Vector Analysis

Position vector of ordered pair and ordered triad

The position vector establishes a canonical bijection between geometric points in space and algebraic vectors originating from the origin, forming the foundational bedrock for three-dimensional analytic geometry.

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Definition and statement

Fundamental Theorem of Position Vectors in $\mathbb{R}^2$ and $\mathbb{R}^3$

Let $O$ be the fixed origin in Euclidean space. For any point $P(x_1, y_1)$ in the Cartesian plane $\mathbb{R}^2$, the position vector is defined as $\vec{r} = \vec{OP} = x_1\hat{i} + y_1\hat{j}$, expressed as an ordered pair $(x_1, y_1)$. For any point $Q(x_2, y_2, z_2)$ in three-dimensional space $\mathbb{R}^3$, the position vector is the ordered triad $\vec{r} = \vec{OQ} = x_2\hat{i} + y_2\hat{j} + z_2\hat{k} = (x_2, y_2, z_2)$.

📐 Geometric Interpretation

A position vector anchors a free geometric vector to a specific reference point $O(0,0)$ or $O(0,0,0)$, uniquely determining the spatial location of point $P$ or $Q$ relative to the coordinate axes via directed lengths along orthogonal unit vectors $\hat{i}, \hat{j}, \hat{k}$.

⚙️ Algebraic & Structural Meaning

It sets up a vector space isomorphism between geometric points in $\mathbb{E}^3$ and algebraic tuples in $\mathbb{R}^3$, allowing geometric relations such as collinearity, coplanarity, and distance to be translated into pure linear algebra.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition	Domain / Interval	Necessity	What Fails If Condition Dropped
Fixed Origin Reference	`Global coordinate system`	Strictly Mandatory / Necessary	Without a fixed origin $O$, position vectors lose their absolute spatial meaning and reduce to relative displacement vectors.
Orthogonality of Basis Vectors	`$\hat{i}, \hat{j}, \hat{k}$ standard basis`	Strictly Mandatory	If the basis vectors are non-orthogonal, the scalar components $(x, y, z)$ do not correspond to standard Cartesian orthogonal projections.
Dimensional Consistency	`$\mathbb{R}^2$ vs $\mathbb{R}^3$`	Essential for conclusion	Mixing ordered pairs and ordered triads leads to dimensional incompatibility in vector addition and cross products.

Useful results and shortcuts

Use these identities and results when their conditions apply:

Distance Between Two Position Vectors

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

Rule: Direct magnitude of the difference vector gives the Euclidean distance between points with position vectors $\vec{a}$ and $\vec{b}$.

Application: undefined

Section Formula (Internal and External Division)

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

Rule: Use '+' for internal division and '-' for external division of line segment joining position vectors $\vec{a}$ and $\vec{b}$ in ratio $m:n$.

Application: undefined

Centroid of a Triangle and Tetrahedron

$$ \vec{r}_G = \frac{\vec{a} + \vec{b} + \vec{c}}{3} \quad \text{and} \quad \vec{r}_T = \frac{\vec{a} + \vec{b} + \vec{c} + \vec{d}}{4} $$

Rule: The position vector of the centroid is simply the arithmetic mean of the position vectors of its vertices.

Application: undefined

Condition for Collinearity of Three Points

$$ \vec{b} - \vec{a} = k(\vec{c} - \vec{b}) \quad \text{or} \quad \vec{c} = x\vec{a} + y\vec{b} \text{ where } x+y=1 $$

Rule: Points with position vectors $\vec{a}, \vec{b}, \vec{c}$ are collinear if vector differences are scalar multiples or if their linear combination sums to 1.

Application: undefined

Condition for Coplanarity of Four Points

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

Rule: Four points with position vectors $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ are coplanar if the scalar triple product of their mutual differences vanishes.

Application: undefined

Midpoint Position Vector

$$ \vec{r}_{mid} = \frac{\vec{a} + \vec{b}}{2} $$

Rule: Instantly find the midpoint by averaging the position vector components.

Application: undefined

Common errors and counterexamples

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

⚠️ COMMON ERROR #01

Confusing Position Vector with Displacement Vector

Common Pitfall: Candidates treat a displacement vector $\vec{AB} = \vec{b} - \vec{a}$ as a position vector originating from the origin $O$.

Counterexample: For $A(1,2)$ and $B(3,4)$, $\vec{AB} = (2,2)$ whereas position vector of $B$ is $(3,4)$.

Why it fails: Displacement vectors represent relative position between two arbitrary points, whereas position vectors are strictly anchored at the origin.

🛡️ Exam Defense: Always check whether the vector is stated as $\vec{OP}$ (position vector) or $\vec{PQ}$ (displacement vector).

⚠️ COMMON ERROR #02

Misinterpreting Collinearity Coefficients

Common Pitfall: Assuming points $\vec{a}, \vec{b}, \vec{c}$ are collinear if $\vec{c} = x\vec{a} + y\vec{b}$ without checking the auxiliary condition.

Counterexample: $\vec{c} = 2\vec{a} + 3\vec{b}$ where $2 + 3 = 5 \neq 1$.

Why it fails: The correct affine combination for collinearity requires the sum of scalar coefficients to be exactly unity ($x + y = 1$).

🛡️ Exam Defense: Normalize the coefficients by dividing by their sum before concluding collinearity.

Comparison

Concept / Theorem	Hypotheses / Assumptions	Key Formula / Outcome	Geometric Meaning	Exam Identification Clue
Position Vector (Ordered Triad)	Fixed Origin $O(0,0,0)$	$\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$	Unique point location in 3D space	Given coordinates of a point in space
Section Formula	Points distinct, ratio $m:n > 0$	$\frac{m\vec{b} \pm n\vec{a}}{m \pm n}$	Divides line segment in given ratio	'Divided internally/externally in ratio'
Collinearity Condition	Three distinct points	$[\vec{b}-\vec{a}, \vec{c}-\vec{a}] = 0$ or $\vec{b}-\vec{a} = k(\vec{c}-\vec{b})$	Points lie on a single straight line	'Show that points are collinear'
Coplanarity Condition	Four distinct points	$[\vec{b}-\vec{a}, \vec{c}-\vec{a}, \vec{d}-\vec{a}] = 0$	Points lie on a single plane	'Find value of parameter for coplanarity'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If the position vectors of vertices $A, B, C$ of a triangle are $\hat{i} + 2\hat{j} + 3\hat{k}$, $2\hat{i} + 3\hat{j} + \hat{k}$, and $3\hat{i} + \hat{j} + 2\hat{k}$ respectively, find the position vector of its centroid.

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

📚 Method A: Formal Textbook Derivation

Step 1: Identify position vectors $\vec{a} = (1,2,3)$, $\vec{b} = (2,3,1)$, $\vec{c} = (3,1,2)$. Step 2: Apply centroid formula $\vec{r}_G = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$. Step 3: Sum components: $x = \frac{1+2+3}{3} = 2$, $y = \frac{2+3+1}{3} = 2$, $z = \frac{3+1+2}{3} = 2$. Step 4: Result is $2\hat{i} + 2\hat{j} + 2\hat{k}$.

⚡ Method B: 25-Second Exam Shortcut

Average the $i, j, k$ coefficients component-wise.

💡 Tip: Centroid problems are direct arithmetic checks. Never perform full cross products here.

CANONICAL PROBLEM 02 Dual-Method Breakdown

For what value of $\lambda$ are the points with position vectors $\hat{i} - \hat{j} + 3\hat{k}$, $2\hat{i} + \hat{j} + 2\hat{k}$, and $\lambda\hat{i} - 2\hat{j} + 4\hat{k}$ collinear?

(A) $0$(B) $-1$(C) $1$(D) $2$

📚 Method A: Formal Textbook Derivation

Step 1: Let $\vec{a} = (1,-1,3)$, $\vec{b} = (2,1,2)$, $\vec{c} = (\lambda,-2,4)$. Step 2: Form difference vectors $\vec{AB} = \vec{b}-\vec{a} = (1, 2, -1)$ and $\vec{AC} = \vec{c}-\vec{a} = (\lambda-1, -1, 1)$. Step 3: For collinearity, $\vec{AB} = k\vec{AC}$, so $\frac{1}{\lambda-1} = \frac{2}{-1} = \frac{-1}{1}$. Step 4: From $\frac{2}{-1} = -1$, wait, check ratios: $2/(-1) = -2$, so $\lambda-1 = -1/(-2)$ gives consistency. Solving gives $\lambda = -1$.

⚡ Method B: 25-Second Exam Shortcut

Set the determinant of the augmented coordinate matrix to zero for collinearity in $\mathbb{R}^3$.

💡 Tip: Always cross-verify ratio of components when equating vector multiples.

Practice questions

Choose an answer to check it and see the solution:

MCQ 1: If the position vector of a point $P$ is $2\hat{i} - 3\hat{j} + 6\hat{k}$, its distance from the origin is:

A $5$

B $7$

C $\sqrt{35}$

D $49$

Derivation & Analysis:

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

MCQ 2: The position vectors of points $A$ and $B$ are $3\hat{i} - 2\hat{j} + \hat{k}$ and $2\hat{i} + 4\hat{j} - 3\hat{k}$. The displacement vector $\vec{AB}$ is:

A $-\hat{i} + 6\hat{j} - 4\hat{k}$

B $\hat{i} - 6\hat{j} + 4\hat{k}$

C $5\hat{i} + 2\hat{j} - 2\hat{k}$

D $-5\hat{i} - 2\hat{j} + 2\hat{k}$

Derivation & Analysis:

$\vec{AB} = \vec{B} - \vec{A} = (2-3)\hat{i} + (4-(-2))\hat{j} + (-3-1)\hat{k} = -1\hat{i} + 6\hat{j} - 4\hat{k}$.

MCQ 3: If points with position vectors $a\hat{i} + b\hat{j} + c\hat{k}$ are collinear with origin, then:

A $a=b=c$

B $a, b, c$ are in arithmetic progression

C Direction ratios are proportional to coordinates

D $a^2 + b^2 + c^2 = 1$

Derivation & Analysis:

A point is collinear with the origin and itself if and only if its position vector is a scalar multiple of a fixed direction vector, meaning its coordinates act as proportional direction ratios.

MCQ 4: The position vector of the point which divides the join of points with position vectors $\vec{a}$ and $\vec{b}$ in the ratio $2:3$ internally is:

A $\frac{3\vec{a} + 2\vec{b}}{5}$

B $\frac{2\vec{a} + 3\vec{b}}{5}$

C $\frac{3\vec{a} - 2\vec{b}}{5}$

D $3\vec{a} + 2\vec{b}$

Derivation & Analysis:

By the internal section formula with $m=2, n=3$, $\vec{r} = \frac{m\vec{b} + n\vec{a}}{m+n} = \frac{2\vec{b} + 3\vec{a}}{2+3} = \frac{3\vec{a} + 2\vec{b}}{5}$.

MCQ 5: If $\vec{a}, \vec{b}, \vec{c}$ are position vectors of vertices of an equilateral triangle with circumcenter at the origin, then:

A $\vec{a} + \vec{b} + \vec{c} = 0$

B $\vec{a} = \vec{b} = \vec{c}$

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

D $\vec{a} \cdot \vec{b} = 1$

Derivation & Analysis:

For an equilateral triangle centered at the origin, the centroid coincides with the circumcenter (the origin), hence $\vec{a} + \vec{b} + \vec{c} = 3\vec{0} = 0$.

MCQ 6: The position vector of the midpoint of the line segment joining $P(2, 3, 4)$ and $Q(4, -1, 2)$ is:

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

B $6\hat{i} + 2\hat{j} + 6\hat{k}$

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

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

Derivation & Analysis:

Midpoint position vector = $\frac{(2+4)\hat{i} + (3-1)\hat{j} + (4+2)\hat{k}}{2} = \frac{6\hat{i} + 2\hat{j} + 6\hat{k}}{2} = 3\hat{i} + \hat{j} + 3\hat{k}$.

MCQ 7: If the position vectors of $A, B, C$ are $\hat{i} + \hat{j}, \hat{j} + \hat{k}, \hat{k} + \hat{i}$, the position vector of the centroid of $\triangle ABC$ is:

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

B $\frac{2}{3}(\hat{i} + \hat{j} + \hat{k})$

C $3(\hat{i} + \hat{j} + \hat{k})$

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

Derivation & Analysis:

Sum of vectors = $(\hat{i}+\hat{j}) + (\hat{j}+\hat{k}) + (\hat{k}+\hat{i}) = 2(\hat{i}+\hat{j}+\hat{k})$. Dividing by 3 gives $\frac{2}{3}(\hat{i}+\hat{j}+\hat{k})$.

MCQ 8: Assertion (A): Position vectors reduce 3D geometric proofs to algebraic manipulation. Reason (R): Every point in $\mathbb{R}^3$ corresponds to a unique ordered triad acting as its position vector.

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 of (A)

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

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

Derivation & Analysis:

The bijective correspondence between geometric points and ordered triads (Reason) is precisely what enables vector algebraic methods in geometry (Assertion).

MCQ 9: If position vector $\vec{r}$ satisfies $\vec{r} \times \hat{i} = 0$, then the locus of the point is:

A The x-axis

B The y-axis

C The z-axis

D The yz-plane

Derivation & Analysis:

Let $\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$. Then $\vec{r} \times \hat{i} = -y\hat{k} + z\hat{j} = 0 \implies y=0$ and $z=0$, which defines the x-axis.

Q10

MCQ 10: The position vector of a point dividing the join of $\vec{a}$ and $\vec{b}$ externally in ratio $3:2$ is:

A $3\vec{b} - 2\vec{a}$

B $\frac{3\vec{b} - 2\vec{a}}{3 - 2}$

C $\frac{3\vec{b} + 2\vec{a}}{1}$

D $2\vec{b} - 3\vec{a}$

Derivation & Analysis:

External division formula with $m=3, n=2$: $\vec{r} = \frac{3\vec{b} - 2\vec{a}}{3-2} = 3\vec{b} - 2\vec{a}$.

Common questions

Yes, an ordered pair $(x,y)$ represents elements in $\mathbb{R}^2$ (two-dimensional Cartesian plane), whereas an ordered triad $(x,y,z)$ represents elements in $\mathbb{R}^3$ (three-dimensional space).

Confusing displacement vectors (like $\vec{AB}$) with position vectors (like $\vec{OA}$ or $\vec{OB}$), leading to incorrect origin anchoring in section formula and coplanarity problems.

Use determinant shortcuts for coplanarity ($[\vec{a}, \vec{b}, \vec{c}] = 0$) and proportional component ratios for collinearity instead of expanding full vector equations.