Vector Algebra Class 12 Maths Notes

Key Vector Terminology

Scalar

A quantity that has only magnitude but no direction. Examples: mass, length, time, temperature.

Vector

A quantity that has both magnitude and direction. It is represented by a directed line segment. Examples: velocity, force, displacement.

Position Vector

A vector that represents the position of a point P in space with respect to a fixed origin O. It is denoted by \(\vec{OP}\).

Unit Vector

A vector with a magnitude of 1. A unit vector in the direction of a vector \(\vec{a}\) is denoted by \(\hat{a}\) and is calculated as \(\hat{a} = \frac{\vec{a}}{|\vec{a}|}\).

Collinear Vectors

Two or more vectors are collinear if they are parallel to the same line, irrespective of their magnitudes and directions. For two vectors \(\vec{a}\) and \(\vec{b}\), they are collinear if \(\vec{a} = \lambda \vec{b}\) for some non-zero scalar \(\lambda\).

Equal Vectors

Two vectors \(\vec{a}\) and \(\vec{b}\) are equal if they have the same magnitude and the same direction, regardless of their initial points.

Zero Vector

A vector with zero magnitude and an arbitrary direction. It is denoted by \(\vec{0}\).

Scalar (Dot) Product

The dot product of two vectors \(\vec{a}\) and \(\vec{b}\) is a scalar quantity defined as \(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos{\theta}\), where \(\theta\) is the angle between them.

Vector (Cross) Product

The cross product of two vectors \(\vec{a}\) and \(\vec{b}\) is a vector quantity defined as \(\vec{a} \times \vec{b} = (|\vec{a}| |\vec{b}| \sin{\theta}) \hat{n}\), where \(\theta\) is the angle between them and \(\hat{n}\) is a unit vector perpendicular to the plane containing \(\vec{a}\) and \(\vec{b}\).

Understanding Vectors and Their Components

A vector is fundamentally a directed line segment. It has a starting point (initial point) and an ending point (terminal point). The length of this segment represents the magnitude of the vector, and the arrow indicates its direction. For instance, if a vector \(\vec{AB}\) represents a displacement, its length is the distance moved, and its direction is the path taken from A to B.

In a 3D Cartesian coordinate system, we can define a position vector for any point P(x, y, z) with respect to the origin O(0, 0, 0). This vector, \(\vec{OP}\), can be broken down into its components along the x, y, and z axes. We use the standard unit vectors \(\hat{i}\), \(\hat{j}\), and \(\hat{k}\) for the x, y, and z directions, respectively. Thus, the vector \(\vec{OP}\) can be written in its component form as \(\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}\). Here, x, y, and z are the scalar components of the vector. The magnitude (or modulus) of this vector is its length, calculated using the distance formula: \(|\vec{r}| = \sqrt{x^2 + y^2 + z^2}\). This component form simplifies all vector operations, like addition, subtraction, and multiplication, turning geometric problems into algebraic ones.

Key Formulas & Properties

• {"point":"Magnitude of a Vector: For \\(\\vec{a} = a_1\\hat{i} + a_2\\hat{j} + a_3\\hat{k}\\), the magnitude is \\(|\\vec{a}| = \\sqrt{a_1^2 + a_2^2 + a_3^2}\\)."}
• {"point":"Unit Vector Formula: The unit vector \\(\\hat{a}\\) in the direction of \\(\\vec{a}\\) is \\(\\hat{a} = \\frac{\\vec{a}}{|\\vec{a}|}\\)."}
• {"point":"Section Formula (Internal Division): The position vector of a point R dividing the line segment joining points P and Q (with position vectors \\(\\vec{p}\\) and \\(\\vec{q}\\)) in the ratio m:n is \\(\\vec{r} = \\frac{m\\vec{q} + n\\vec{p}}{m+n}\\)."}
• {"point":"Dot Product (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} \\cdot \\vec{b} = a_1b_1 + a_2b_2 + a_3b_3\\)."}
• {"point":"Angle between two vectors: \\(\\cos{\\theta} = \\frac{\\vec{a} \\cdot \\vec{b}}{|\\vec{a}| |\\vec{b}|}\\). Vectors are perpendicular if \\(\\vec{a} \\cdot \\vec{b} = 0\\). They are parallel or collinear if \\(\\vec{a} \\times \\vec{b} = \\vec{0}\\)."}
• {"point":"Projection of \\(\\vec{a}\\) on \\(\\vec{b}\\): This is a scalar value given by \\(\\frac{\\vec{a} \\cdot \\vec{b}}{|\\vec{b}|}\\)."}
• {"point":"Cross Product (Determinant Form): \\(\\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}\\)."}
• {"point":"Geometric Meaning of Cross Product: The magnitude \\(|\\vec{a} \\times \\vec{b}|\\) represents the area of the parallelogram with adjacent sides \\(\\vec{a}\\) and \\(\\vec{b}\\)."}
• {"point":"Area of a Triangle: The area of a triangle with adjacent sides \\(\\vec{a}\\) and \\(\\vec{b}\\) is \\(\\frac{1}{2} |\\vec{a} \\times \\vec{b}|\\)."}

Scalar (Dot) Product vs. Vector (Cross) Product

Solved Examples

• Find the unit vector in the direction of the vector \(\vec{a} = 2\hat{i} + 3\hat{j} + \hat{k}\). First, find the magnitude: \(|\vec{a}| = \sqrt{2^2 + 3^2 + 1^2} = \sqrt{4 + 9 + 1} = \sqrt{14}\). The unit vector is \(\hat{a} = \frac{\vec{a}}{|\vec{a}|} = \frac{2\hat{i} + 3\hat{j} + \hat{k}}{\sqrt{14}} = \frac{2}{\sqrt{14}}\hat{i} + \frac{3}{\sqrt{14}}\hat{j} + \frac{1}{\sqrt{14}}\hat{k}\).
• Find the projection of vector \(\vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k}\) on the vector \(\vec{b} = \hat{i} + 2\hat{j} + \hat{k}\). Projection of \(\vec{a}\) on \(\vec{b}\) is \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\). \(\vec{a} \cdot \vec{b} = (2)(1) + (3)(2) + (2)(1) = 2 + 6 + 2 = 10\). \(|\vec{b}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{6}\). Projection = \(\frac{10}{\sqrt{6}}\).
• Find the area of the parallelogram whose adjacent sides are given by the vectors \(\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}\) and \(\vec{b} = \hat{i} - \hat{j} + \hat{k}\). Area = \(|\vec{a} \times \vec{b}|\). First, calculate \(\vec{a} \times \vec{b}\) using the determinant: \(\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix} = \hat{i}(1 - (-4)) - \hat{j}(3 - 4) + \hat{k}(-3 - 1) = 5\hat{i} + \hat{j} - 4\hat{k}\). Now, find its magnitude: \(|\vec{a} \times \vec{b}| = \sqrt{5^2 + 1^2 + (-4)^2} = \sqrt{25 + 1 + 16} = \sqrt{42}\). The area is \(\sqrt{42}\) square units.

Exam Tip: Dot vs. Cross Product Traps

A very common error is mixing up the properties and results of dot and cross products. Remember: Dot Product gives a scalar, Cross Product gives a vector. If a question asks for 'work done' (force · displacement) or 'angle between vectors', you need the dot product. If it asks for 'area of a parallelogram/triangle' or 'a vector perpendicular to two given vectors', you must use the cross product. Pay close attention to the order in the cross product: \(\vec{a} \times \vec{b}\) is not the same as \(\vec{b} \times \vec{a}\). Reversing the order introduces a negative sign. This is a frequent source of sign errors in exams.

Practice Questions with Solutions

• Q: What is the magnitude of the vector \(2\hat{i} - 7\hat{j} - 3\hat{k}\)? A: \(\sqrt{2^2 + (-7)^2 + (-3)^2} = \sqrt{4 + 49 + 9} = \sqrt{62}\).
• Q: If \(|\vec{a}| = \sqrt{26}\), \(|\vec{b}| = 7\), and \(|\vec{a} \times \vec{b}| = 35\), find \(\vec{a} \cdot \vec{b}\). A: We know \(|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin{\theta}\), so \(35 = \sqrt{26} \times 7 \times \sin{\theta} \implies \sin{\theta} = 5/\sqrt{26}\). Then \(\cos{\theta} = \sqrt{1 - \sin^2{\theta}} = \sqrt{1 - 25/26} = 1/\sqrt{26}\). So, \(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos{\theta} = \sqrt{26} \times 7 \times (1/\sqrt{26}) = 7\).
• Q: What does it mean if \(\vec{a} \cdot \vec{b} = 0\) for two non-zero vectors \(\vec{a}\) and \(\vec{b}\)? A: It means the vectors \(\vec{a}\) and \(\vec{b}\) are perpendicular (orthogonal) to each other, as \(\cos{\theta} = 0\), which implies \(\theta = 90^\circ\).
• Q: When are two vectors \(\vec{a}\) and \(\vec{b}\) collinear? A: When they are parallel to the same line. Algebraically, \(\vec{a} = \lambda \vec{b}\) for some non-zero scalar \(\lambda\). This is equivalent to their cross product being the zero vector: \(\vec{a} \times \vec{b} = \vec{0}\).

Frequently Asked Questions