Chapter 10: Vector Algebra

Dear Class 12 Student! Welcome to the three-dimensional world of Vector Algebra. This chapter is the critical foundation not only for 3D Geometry in Mathematics but also for Mechanics and Electromagnetism in Physics. We will move beyond simple magnitudes and introduce "Direction". For the Boards, Direction Cosines and Cross/Dot products are heavily weighted. For JEE, the Scalar and Vector Triple Products are the ultimate deciders. Let's direct our focus!

1. Basic Concepts

Scalars: Physical quantities that have only magnitude (e.g., Mass, Volume, Temperature, Speed, Distance).
Vectors: Physical quantities that have both magnitude and a specific direction, and obey the laws of vector addition (e.g., Displacement, Velocity, Force, Acceleration).

Representation and Position Vector Representation: A vector is graphically represented by a directed line segment with an arrow. Denoted as $\vec{AB}$ (tail at A, head at B) or simply $\vec{a}$.
Magnitude (Length): The length of the line segment represents the magnitude, denoted by $|\vec{a}|$ or $a$. It is always a non-negative real number.

Position Vector: If $O(0,0,0)$ is the origin, the position of any point $P(x, y, z)$ in 3D space is given by the position vector $\vec{r} = \vec{OP}$.
$$\vec{r} = x\hat{i} + y\hat{j} + z\hat{k}$$ Where $\hat{i}, \hat{j}, \hat{k}$ are orthogonal unit vectors along the x, y, and z axes respectively.
Magnitude of $\vec{r}$: $\quad |\vec{r}| = \sqrt{x^2 + y^2 + z^2}$

Practice Problem 1: Magnitude Question: Find the magnitude of the vector $\vec{a} = 2\hat{i} - 3\hat{j} + 6\hat{k}$.

Solution:
The magnitude of $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$ is given by $\sqrt{x^2 + y^2 + z^2}$.
$|\vec{a}| = \sqrt{(2)^2 + (-3)^2 + (6)^2}$
$|\vec{a}| = \sqrt{4 + 9 + 36} = \sqrt{49} = \mathbf{7 \text{ units}}$.

2. Direction Cosines and Direction Ratios (Crucial Foundation)

This is the most important concept linking Vector Algebra to 3D Geometry.

Direction Angles ($\alpha, \beta, \gamma$): The angles made by the position vector $\vec{r}$ with the positive directions of the x, y, and z-axes respectively.
Direction Cosines ($l, m, n$): The cosine values of the direction angles.
$l = \cos\alpha, \quad m = \cos\beta, \quad n = \cos\gamma$.

Fundamental Relations 1. The Magic Identity: The sum of the squares of the direction cosines of any line/vector is always exactly 1.
$$l^2 + m^2 + n^2 = 1$$
2. Direction Ratios ($a, b, c$): Any three numbers that are directly proportional to the direction cosines are called direction ratios.
$a = \lambda l, \quad b = \lambda m, \quad c = \lambda n$.
Note: Unlike DCs, the sum of squares of DRs is generally NOT 1 ($a^2 + b^2 + c^2 \neq 1$).

3. Converting DRs to DCs: If $a, b, c$ are direction ratios of a vector, its direction cosines are:
$$l = \frac{a}{\sqrt{a^2+b^2+c^2}}, \quad m = \frac{b}{\sqrt{a^2+b^2+c^2}}, \quad n = \frac{c}{\sqrt{a^2+b^2+c^2}}$$

Practice Problem 2: Direction Cosines Question: Find the direction cosines of the vector $\vec{r} = \hat{i} + 2\hat{j} + 3\hat{k}$.

Solution:
1. The scalar components are the Direction Ratios (DRs): $a = 1, b = 2, c = 3$.
2. Calculate the magnitude of the vector:
$\sqrt{a^2 + b^2 + c^2} = \sqrt{1^2 + 2^2 + 3^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$.
3. Divide each DR by the magnitude to get the Direction Cosines (DCs):
$l = \frac{1}{\sqrt{14}}, \quad m = \frac{2}{\sqrt{14}}, \quad n = \frac{3}{\sqrt{14}}$.
Answer: The DCs are $\left( \frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}} \right)$.

3. Types of Vectors

Zero (Null) Vector ($\vec{0}$): A vector whose initial and terminal points coincide. It has zero magnitude and an arbitrary (undefined) direction.
Unit Vector ($\hat{a}$): A vector whose magnitude is exactly 1 unit. It is primarily used to specify a pure direction. Formula: $\mathbf{\hat{a} = \frac{\vec{a}}{|\vec{a}|}}$.
Co-initial Vectors: Two or more vectors starting from the exact same initial point.
Collinear (Parallel) Vectors: Vectors that are parallel to the same line, irrespective of their magnitudes or directions.
Condition: $\vec{a}$ and $\vec{b}$ are collinear if $\mathbf{\vec{a} = \lambda \vec{b}}$ (where $\lambda$ is a non-zero scalar).
Equal Vectors: Two vectors having the exact same magnitude AND the exact same direction.
Negative of a Vector: A vector with the same magnitude but perfectly opposite direction. ($\vec{BA} = -\vec{AB}$).

Practice Problem 3: Unit Vector Question: Find a unit vector in the direction of $\vec{a} = 3\hat{i} - 4\hat{j}$.

Solution:
1. Find the magnitude: $|\vec{a}| = \sqrt{3^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$.
2. Use the unit vector formula: $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$.
3. $\hat{a} = \frac{3\hat{i} - 4\hat{j}}{5} = \mathbf{\frac{3}{5}\hat{i} - \frac{4}{5}\hat{j}}$.

4. Algebra of Vectors

Triangle Law of Vector Addition

If two vectors are represented by two sides of a triangle in continuous sequence (head-to-tail), their sum (resultant) is represented by the third side taken in the opposite order (tail-to-head).

A. Vector Addition & Scalar Multiplication

Parallelogram Law: If two vectors act simultaneously at a point and are represented by adjacent sides of a parallelogram, their sum is the diagonal passing through that same point.
Properties of Addition: Commutative ($\vec{a} + \vec{b} = \vec{b} + \vec{a}$) and Associative ($(\vec{a}+\vec{b})+\vec{c} = \vec{a}+(\vec{b}+\vec{c})$).
Multiplication by a Scalar: Multiplying $\vec{a}$ by a scalar $\lambda$ yields a vector $\lambda\vec{a}$ whose magnitude is $|\lambda|$ times $|\vec{a}|$. If $\lambda$ is negative, the direction reverses.

B. Vector Joining Two Points

If point $P_1$ has coordinates $(x_1, y_1, z_1)$ and $P_2$ has coordinates $(x_2, y_2, z_2)$, then the vector drawn from $P_1$ to $P_2$ is:

$$\vec{P_1 P_2} = (x_2 - x_1)\hat{i} + (y_2 - y_1)\hat{j} + (z_2 - z_1)\hat{k}$$

Practice Problem 4: Vector Joining Points Question: Find the vector joining the points $P(2, 3, 0)$ and $Q(-1, -2, -4)$ directed from $P$ to $Q$.

Solution:
1. The vector is $\vec{PQ}$. We subtract coordinates of Initial point ($P$) from Terminal point ($Q$).
2. $\vec{PQ} = (x_Q - x_P)\hat{i} + (y_Q - y_P)\hat{j} + (z_Q - z_P)\hat{k}$
3. $\vec{PQ} = (-1 - 2)\hat{i} + (-2 - 3)\hat{j} + (-4 - 0)\hat{k}$
4. $\vec{PQ} = \mathbf{-3\hat{i} - 5\hat{j} - 4\hat{k}}$.

C. Section Formula Let $\vec{a}$ and $\vec{b}$ be the position vectors of points $A$ and $B$. The position vector $\vec{r}$ of a point $R$ dividing the line segment $AB$ in the ratio $m : n$ is:

Internal Division: $\quad \mathbf{\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n}}$

External Division: $\quad \mathbf{\vec{r} = \frac{m\vec{b} - n\vec{a}}{m - n}}$

Midpoint Formula: (Special case where $m = n = 1$)
$\quad \mathbf{\vec{r} = \frac{\vec{a} + \vec{b}}{2}}$

Practice Problem 5: Section Formula Question: Find the position vector of a point $R$ which divides the line joining two points $P$ and $Q$ whose position vectors are $\hat{i} + 2\hat{j} - \hat{k}$ and $-\hat{i} + \hat{j} + \hat{k}$ respectively, in the ratio $2:1$ internally.

Solution:
1. Given $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ (for P), $\vec{b} = -\hat{i} + \hat{j} + \hat{k}$ (for Q). Ratio $m=2, n=1$.
2. Internal division formula: $\vec{r} = \frac{m\vec{b} + n\vec{a}}{m + n}$.
3. $\vec{r} = \frac{2(-\hat{i} + \hat{j} + \hat{k}) + 1(\hat{i} + 2\hat{j} - \hat{k})}{2 + 1}$
4. $\vec{r} = \frac{-2\hat{i} + 2\hat{j} + 2\hat{k} + \hat{i} + 2\hat{j} - \hat{k}}{3}$
5. $\vec{r} = \mathbf{\frac{-\hat{i} + 4\hat{j} + \hat{k}}{3}}$.

5. Scalar (Dot) Product of Two Vectors

Definition and Properties The scalar product of two non-zero vectors $\vec{a}$ and $\vec{b}$ is defined as:
$$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$$ Where $\theta$ is the angle between them ($0 \le \theta \le \pi$). The final result is a purely real number (SCALAR).

Key Properties:
- Commutative: $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$.
- Distributive: $\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}$.
- Self Dot Product: $\mathbf{\vec{a} \cdot \vec{a} = |\vec{a}|^2}$.
- Orthogonal Unit Vectors: $\hat{i}\cdot\hat{i} = \hat{j}\cdot\hat{j} = \hat{k}\cdot\hat{k} = 1$ because $\cos(0^\circ) = 1$.
- Cross-axis Dot Product: $\hat{i}\cdot\hat{j} = \hat{j}\cdot\hat{k} = \hat{k}\cdot\hat{i} = 0$ because $\cos(90^\circ) = 0$.
- Component Form: $\vec{a} \cdot \vec{b} = (a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \cdot (b_1\hat{i} + b_2\hat{j} + b_3\hat{k}) = \mathbf{a_1 b_1 + a_2 b_2 + a_3 b_3}$.

Crucial Condition: Two non-zero vectors are Perpendicular (Orthogonal) if and only if their dot product is zero ($\mathbf{\vec{a} \cdot \vec{b} = 0}$).

Angle between Vectors: $\mathbf{\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}}$.

Practice Problem 6: Angle Calculation Question: Find the angle between the vectors $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$.

Solution:
1. Find the dot product: $\vec{a} \cdot \vec{b} = (1)(1) + (1)(-1) + (-1)(1) = 1 - 1 - 1 = -1$.
2. Find magnitude of $\vec{a}$: $|\vec{a}| = \sqrt{1^2 + 1^2 + (-1)^2} = \sqrt{3}$.
3. Find magnitude of $\vec{b}$: $|\vec{b}| = \sqrt{1^2 + (-1)^2 + 1^2} = \sqrt{3}$.
4. Use cosine formula: $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} = \frac{-1}{\sqrt{3} \cdot \sqrt{3}} = -\frac{1}{3}$.
5. Angle $\theta = \mathbf{\cos^{-1}\left(-\frac{1}{3}\right)}$.

Projection of a Vector (High Board Priority) The projection of vector $\vec{a}$ onto vector $\vec{b}$ is the "shadow" cast by $\vec{a}$ along the line of $\vec{b}$.

1. Scalar Projection of $\vec{a}$ on $\vec{b}$:
This is just the length of the shadow. Formula: $\mathbf{\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}}$ (or simply $\vec{a} \cdot \hat{b}$).

2. Vector Projection of $\vec{a}$ on $\vec{b}$:
This gives the shadow both magnitude and direction along $\vec{b}$. Formula: $\mathbf{\left( \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \right) \vec{b}}$.

Practice Problem 7: Scalar Projection Question: Find the projection of the vector $\vec{a} = 2\hat{i} + 3\hat{j} + 2\hat{k}$ on the vector $\vec{b} = \hat{i} + 2\hat{j} + \hat{k}$.

Solution:
1. Projection formula is $\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$.
2. Find dot product $\vec{a} \cdot \vec{b} = (2)(1) + (3)(2) + (2)(1) = 2 + 6 + 2 = 10$.
3. Find magnitude of $\vec{b}$: $|\vec{b}| = \sqrt{1^2 + 2^2 + 1^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$.
4. Projection $= \mathbf{\frac{10}{\sqrt{6}}}$. (Can be rationalized to $\frac{5\sqrt{6}}{3}$).

6. Vector (Cross) Product of Two Vectors

Definition and Properties The cross product of two non-zero vectors $\vec{a}$ and $\vec{b}$ produces a new VECTOR that is perfectly perpendicular (orthogonal) to both $\vec{a}$ and $\vec{b}$.
$$\vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin\theta \, \hat{n}$$ Where $\hat{n}$ is a unit vector perpendicular to the plane containing $\vec{a}$ and $\vec{b}$, determined by the Right-Hand Thumb Rule.

Key Properties:
- Non-commutative: $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$. Reversing the order flips the direction of the resultant vector.
- Self Cross Product: $\mathbf{\vec{a} \times \vec{a} = \vec{0}}$ (because $\sin(0^\circ) = 0$).
- Collinear Condition: Two non-zero vectors are parallel/collinear if and only if $\mathbf{\vec{a} \times \vec{b} = \vec{0}}$.
- Orthogonal Unit Vectors: $\hat{i}\times\hat{i} = \hat{j}\times\hat{j} = \hat{k}\times\hat{k} = \vec{0}$.
- Cyclic Order: $\hat{i}\times\hat{j} = \hat{k}$, $\quad \hat{j}\times\hat{k} = \hat{i}$, $\quad \hat{k}\times\hat{i} = \hat{j}$. (Reverse order yields negative, e.g., $\hat{j}\times\hat{i} = -\hat{k}$).

Component Form (Determinant Method): To calculate the cross product easily without doing 9 separate multiplications, use a determinant:
$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$$

Practice Problem 8: Cross Product Question: Find a unit vector perpendicular to both vectors $\vec{a} = 2\hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + 2\hat{k}$.

Solution:
1. A vector perpendicular to both is their cross product $\vec{c} = \vec{a} \times \vec{b}$.
2. Calculate $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & -1 & 2 \end{vmatrix}$.
$= \hat{i}(2 - (-1)) - \hat{j}(4 - 1) + \hat{k}(-2 - 1)$
$= \hat{i}(3) - \hat{j}(3) + \hat{k}(-3) = 3\hat{i} - 3\hat{j} - 3\hat{k}$.
3. We need a unit vector. So we must divide $\vec{c}$ by its magnitude.
$|\vec{c}| = \sqrt{3^2 + (-3)^2 + (-3)^2} = \sqrt{9 + 9 + 9} = \sqrt{27} = 3\sqrt{3}$.
4. Unit vector $\hat{n} = \frac{\vec{c}}{|\vec{c}|} = \frac{3\hat{i} - 3\hat{j} - 3\hat{k}}{3\sqrt{3}} = \mathbf{\frac{1}{\sqrt{3}}\hat{i} - \frac{1}{\sqrt{3}}\hat{j} - \frac{1}{\sqrt{3}}\hat{k}}$.
(Note: Multiplying this answer by -1 also gives a valid opposite-facing unit vector).

Geometrical Applications of Cross Product (High Priority) The magnitude of the cross product gives us areas of basic geometric shapes:

1. Area of a Parallelogram:
- If adjacent sides are given by vectors $\vec{a}$ and $\vec{b}$: $\mathbf{\text{Area} = |\vec{a} \times \vec{b}|}$.
- If diagonals are given by vectors $\vec{d_1}$ and $\vec{d_2}$: $\mathbf{\text{Area} = \frac{1}{2} |\vec{d_1} \times \vec{d_2}|}$.

2. Area of a Triangle:
- If adjacent sides are given by vectors $\vec{a}$ and $\vec{b}$: $\mathbf{\text{Area} = \frac{1}{2} |\vec{a} \times \vec{b}|}$.

Practice Problem 9: Area of Triangle Question: Find the area of a triangle having vertices $A(1, 1, 1)$, $B(1, 2, 3)$, and $C(2, 3, 1)$.

Solution:
1. First, find two adjacent side vectors, say $\vec{AB}$ and $\vec{AC}$.
$\vec{AB} = (1-1)\hat{i} + (2-1)\hat{j} + (3-1)\hat{k} = 0\hat{i} + 1\hat{j} + 2\hat{k}$.
$\vec{AC} = (2-1)\hat{i} + (3-1)\hat{j} + (1-1)\hat{k} = 1\hat{i} + 2\hat{j} + 0\hat{k}$.
2. Find their cross product $\vec{AB} \times \vec{AC}$:
$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 0 & 1 & 2 \\ 1 & 2 & 0 \end{vmatrix} = \hat{i}(0 - 4) - \hat{j}(0 - 2) + \hat{k}(0 - 1) = -4\hat{i} + 2\hat{j} - \hat{k}$.
3. Find the magnitude of the cross product:
$|\vec{AB} \times \vec{AC}| = \sqrt{(-4)^2 + 2^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21}$.
4. Area of Triangle $= \frac{1}{2} |\vec{AB} \times \vec{AC}| = \mathbf{\frac{\sqrt{21}}{2} \text{ sq. units}}$.

7. Scalar Triple Product (R.D. Sharma / JEE Focus)

Note: While often rationalized in basic NCERT exams, the Scalar Triple Product (STP) is an absolute non-negotiable requirement for JEE Mains and Advanced.

Definition: The dot product of one vector with the cross product of two other vectors is called the scalar triple product. It is denoted by a box notation.
$$[\vec{a} \, \vec{b} \, \vec{c}] = \vec{a} \cdot (\vec{b} \times \vec{c})$$ The final result is a purely real number (SCALAR).

Calculation and Geometry of STP Component Form (Determinant Method): Instead of doing a cross product and then a dot product, simply evaluate a $3 \times 3$ determinant formed by the coefficients of the three vectors.
$$[\vec{a} \, \vec{b} \, \vec{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$$

Geometrical Interpretation: The absolute value of the scalar triple product represents the Volume of a Parallelepiped whose three adjacent edges are represented by the vectors $\vec{a}, \vec{b}, \text{ and } \vec{c}$.
$\text{Volume} = |[\vec{a} \, \vec{b} \, \vec{c}]|$.

Crucial Coplanarity Condition: Three non-zero vectors $\vec{a}, \vec{b}, \text{ and } \vec{c}$ lie in the exact same plane (i.e., they are coplanar) IF AND ONLY IF their scalar triple product is exactly zero. (Because a flat plane has zero volume!).
$$[\vec{a} \, \vec{b} \, \vec{c}] = 0 \iff \text{Vectors are Coplanar.}$$

Practice Problem 10: Coplanarity via STP Question: Show that the vectors $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$, $\vec{b} = -2\hat{i} + 3\hat{j} - 4\hat{k}$, and $\vec{c} = \hat{i} - 3\hat{j} + 5\hat{k}$ are coplanar.

Solution:
Three vectors are coplanar if their Scalar Triple Product is zero.
$[\vec{a} \, \vec{b} \, \vec{c}] = \begin{vmatrix} 1 & -2 & 3 \\ -2 & 3 & -4 \\ 1 & -3 & 5 \end{vmatrix}$
Expanding along Row 1:
$= 1(15 - 12) - (-2)(-10 - (-4)) + 3(6 - 3)$
$= 1(3) + 2(-10 + 4) + 3(3)$
$= 3 + 2(-6) + 9$
$= 3 - 12 + 9 = 12 - 12 = \mathbf{0}$.
Since the STP is zero, the three given vectors are perfectly coplanar.

8. Vector Triple Product (Strictly JEE Advanced)

Definition: The cross product of a vector with the cross product of two other vectors. It results in a new VECTOR.
$\vec{a} \times (\vec{b} \times \vec{c})$.
Geometrically: The resulting vector is coplanar with vectors $\vec{b}$ and $\vec{c}$, and is perfectly perpendicular to vector $\vec{a}$.

The BAC - CAB Rule Expanding a vector triple product using a determinant is tedious. Use the fundamental BAC-CAB expansion formula:
$$\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$$ Note: The terms in parentheses are scalar dot products, which then multiply the vectors outside.

Practice Problem 11: VTP Expansion Question: Simplify the expression $\hat{i} \times (\hat{j} \times \hat{k}) + \hat{j} \times (\hat{k} \times \hat{i}) + \hat{k} \times (\hat{i} \times \hat{j})$.

Solution:
We can solve this rapidly using the cyclic properties of the cross product.
1. First term: $\hat{j} \times \hat{k} = \hat{i}$. So, $\hat{i} \times (\hat{j} \times \hat{k}) = \hat{i} \times \hat{i} = \vec{0}$.
2. Second term: $\hat{k} \times \hat{i} = \hat{j}$. So, $\hat{j} \times (\hat{k} \times \hat{i}) = \hat{j} \times \hat{j} = \vec{0}$.
3. Third term: $\hat{i} \times \hat{j} = \hat{k}$. So, $\hat{k} \times (\hat{i} \times \hat{j}) = \hat{k} \times \hat{k} = \vec{0}$.
Adding them all together: $\vec{0} + \vec{0} + \vec{0} = \mathbf{\vec{0}}$.