How do I determine if three vectors are linearly independent or coplanar?

Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar (linearly dependent) if and only if their scalar triple product $[\vec{a} \ \vec{b} \ \vec{c}] = 0$. Algebraically, this implies the $3 \times 3$ determinant of their component matrix evaluates to $0$.

What is the key functional difference between scalar projection and vector projection?

Scalar projection of $\vec{a}$ on $\vec{b}$ is a real scalar value given by $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$. Vector projection attaches direction to this scalar using unit vector $\hat{b}$, giving $\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b}$.

What is the fastest strategy to solve vector equations with multiple cross products?

Take the dot product or cross product of both sides with one of the known vectors (such as $\vec{a}$ or $\vec{b}$). Apply the Vector Triple Product expansion $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$ to convert cross products into solvable linear scalar equations.

Vector Algebra Notes, Formulas, Traps & PYQs | ExamiFly

Beta

Please wait while we prepare everything for you

Vector Algebra Notes, Formulas, Traps & PYQs | ExamiFly | Examifly

Key Formulas & High-Yield Equations

Dot Product and Projection Formula

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

Valid for non-zero vectors

\vec{a}

and

\vec{b}

. The scalar projection of

\vec{a}

\vec{b}

\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}

, while vector projection is

\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2}\right)\vec{b}

Cross Product & Vector Area

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

Where

\hat{n}

is the unit normal vector determined by the right-hand thumb rule. Magnitude

|\vec{a} \times \vec{b}|

equals the area of a parallelogram with sides

\vec{a}

and

\vec{b}

Scalar Triple Product (Box Product)

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

Represents the scalar volume of a parallelepiped with coterminous edges

\vec{a}, \vec{b}, \vec{c}

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

strictly implies coplanarity of vectors.

Vector Triple Product (VTP Rule)

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

Valid for any three vectors in

\mathbb{R}^3

. Note that cross product is non-associative:

\vec{a} \times (\vec{b} \times \vec{c}) \neq (\vec{a} \times \vec{b}) \times \vec{c}

Section Formula for Position Vectors

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

Divides the line segment joining position vectors

\vec{a}

and

\vec{b}

internally in ratio

m:n

. For external division, replace

n

with

-n

where

m \neq n

Solved High-Yield Sample Questions

Practice Mode

Question #1JEE Main 2024

If $\vec{a} = 2\hat{i} + \hat{j} + \hat{k}$ , $\vec{b} = \hat{i} + 3\hat{j} + \hat{k}$ , and $\vec{c} = \hat{i} + \hat{j} + 3\hat{k}$ , then find the volume of the parallelepiped formed by $\vec{a}$ , $\vec{b}$ , and $\vec{c}$ as coterminous edges.

Solution Locked

Step-by-Step Solution & Concept

Step 1

The volume $V$ of a parallelepiped with coterminous vectors $\vec{a}, \vec{b}, \vec{c}$ is given by $V = |[\vec{a} \ \vec{b} \ \vec{c}]|$ .

Step 2

Set up the determinant $V = \begin{vmatrix} 2 & 1 & 1 \\ 1 & 3 & 1 \\ 1 & 1 & 3 \end{vmatrix}$ and evaluate along the first row.

Step 3

Calculate $V = 2(3 \times 3 - 1) - 1(1 \times 3 - 1) + 1(1 - 3) = 2(8) - 1(2) + 1(-2) = 16 - 2 - 2 = 12$ cubic units.

Select an option & click Check Answer to reveal solution

Question #2JEE Main 2023

Let $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} + \hat{k}$ . If $\vec{c}$ is a vector such that $\vec{a} \cdot \vec{c} = 3$ and $\vec{a} \times \vec{c} = \vec{b}$ , then calculate the value of $\vec{a} \cdot (\vec{b} \times \vec{c})$ .

Solution Locked

Step-by-Step Solution & Concept

Step 1

Use scalar triple product properties: $\vec{a} \cdot (\vec{b} \times \vec{c}) = [\vec{a} \ \vec{b} \ \vec{c}] = -[\vec{a} \ \vec{c} \ \vec{b}]$ .

Step 2

Rewrite the expression as $-(\vec{a} \times \vec{c}) \cdot \vec{b}$ .

Step 3

Substitute given $\vec{a} \times \vec{c} = \vec{b}$ into the expression: $-(\vec{b} \cdot \vec{b}) = -|\vec{b}|^2 = -(2^2 + 1^2 + 1^2) = -6$ .

Select an option & click Check Answer to reveal solution

If you want to explore more chapters of Mathematics then:

Click Here

Mastering Vector Algebra: Formulas, Traps & Exam Strategy | ExamiFly

Chapter Overview

Core Subtopics & Syllabus Breakdown

Key Formulas & High-Yield Equations

Dot Product and Projection Formula

Cross Product & Vector Area

Scalar Triple Product (Box Product)

Vector Triple Product (VTP Rule)

Section Formula for Position Vectors

Common Mistakes & Traps to Avoid

Past Year Question (PYQ) Trends

Solved High-Yield Sample Questions

Frequently Asked Questions

Master Vector Algebra with 250+ Curated JEE Drills & AI Error-Correction