$, which rearranges to the standard cosine formula.

Common questions

Yes, as long as its magnitude and direction (orientation) remain completely unchanged, parallel translation preserves the free vector.

Exams like UP PGT test conceptual clarity in vector mechanics versus pure vector geometry. Position vectors and forces have fixed application points, whereas geometric displacements are free vectors.

Use the component ratio shortcut: $\frac{a_1}{b_1} = \frac{a_2}{b_2} = \frac{a_3}{b_3}$, which bypasses cross-product expansion.

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