Shortcut Methods

JEE Main

Modulus of a vector:

$| \vec{a} | = \sqrt{a_{x}^{2} + a_{y}^{2} + a_{z}^{2}}$

Unit vectors:

$\hat{a} = \frac{\vec{a}}{| \vec{a} |}$

$\hat{i} = \frac{\vec{i}}{| \vec{i} |} = (1, 0, 0)$

$\hat{j} = \frac{\vec{j}}{| \vec{j} |} = (0, 1, 0)$

$\hat{k} = \frac{\vec{k}}{| \vec{k} |} = (0, 0, 1)$

Direction Cosines:

The direction cosines of a vector (\vec{a}) with respect to the positive x-axis, y-axis, and z-axis respectively are: $\cos α = \frac{a_{x}}{| \vec{a} |}, \cos β = \frac{a_{y}}{| \vec{a} |}, \cos γ = \frac{a_{z}}{| \vec{a} |}$

where (a_{x}), (a_{y}), and (a_{z}) are the components of the vector in the x, y, and z directions, respectively.

Addition of vectors:

$\vec{a} + \vec{b} = (a_{x} + b_{x}) \hat{i} + (a_{y} + b_{y}) \hat{j} + (a_{z} + b_{z}) \hat{k}$

Subtraction of vectors:

$\vec{a} - \vec{b} = (a_{x} - b_{x}) \hat{i} + (a_{y} - b_{y}) \hat{j} + (a_{z} - b_{z}) \hat{k}$

Scalar product of vectors:

$\vec{a} \cdot \vec{b} = a_{x} b_{x} + a_{y} b_{y} + a_{z} b_{z}$

Vector product of vectors:

$\vec{a} \times \vec{b} = | \begin{matrix} i & \hat{j} & \hat{k} \\ a_{x} & a_{y} & a_{z} \\ b_{x} & b_{y} & b_{z} \end{matrix} | = (a_{y} b_{z} - a_{z} b_{y}) \hat{i} + (a_{z} b_{x} - a_{x} b_{z}) \hat{j} + (a_{x} b_{y} - a_{y} b_{x}) \hat{k}$

Angle between two vectors:

$θ = \cos^{- 1} (\frac{\vec{a} \cdot \vec{b}}{| \vec{a} | | \vec{b} |})$

CBSE Board Exam

Modulus of a vector: $| \vec{a} | = \sqrt{a_{1}^{2} + a_{2}^{2} + a_{3}^{2}}$
Unit vector: $\hat{a} = \frac{\vec{a}}{| \vec{a} |}$
Direction cosines: $\cos α = \frac{a_{1}}{| \vec{a} |,} \cos β = \frac{a_{2}}{| \vec{a} |}, \cos γ = \frac{a_{3}}{| \vec{a} |}$
Addition of vectors: $\vec{a} + \vec{b} = (a_{1} + b_{1}) \hat{i} + (a_{2} + b_{2}) \hat{j} + (a_{3} + b_{3}) \hat{k}$
Subtraction of vectors: $\vec{a} - \vec{b} = (a_{1} - b_{1}) \hat{i} + (a_{2} - b_{2}) \hat{j} + (a_{3} - b_{3}) \hat{k}$
Scalar product of vectors: $\vec{a} \cdot \vec{b} = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$
Angle between two vectors: $θ = \cos^{- 1} (\frac{\vec{a} \cdot \vec{b}}{| \vec{a} | | \vec{b} |}) α$

Shortcut Methods

JEE Main

CBSE Board Exam

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Mentorship

NCERT Exercise Video Solution

Shortcut Methods

JEE Main

CBSE Board Exam

Table of Contents

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Mentorship

NCERT Exercise Video Solution

Menu