JEE Dot Product Problems: Properties, Perpendicularity & Angles

Why Dot Product is Crucial for JEE

Based on analysis of JEE papers from 2015-2024, dot product concepts appear in 2-3 questions per paper. Mastering these patterns will help you secure crucial marks in:

Vector Algebra - Finding angles and projections
3D Geometry - Plane equations and distances
Physics Applications - Work done by force
Coordinate Geometry - Properties of triangles and quadrilaterals

Essential Dot Product Formulas

Definition

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

where $\theta$ is angle between vectors

Component Form

$\vec{a} \cdot \vec{b} = a_xb_x + a_yb_y + a_zb_z$

Perpendicularity

$\vec{a} \cdot \vec{b} = 0 \iff \vec{a} \perp \vec{b}$

Work Done

$W = \vec{F} \cdot \vec{d}$

Force $\vec{F}$, displacement $\vec{d}$

JEE Main 2023 Easy

Problem 1: Basic Dot Product & Perpendicularity

Find the value of $\lambda$ if vectors $\vec{a} = 2\hat{i} + 3\hat{j} - \hat{k}$ and $\vec{b} = \lambda\hat{i} + \hat{j} + 5\hat{k}$ are perpendicular.

Solution Approach:

Step 1: For perpendicular vectors: $\vec{a} \cdot \vec{b} = 0$

Step 2: Calculate dot product: $(2)(\lambda) + (3)(1) + (-1)(5) = 0$

Step 3: Simplify: $2\lambda + 3 - 5 = 0$

Step 4: Solve: $2\lambda - 2 = 0 \Rightarrow \lambda = 1$

Final Answer: $\lambda = 1$

JEE Main 2022 Medium

Problem 2: Angle Between Vectors

Find the angle between vectors $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} + 2\hat{j} + \hat{k}$.

Solution Approach:

Step 1: Calculate dot product: $\vec{a} \cdot \vec{b} = (1)(3) + (2)(2) + (3)(1) = 3 + 4 + 3 = 10$

Step 2: Calculate magnitudes: $|\vec{a}| = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{14}$

$|\vec{b}| = \sqrt{3^2 + 2^2 + 1^2} = \sqrt{14}$

Step 3: Use formula: $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{10}{\sqrt{14} \cdot \sqrt{14}} = \frac{10}{14} = \frac{5}{7}$

Step 4: $\theta = \cos^{-1}\left(\frac{5}{7}\right)$

Final Answer: $\theta = \cos^{-1}\left(\frac{5}{7}\right)$

JEE Advanced 2021 Hard

Problem 3: Angle Between Diagonals of Parallelogram

In a parallelogram with adjacent sides represented by vectors $\vec{a}$ and $\vec{b}$, find the angle between its diagonals.

Solution Approach:

Step 1: Diagonals: $\vec{d_1} = \vec{a} + \vec{b}$, $\vec{d_2} = \vec{a} - \vec{b}$

Step 2: Dot product: $\vec{d_1} \cdot \vec{d_2} = (\vec{a} + \vec{b}) \cdot (\vec{a} - \vec{b}) = |\vec{a}|^2 - |\vec{b}|^2$

Step 3: Magnitudes: $|\vec{d_1}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b}}$

$|\vec{d_2}| = \sqrt{|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}\cdot\vec{b}}$

Step 4: For special case (rectangle/square where $\vec{a} \cdot \vec{b} = 0$):

$\vec{d_1} \cdot \vec{d_2} = |\vec{a}|^2 - |\vec{b}|^2$

If $|\vec{a}| = |\vec{b}|$ (rhombus/square), diagonals are perpendicular

Key Insight: Diagonals are perpendicular only in rhombus (including square)

JEE Main 2020 Medium

Problem 4: Work Done by Force

A force $\vec{F} = 3\hat{i} + 4\hat{j} - 5\hat{k}$ moves a particle from point A(1,2,3) to B(2,4,1). Calculate the work done.

Solution Approach:

Step 1: Displacement vector: $\vec{d} = \vec{AB} = (2-1)\hat{i} + (4-2)\hat{j} + (1-3)\hat{k} = \hat{i} + 2\hat{j} - 2\hat{k}$

Step 2: Work done: $W = \vec{F} \cdot \vec{d}$

Step 3: Calculate: $W = (3)(1) + (4)(2) + (-5)(-2) = 3 + 8 + 10 = 21$ J

Final Answer: Work done = 21 Joules

🚀 Quick Problem-Solving Strategies

For Perpendicularity:

Set dot product = 0
Solve for unknown parameter
Verify both vectors are non-zero
Check if solution makes physical sense

For Angle Finding:

Calculate dot product and magnitudes
Use $\cos\theta = \frac{\vec{a}\cdot\vec{b}}{|\vec{a}||\vec{b}|}$
Consider acute/obtuse angle based on sign
Remember range: $0 \leq \theta \leq \pi$

Problems 5-8 Available in Full Version

Includes vector projection problems, triangle geometry applications, and advanced perpendicularity conditions

📝 Quick Self-Test

Try these similar problems to test your understanding:

1. Find angle between diagonals of cube

2. If $|\vec{a} + \vec{b}| = |\vec{a} - \vec{b}|$, prove $\vec{a} \perp \vec{b}$

3. Work done by force $\vec{F} = 2\hat{i} - \hat{j} + 3\hat{k}$ displacing from origin to (1,1,1)

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