Vector Algebra: CBSE Class 12 Maths NCERT Guide

Fundamental Concepts of Vectors

Scalar

A quantity that has only magnitude. Examples: length, mass, temperature, time.

Vector

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

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 \(\vec{OP}\).

Magnitude of a Vector

The length of the vector. For a vector \(\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}\), its magnitude is \(|\vec{a}| = \sqrt{x^2 + y^2 + z^2}\).

Unit Vector

A vector with a magnitude of 1. A unit vector in the direction of \(\vec{a}\) 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. Mathematically, \(\vec{a}\) and \(\vec{b}\) are collinear if \(\vec{b} = \lambda\vec{a}\) for some non-zero scalar \(\lambda\).

Vector Operations: Addition and Scalar Multiplication

Understanding how to manipulate vectors is key. The two fundamental operations are addition and scalar multiplication.

Vector Addition:
How do we add two forces or two displacements? We use vector addition. There are two main geometric rules:

1. Triangle Law: To add \(\vec{a}\) and \(\vec{b}\), place the tail of \(\vec{b}\) at the head of \(\vec{a}\). The resultant vector, \(\vec{a} + \vec{b}\), is the vector drawn from the tail of \(\vec{a}\) to the head of \(\vec{b}\). It forms the third side of the triangle.
2. Parallelogram Law: If two vectors \(\vec{a}\) and \(\vec{b}\) are represented by the two adjacent sides of a parallelogram starting from the same point, then their sum \(\vec{a} + \vec{b}\) is represented by the diagonal of the parallelogram starting from that same point.

Algebraically, if \(\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\) and \(\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\), their sum is simply: \(\vec{a} + \vec{b} = (a_1 + b_1)\hat{i} + (a_2 + b_2)\hat{j} + (a_3 + b_3)\hat{k}\).

Scalar Multiplication:
Multiplying a vector \(\vec{a}\) by a scalar (a real number) \(\lambda\) gives a new vector \(\lambda\vec{a}\). The magnitude of this new vector is \(|\lambda|\) times the magnitude of \(\vec{a}\). Its direction is the same as \(\vec{a}\) if \(\lambda > 0\), and opposite to \(\vec{a}\) if \(\lambda < 0\). Algebraically, \(\lambda\vec{a} = \lambda(a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) = (\lambda a_1)\hat{i} + (\lambda a_2)\hat{j} + (\lambda a_3)\hat{k}\).

Worked Examples: Dot and Cross Products

• Example 1: Find the angle between two vectors \(\vec{a} = 2\hat{i} + 2\hat{j} - \hat{k}\) and \(\vec{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}\). Step 1: Calculate the dot product \(\vec{a} \cdot \vec{b}\). The dot product is the sum of the products of the corresponding components. \(\vec{a} \cdot \vec{b} = (2)(6) + (2)(-3) + (-1)(2)\) \(\vec{a} \cdot \vec{b} = 12 - 6 - 2 = 4\) Step 2: Calculate the magnitudes of \(\vec{a}\) and \(\vec{b}\). \(|\vec{a}| = \sqrt{2^2 + 2^2 + (-1)^2} = \sqrt{4+4+1} = \sqrt{9} = 3\) \(|\vec{b}| = \sqrt{6^2 + (-3)^2 + 2^2} = \sqrt{36+9+4} = \sqrt{49} = 7\) Step 3: Use the dot product formula to find the angle \(\theta\). We know that \(\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta\). \(\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}\) \(\cos\theta = \frac{4}{(3)(7)} = \frac{4}{21}\) Final Answer: The angle between the vectors is \(\theta = \cos^{-1}\left(\frac{4}{21}\right)\).
• Example 2: Find a unit vector perpendicular to both \(\vec{a} = \hat{i} - 7\hat{j} + 7\hat{k}\) and \(\vec{b} = 3\hat{i} - 2\hat{j} + 2\hat{k}\). Step 1: Calculate the cross product \(\vec{a} \times \vec{b}\). The cross product gives a vector perpendicular to both \(\vec{a}\) and \(\vec{b}\). We use the determinant method. \(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -7 & 7 \\ 3 & -2 & 2 \end{vmatrix}\) \(= \hat{i}((-7)(2) - (7)(-2)) - \hat{j}((1)(2) - (7)(3)) + \hat{k}((1)(-2) - (-7)(3))\) \(= \hat{i}(-14 - (-14)) - \hat{j}(2 - 21) + \hat{k}(-2 - (-21))\) \(= \hat{i}(0) - \hat{j}(-19) + \hat{k}(19)\) Let's call this new vector \(\vec{c} = 19\hat{j} + 19\hat{k}\). Step 2: Find the magnitude of the resultant vector \(\vec{c}\). \(|\vec{c}| = |\vec{a} \times \vec{b}| = \sqrt{0^2 + 19^2 + 19^2} = \sqrt{2 \times 19^2} = 19\sqrt{2}\) Step 3: Calculate the unit vector. A unit vector perpendicular to \(\vec{a}\) and \(\vec{b}\) is given by \(\hat{c} = \frac{\vec{c}}{|\vec{c}|}\). \(\hat{c} = \frac{19\hat{j} + 19\hat{k}}{19\sqrt{2}} = \frac{19(\hat{j} + \hat{k})}{19\sqrt{2}} = \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}\) Final Answer: A unit vector perpendicular to both given vectors is \(\frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}\).

Exam Traps and Key Formulae

Vector algebra questions are generally straightforward if your concepts are clear, but some traps can cost you marks.

• Dot vs. Cross Product: The most common error is confusing the results. The dot product (\(\vec{a} \cdot \vec{b}\)) is a scalar (a number). The cross product (\(\vec{a} \times \vec{b}\)) is a vector which is perpendicular to both \(\vec{a}\) and \(\vec{b}\). Never write \(\vec{a} \cdot \vec{b} = 5\hat{i}\).
• Conditions for Perpendicular & Parallel:
• Two non-zero vectors \(\vec{a}\) and \(\vec{b}\) are perpendicular (orthogonal) if and only if their dot product is zero: \(\vec{a} \cdot \vec{b} = 0\).
• Two non-zero vectors \(\vec{a}\) and \(\vec{b}\) are parallel (collinear) if and only if their cross product is the zero vector: \(\vec{a} \times \vec{b} = \vec{0}\).
• Projection Formula: The projection of vector \(\vec{a}\) onto vector \(\vec{b}\) is a scalar value given by \(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\). Don't mix up the denominator; it's the magnitude of the vector you are projecting onto.
• Area of Parallelogram/Triangle: If adjacent sides of a parallelogram are \(\vec{a}\) and \(\vec{b}\), the area is \(|\vec{a} \times \vec{b}|\). The area of a triangle with these adjacent sides is \(\frac{1}{2}|\vec{a} \times \vec{b}|\). Remember to calculate the magnitude of the cross product, not just the cross product vector itself.

Practice Questions with Solutions

• Q: Find the unit vector in the direction of the sum of the vectors \(\vec{a} = 2\hat{i} - \hat{j} + 2\hat{k}\) and \(\vec{b} = -\hat{i} + \hat{j} + 3\hat{k}\). A: Step 1: Find the sum of the vectors. Let \(\vec{c} = \vec{a} + \vec{b}\). \(\vec{c} = (2\hat{i} - \hat{j} + 2\hat{k}) + (-\hat{i} + \hat{j} + 3\hat{k}) = (2-1)\hat{i} + (-1+1)\hat{j} + (2+3)\hat{k} = \hat{i} + 0\hat{j} + 5\hat{k} = \hat{i} + 5\hat{k}\). Step 2: Calculate the magnitude of the resultant vector \(\vec{c}\). \(|\vec{c}| = \sqrt{1^2 + 5^2} = \sqrt{1+25} = \sqrt{26}\). Step 3: Find the unit vector \(\hat{c}\) using the formula \(\hat{c} = \frac{\vec{c}}{|\vec{c}|}\). \(\hat{c} = \frac{\hat{i} + 5\hat{k}}{\sqrt{26}} = \frac{1}{\sqrt{26}}\hat{i} + \frac{5}{\sqrt{26}}\hat{k}\). Final answer: The required unit vector is \(\frac{1}{\sqrt{26}}\hat{i} + \frac{5}{\sqrt{26}}\hat{k}\).
• Q: Find the projection of the vector \(\vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k}\) on the vector \(\vec{b} = \hat{i} + 2\hat{j} + \hat{k}\). A: Step 1: Recall the formula for the projection of \(\vec{a}\) on \(\vec{b}\): \(\text{Proj}_\vec{b} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}\). Step 2: Calculate the dot product \(\vec{a} \cdot \vec{b}\). \(\vec{a} \cdot \vec{b} = (2)(1) + (3)(2) + (2)(1) = 2 + 6 + 2 = 10\). Step 3: Calculate the magnitude of \(\vec{b}\). \(|\vec{b}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{1+4+1} = \sqrt{6}\). Step 4: Substitute the values into the projection formula. \(\text{Proj}_\vec{b} \vec{a} = \frac{10}{\sqrt{6}}\). Final answer: The projection of \(\vec{a}\) on \(\vec{b}\) is \(\frac{10}{\sqrt{6}}\).
• Q: If \(\vec{a}\) is a unit vector and \((\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = 15\), find \(|\vec{x}|\). A: Step 1: Expand the dot product expression using distributive property. \((\vec{x} - \vec{a}) \cdot (\vec{x} + \vec{a}) = \vec{x} \cdot \vec{x} + \vec{x} \cdot \vec{a} - \vec{a} \cdot \vec{x} - \vec{a} \cdot \vec{a}\). Step 2: Simplify the expression. Since dot product is commutative (\(\vec{x} \cdot \vec{a} = \vec{a} \cdot \vec{x}\)), the middle terms cancel out. Also, \(\vec{v} \cdot \vec{v} = |\vec{v}|^2\). So, the expression becomes \(|\vec{x}|^2 - |\vec{a}|^2 = 15\). Step 3: Use the given information. We are given that \(\vec{a}\) is a unit vector, so \(|\vec{a}| = 1\). \(|\vec{x}|^2 - 1^2 = 15 \implies |\vec{x}|^2 - 1 = 15\). Step 4: Solve for \(|\vec{x}|\). \(|\vec{x}|^2 = 16 \implies |\vec{x}| = 4\) (Magnitude cannot be negative). Final answer: \(|\vec{x}| = 4\).
• Q: Find the area of the parallelogram whose adjacent sides are determined by the vectors \(\vec{a} = 3\hat{i} + \hat{j} + 4\hat{k}\) and \(\vec{b} = \hat{i} - \hat{j} + \hat{k}\). A: Step 1: Recall that the area of a parallelogram with adjacent sides \(\vec{a}\) and \(\vec{b}\) is given by \(|\vec{a} \times \vec{b}|\). Step 2: Calculate the cross product \(\vec{a} \times \vec{b}\). \(\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 1 & 4 \\ 1 & -1 & 1 \end{vmatrix}\) \(= \hat{i}((1)(1) - (4)(-1)) - \hat{j}((3)(1) - (4)(1)) + \hat{k}((3)(-1) - (1)(1))\) \(= \hat{i}(1 + 4) - \hat{j}(3 - 4) + \hat{k}(-3 - 1)\) \(= 5\hat{i} + \hat{j} - 4\hat{k}\). Step 3: Calculate the magnitude of the cross product vector. \(|\vec{a} \times \vec{b}| = \sqrt{5^2 + 1^2 + (-4)^2} = \sqrt{25 + 1 + 16} = \sqrt{42}\). Final answer: The area of the parallelogram is \(\sqrt{42}\) square units.

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