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Vector Algebra Reading Time: 15 min 6 Key Concepts

JEE Problem Patterns: Mastering the Cross Product

Essential properties, collinearity conditions, and area calculations with solved JEE problems.

3-5
Questions in JEE
100%
Concept Coverage
6
Key Applications
25min
Mastery Time

Why Cross Product Matters in JEE

The cross product (vector product) is one of the most important operations in vector algebra for JEE. It appears in 3-5 questions every year across these applications:

  • Finding perpendicular vectors to given vectors
  • Testing collinearity of points or vectors
  • Calculating areas of triangles and parallelograms
  • Physics applications in torque, angular momentum
  • Coordinate geometry problems in 3D space

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Definition & Formula Key Properties Collinearity Area Calculations JEE Problems Practice

1. Cross Product Definition & Formula

Mathematical Definition

The cross product of two vectors $\vec{a}$ and $\vec{b}$ is a vector $\vec{c}$ that is perpendicular to both $\vec{a}$ and $\vec{b}$, with magnitude equal to the area of the parallelogram formed by $\vec{a}$ and $\vec{b}$.

Formula in Component Form

If $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, then:

$$ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ \end{vmatrix} $$

Which expands to:

$$ \vec{a} \times \vec{b} = (a_2b_3 - a_3b_2)\hat{i} - (a_1b_3 - a_3b_1)\hat{j} + (a_1b_2 - a_2b_1)\hat{k} $$

Geometric Interpretation

Magnitude:

$$ |\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta $$

Where $\theta$ is the angle between $\vec{a}$ and $\vec{b}$

Direction:

Given by the right-hand rule: Point fingers in direction of $\vec{a}$, curl towards $\vec{b}$, thumb gives direction of $\vec{a} \times \vec{b}$.

2. Essential Properties of Cross Product

Anti-commutative Property

$$ \vec{a} \times \vec{b} = -(\vec{b} \times \vec{a}) $$

Order matters! Changing order reverses the direction.

Distributive Property

$$ \vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c} $$

Cross product distributes over vector addition.

Scalar Multiplication

$$ (k\vec{a}) \times \vec{b} = k(\vec{a} \times \vec{b}) = \vec{a} \times (k\vec{b}) $$

Scalars can be factored out.

Parallel Vectors

$$ \vec{a} \times \vec{b} = \vec{0} \iff \vec{a} \parallel \vec{b} $$

Cross product is zero if vectors are parallel.

⚠️ Important Notes

  • Cross product is not associative: $\vec{a} \times (\vec{b} \times \vec{c}) \neq (\vec{a} \times \vec{b}) \times \vec{c}$
  • Cross product of a vector with itself is zero: $\vec{a} \times \vec{a} = \vec{0}$
  • Cross product is defined only in 3D space

3. Condition for Collinearity

Fundamental Collinearity Condition

Three points A, B, C are collinear if and only if the vectors $\vec{AB}$ and $\vec{AC}$ are parallel, which means:

$$ \vec{AB} \times \vec{AC} = \vec{0} $$

Or equivalently, $\vec{AB} = k\vec{AC}$ for some scalar $k$.

JEE Main 2022 Medium

Problem: Checking Collinearity

Determine if points A(1, 2, 3), B(4, 5, 6), and C(7, 8, 9) are collinear.

Solution:

Step 1: Find vectors $\vec{AB}$ and $\vec{AC}$

$\vec{AB} = (4-1, 5-2, 6-3) = (3, 3, 3)$

$\vec{AC} = (7-1, 8-2, 9-3) = (6, 6, 6)$

Step 2: Check if $\vec{AB} \times \vec{AC} = \vec{0}$

$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 3 & 3 \\ 6 & 6 & 6 \\ \end{vmatrix}$

$= \hat{i}(3\cdot6 - 3\cdot6) - \hat{j}(3\cdot6 - 3\cdot6) + \hat{k}(3\cdot6 - 3\cdot6)$

$= \hat{i}(0) - \hat{j}(0) + \hat{k}(0) = \vec{0}$

Step 3: Since $\vec{AB} \times \vec{AC} = \vec{0}$, the points are collinear.

4. Area Calculations Using Cross Product

Area of Parallelogram

$$ \text{Area} = |\vec{a} \times \vec{b}| $$

Where $\vec{a}$ and $\vec{b}$ are adjacent sides of the parallelogram.

Area of Triangle

$$ \text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}| $$

Where $\vec{a}$ and $\vec{b}$ are two sides of the triangle.

JEE Main 2021 Medium

Problem: Area of Triangle

Find the area of triangle with vertices A(1, 1, 1), B(1, 2, 3), and C(2, 3, 1).

Solution:

Step 1: Find vectors $\vec{AB}$ and $\vec{AC}$

$\vec{AB} = (1-1, 2-1, 3-1) = (0, 1, 2)$

$\vec{AC} = (2-1, 3-1, 1-1) = (1, 2, 0)$

Step 2: Calculate $\vec{AB} \times \vec{AC}$

$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0 \\ \end{vmatrix}$

$= \hat{i}(1\cdot0 - 2\cdot2) - \hat{j}(0\cdot0 - 2\cdot1) + \hat{k}(0\cdot2 - 1\cdot1)$

$= \hat{i}(-4) - \hat{j}(-2) + \hat{k}(-1) = (-4, 2, -1)$

Step 3: Find magnitude of cross product

$|\vec{AB} \times \vec{AC}| = \sqrt{(-4)^2 + 2^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21}$

Step 4: Calculate area of triangle

$\text{Area} = \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{1}{2} \sqrt{21}$

5. JEE-Level Practice Problems

JEE Advanced 2020 Hard

Problem 1: Vector Triple Product

If $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \times \vec{c}$, then which of the following is true?

Solution Approach:

Step 1: Use vector triple product identity

$\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$

$(\vec{a} \times \vec{b}) \times \vec{c} = - \vec{c} \times (\vec{a} \times \vec{b}) = -[(\vec{c} \cdot \vec{b})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}]$

Step 2: Set them equal and simplify

This leads to conditions on the relationships between vectors

Step 3: The equality holds if $\vec{a}$ and $\vec{c}$ are parallel

JEE Main 2019 Medium

Problem 2: Finding Perpendicular Vector

Find a unit vector perpendicular to both $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + 2\hat{k}$.

Solution:

Step 1: Calculate $\vec{a} \times \vec{b}$

$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -1 \\ 2 & 1 & 2 \\ \end{vmatrix}$

$= \hat{i}(4 - (-1)) - \hat{j}(2 - (-2)) + \hat{k}(1 - 4)$

$= 5\hat{i} - 4\hat{j} - 3\hat{k}$

Step 2: Find magnitude

$|\vec{a} \times \vec{b}| = \sqrt{5^2 + (-4)^2 + (-3)^2} = \sqrt{25 + 16 + 9} = \sqrt{50} = 5\sqrt{2}$

Step 3: Unit vector = $\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$

$\hat{n} = \frac{5\hat{i} - 4\hat{j} - 3\hat{k}}{5\sqrt{2}} = \frac{1}{\sqrt{2}}\left(\hat{i} - \frac{4}{5}\hat{j} - \frac{3}{5}\hat{k}\right)$

🚀 Quick Revision Formulas

Cross Product Formulas

  • $\vec{a} \times \vec{b} = -\vec{b} \times \vec{a}$
  • $\vec{a} \times \vec{a} = \vec{0}$
  • $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$
  • $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta$

Applications

  • Collinearity: $\vec{a} \times \vec{b} = \vec{0}$
  • Area of parallelogram: $|\vec{a} \times \vec{b}|$
  • Area of triangle: $\frac{1}{2}|\vec{a} \times \vec{b}|$
  • Unit perpendicular: $\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$

📝 Practice Exercise

Test your understanding with these problems:

1. Find $\vec{a} \times \vec{b}$ if $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} - 2\hat{k}$

2. Check if points P(1,0,1), Q(0,1,1), R(1,1,0) are collinear

3. Find area of triangle with vertices (1,0,0), (0,1,0), (0,0,1)

4. Find a vector of magnitude 3 perpendicular to both $\hat{i} + \hat{j}$ and $\hat{i} + \hat{k}$

Answers: 1. $-\hat{i} + 5\hat{j} + 7\hat{k}$, 2. Not collinear, 3. $\frac{\sqrt{3}}{2}$, 4. $\pm\sqrt{3}(\hat{i} - \hat{j} - \hat{k})$

Mastered Cross Product?

Continue your vector algebra journey with these topics

Dot Product & Applications Vector Triple Product