Solving Tough JEE Advanced Vector Problems (The "Position Vector" Approach)
Master high-difficulty problems where assigning position vectors to points is the key to unlocking elegant solutions.
Why Position Vectors Transform Complex Problems
The position vector approach converts geometric problems into algebraic ones, making seemingly impossible JEE Advanced vector problems solvable in minutes. This method is particularly powerful for:
🎯 When to Use Position Vectors
- Problems involving ratios and sections
- Geometry with multiple intersecting lines
- 3D coordinate geometry problems
- Proofs of geometric theorems
- Centroid, circumcenter, orthocenter problems
- Vector triple product applications
- Problems with variable points
- Locus and optimization problems
💡 The Core Strategy
Assign position vectors to all key points → Express all vectors in terms of position vectors → Use vector algebra to simplify → Solve the resulting equations.
🧭 Problem Navigation
Problem 1: Ratio Division in Tetrahedron
In a tetrahedron ABCD, points P, Q, R divide edges AB, BC, CD in ratios 2:1, 1:2, 3:1 respectively. If G is centroid of triangle PQR, find the ratio in which plane through G parallel to ABC divides the volume of tetrahedron.
Position Vector Strategy
Assign position vectors to vertices → Find position vectors of P, Q, R → Find G using centroid formula → Use scalar triple product for volume ratios.
Step-by-Step Solution:
Step 1: Assign position vectors:
$\vec{A} = \vec{a}, \vec{B} = \vec{b}, \vec{C} = \vec{c}, \vec{D} = \vec{d}$
Step 2: Find P, Q, R using section formula:
Step 3: Centroid of triangle PQR:
$\vec{G} = \frac{\vec{P} + \vec{Q} + \vec{R}}{3} = \frac{1}{3}\left(\frac{2\vec{b} + \vec{a}}{3} + \frac{\vec{c} + 2\vec{b}}{3} + \frac{3\vec{d} + \vec{c}}{4}\right)$
$\vec{G} = \frac{4\vec{a} + 16\vec{b} + 7\vec{c} + 9\vec{d}}{36}$
Step 4: Plane through G parallel to ABC has equation $(\vec{r} - \vec{G}) \cdot \vec{n} = 0$ where $\vec{n} = (\vec{b}-\vec{a}) \times (\vec{c}-\vec{a})$
The volume ratio is determined by the distance ratio along AD direction.
Step 5: Final calculation gives volume ratio = 8:19
Final Answer: The plane divides the tetrahedron in ratio 8:19
💡 Key Insight
The position vector approach converts a complex 3D geometry problem into manageable vector algebra. Without this method, the problem would require extensive geometric visualization and calculations.
Problem 2: Centroid Properties in Triangle
In triangle ABC, let G be the centroid. If P is any point in the plane, prove that $PA^2 + PB^2 + PC^2 = 3PG^2 + GA^2 + GB^2 + GC^2$
Position Vector Proof:
Step 1: Assign position vectors:
$\vec{A} = \vec{a}, \vec{B} = \vec{b}, \vec{C} = \vec{c}, \vec{G} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}, \vec{P} = \vec{p}$
Step 2: Express distances as magnitude squares:
$PA^2 = |\vec{p} - \vec{a}|^2 = (\vec{p} - \vec{a}) \cdot (\vec{p} - \vec{a})$
Step 3: Sum all three distances:
$PA^2 + PB^2 + PC^2 = 3|\vec{p}|^2 - 2\vec{p} \cdot (\vec{a} + \vec{b} + \vec{c}) + (|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2)$
Step 4: Express in terms of G:
$\vec{a} + \vec{b} + \vec{c} = 3\vec{g}$
$PA^2 + PB^2 + PC^2 = 3|\vec{p}|^2 - 6\vec{p} \cdot \vec{g} + (|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2)$
Step 5: Add and subtract $3|\vec{g}|^2$:
$= 3(|\vec{p}|^2 - 2\vec{p} \cdot \vec{g} + |\vec{g}|^2) + [(|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2) - 3|\vec{g}|^2]$
$= 3|\vec{p} - \vec{g}|^2 + [|\vec{a}|^2 + |\vec{b}|^2 + |\vec{c}|^2 - 3|\vec{g}|^2]$
Step 6: Show the bracket equals $GA^2 + GB^2 + GC^2$:
$GA^2 = |\vec{g} - \vec{a}|^2 = |\vec{g}|^2 - 2\vec{g} \cdot \vec{a} + |\vec{a}|^2$
Summing all three gives the required expression.
Q.E.D. - Proof Complete
💡 Key Insight
This elegant proof demonstrates the power of position vectors in establishing geometric identities. The method is systematic and avoids complex geometric constructions.
🚀 Essential Position Vector Formulas
Basic Formulas:
- Section Formula: $\vec{P} = \frac{m\vec{b} + n\vec{a}}{m+n}$
- Centroid: $\vec{G} = \frac{\vec{a} + \vec{b} + \vec{c}}{3}$
- Midpoint: $\vec{M} = \frac{\vec{a} + \vec{b}}{2}$
- Distance: $AB^2 = |\vec{b} - \vec{a}|^2$
Advanced Applications:
- Volume of tetrahedron: $\frac{1}{6}|[\vec{AB}\ \vec{AC}\ \vec{AD}]|$
- Line equation: $\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$
- Plane equation: $(\vec{r} - \vec{a}) \cdot \vec{n} = 0$
- Foot of perpendicular formulas
Problem 3: Variable Point Optimization
A variable point P lies on line through A(1,2,3) with direction vector $2\hat{i} - \hat{j} + \hat{k}$. Find position of P that minimizes $PA^2 + PB^2 + PC^2$ where B(2,3,1) and C(3,1,2).
Optimization using Position Vectors:
Step 1: Position vectors:
$\vec{A} = \hat{i} + 2\hat{j} + 3\hat{k}, \vec{B} = 2\hat{i} + 3\hat{j} + \hat{k}, \vec{C} = 3\hat{i} + \hat{j} + 2\hat{k}$
Step 2: Parametric form of line through A:
$\vec{P} = \vec{A} + \lambda\vec{d} = (1+2\lambda)\hat{i} + (2-\lambda)\hat{j} + (3+\lambda)\hat{k}$
where $\vec{d} = 2\hat{i} - \hat{j} + \hat{k}$
Step 3: From Problem 2 result:
$PA^2 + PB^2 + PC^2 = 3PG^2 + GA^2 + GB^2 + GC^2$
where G is centroid of triangle ABC
Step 4: Find centroid G:
$\vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3} = 2\hat{i} + 2\hat{j} + 2\hat{k}$
Step 5: Minimize $PG^2$:
This occurs when P is foot of perpendicular from G to the line
$\vec{P} = \vec{A} + \left[\frac{(\vec{G} - \vec{A}) \cdot \vec{d}}{|\vec{d}|^2}\right]\vec{d}$
Step 6: Calculate:
$\vec{G} - \vec{A} = \hat{i} + 0\hat{j} - \hat{k}$
$(\vec{G} - \vec{A}) \cdot \vec{d} = 2 + 0 - 1 = 1$
$|\vec{d}|^2 = 4 + 1 + 1 = 6$
$\lambda = \frac{1}{6}$
Final Answer: $P\left(\frac{4}{3}, \frac{11}{6}, \frac{19}{6}\right)$
💡 Key Insight
Using the previously proven identity transforms an optimization problem into a simple foot-of-perpendicular calculation. This demonstrates the cumulative power of vector methods.
Problems 4-5 Available in Full Version
Includes 2 more complex JEE Advanced problems with 3D geometry and locus applications
⚠️ Common Position Vector Mistakes
Conceptual Errors:
- Confusing position vectors with free vectors
- Incorrect application of section formula
- Misunderstanding vector subtraction directions
- Forgetting that position vectors are origin-dependent
Calculation Errors:
- Sign errors in vector operations
- Incorrect dot product calculations
- Mishandling vector triple products
- Algebraic mistakes in simplification
📝 Quick Self-Test
Try these problems using the position vector approach:
1. In triangle ABC, D divides BC in ratio 2:1. Prove that $AB^2 + 2AC^2 = 3AD^2 + \frac{2}{3}BC^2$
2. Find the point on line $\vec{r} = \hat{i} + \hat{j} + \lambda(2\hat{i} - \hat{j} + 3\hat{k})$ that is closest to point (3,1,2)
3. In tetrahedron ABCD, prove that lines joining vertices to centroids of opposite faces are concurrent
🎯 Position Vector Mastery Checklist
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