3.1 Vector Algebra

Detailed Theory: Vector Algebra

1. Basic Concepts of Vectors

1.1 What is a Vector?

A vector is a mathematical object that has both magnitude (size) and direction.

Examples from real life:

• Force (push/pull in a specific direction)

• Velocity (speed with direction)

• Displacement (change in position)

1.2 Scalar vs Vector

Scalar: Only magnitude (no direction)

Examples: $5$ (a number), $10$ meters (distance), $20^\circ C$ (temperature)

Vector: Both magnitude and direction

Examples: $5$ meters North, $10$ m/s East, $15$ N downward

1.3 Representation of Vectors

a) Graphical Representation

A vector is shown as an arrow:

• Length represents magnitude

• Arrowhead shows direction

b) Algebraic Representation

In 2D: $\vec{v} = a\hat{i} + b\hat{j}$

In 3D: $\vec{v} = a\hat{i} + b\hat{j} + c\hat{k}$

where $\hat{i}, \hat{j}, \hat{k}$ are unit vectors along x, y, z axes.

c) Component Form

2D: $\vec{v} = \langle a, b \rangle$

3D: $\vec{v} = \langle a, b, c \rangle$

1.4 Types of Vectors

a) Zero Vector (Null Vector)

Magnitude = 0, direction undefined.

Notation: $\vec{0}$ or $0$

In component form: $\langle 0, 0 \rangle$ or $\langle 0, 0, 0 \rangle$

b) Unit Vector

Magnitude = 1.

Used to indicate direction only.

Finding unit vector: If $\vec{v}$ is a vector, its unit vector is:

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

c) Equal Vectors

Two vectors are equal if they have same magnitude and same direction.

d) Negative of a Vector

Same magnitude but opposite direction.

If $\vec{v} = \langle a, b, c \rangle$, then $-\vec{v} = \langle -a, -b, -c \rangle$

e) Position Vector

Vector from origin to a point.

If point is $P(x, y, z)$, position vector is:

$$\vec{OP} = x\hat{i} + y\hat{j} + z\hat{k}$$

f) Co-initial Vectors

Vectors starting from same point.

g) Collinear Vectors

Vectors parallel to same line (or lying on same line).

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2. Vector Operations

2.1 Addition of Vectors

a) Triangle Law

If two vectors are represented as two sides of a triangle taken in order, their sum is represented by the third side.

Graphically: Place tail of second vector at head of first vector. The sum is vector from tail of first to head of second.

b) Parallelogram Law

If two vectors are represented as adjacent sides of a parallelogram, their sum is the diagonal through the common point.

c) Component-wise Addition

If $\vec{a} = \langle a_1, a_2, a_3 \rangle$ and $\vec{b} = \langle b_1, b_2, b_3 \rangle$, then:

$$\vec{a} + \vec{b} = \langle a_1 + b_1, a_2 + b_2, a_3 + b_3 \rangle$$

d) Properties of Vector Addition

1. Commutative: $\vec{a} + \vec{b} = \vec{b} + \vec{a}$

2. Associative: $(\vec{a} + \vec{b}) + \vec{c} = \vec{a} + (\vec{b} + \vec{c})$

3. Additive Identity: $\vec{a} + \vec{0} = \vec{a}$

4. Additive Inverse: $\vec{a} + (-\vec{a}) = \vec{0}$

2.2 Subtraction of Vectors

a) Geometric Method

$\vec{a} - \vec{b} = \vec{a} + (-\vec{b})$

b) Component Method

If $\vec{a} = \langle a_1, a_2, a_3 \rangle$ and $\vec{b} = \langle b_1, b_2, b_3 \rangle$, then:

$$\vec{a} - \vec{b} = \langle a_1 - b_1, a_2 - b_2, a_3 - b_3 \rangle$$

2.3 Scalar Multiplication

a) Definition

If $k$ is a scalar and $\vec{v}$ is a vector, then $k\vec{v}$ is:

• Magnitude = $|k| \times |\vec{v}|$

• Direction: same as $\vec{v}$ if $k > 0$, opposite if $k < 0$

b) Component Form

If $\vec{v} = \langle v_1, v_2, v_3 \rangle$, then:

$$k\vec{v} = \langle kv_1, kv_2, kv_3 \rangle$$

c) Properties

1. $k(\vec{a} + \vec{b}) = k\vec{a} + k\vec{b}$

2. $(k + l)\vec{a} = k\vec{a} + l\vec{a}$

3. $k(l\vec{a}) = (kl)\vec{a}$

4. $1 \cdot \vec{a} = \vec{a}$

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3. Magnitude and Direction

3.1 Magnitude (Length) of a Vector

For $\vec{v} = \langle a, b, c \rangle$:

2D:

$$|\vec{v}| = \sqrt{a^2 + b^2}$$

3D:

$$|\vec{v}| = \sqrt{a^2 + b^2 + c^2}$$

3.2 Direction Cosines

For 3D vector $\vec{v} = \langle a, b, c \rangle$:

Direction cosines are:

$$l = \cos\alpha = \frac{a}{|\vec{v}|}, \quad m = \cos\beta = \frac{b}{|\vec{v}|}, \quad n = \cos\gamma = \frac{c}{|\vec{v}|}$$

where $\alpha, \beta, \gamma$ are angles with x, y, z axes.

Property:

$$l^2 + m^2 + n^2 = 1$$

3.3 Direction Ratios

Three numbers proportional to direction cosines.

If direction cosines are $l, m, n$, then direction ratios are $a, b, c$ where:

$$l : m : n = a : b : c$$

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4. Dot Product (Scalar Product)

4.1 Definition

The dot product of two vectors $\vec{a}$ and $\vec{b}$ is:

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

where $\theta$ is angle between the vectors.

4.2 Component Form

If $\vec{a} = \langle a_1, a_2, a_3 \rangle$ and $\vec{b} = \langle b_1, b_2, b_3 \rangle$, then:

$$\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$$

4.3 Geometric Interpretation

1. Projection: $\vec{a} \cdot \vec{b} = |\vec{a}| \times (\text{projection of } \vec{b} \text{ on } \vec{a})$

2. Angle between vectors:

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

1. Orthogonality: $\vec{a} \cdot \vec{b} = 0 \iff \vec{a} \perp \vec{b}$ (vectors are perpendicular)

4.4 Properties of Dot Product

1. Commutative: $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$

2. Distributive: $\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}$

3. Scalar multiplication: $(k\vec{a}) \cdot \vec{b} = k(\vec{a} \cdot \vec{b}) = \vec{a} \cdot (k\vec{b})$

4. Self product: $\vec{a} \cdot \vec{a} = |\vec{a}|^2$

5. $\vec{a} \cdot \vec{a} = 0 \iff \vec{a} = \vec{0}$

4.5 Applications

a) Finding Angle Between Vectors

Example: Find angle between $\vec{a} = \hat{i} + 2\hat{j} - \hat{k}$ and $\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$

Solution:

$$\vec{a} \cdot \vec{b} = (1)(2) + (2)(-1) + (-1)(1) = 2 - 2 - 1 = -1$$

$$|\vec{a}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}$$

$$|\vec{b}| = \sqrt{2^2 + (-1)^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$$

$$\cos\theta = \frac{-1}{\sqrt{6} \times \sqrt{6}} = \frac{-1}{6}$$

$$\theta = \cos^{-1}\left(-\frac{1}{6}\right)$$

b) Checking Perpendicularity

Example: Are $\vec{a} = \langle 2, 3, 1 \rangle$ and $\vec{b} = \langle 1, -1, 1 \rangle$ perpendicular?

Solution:

$$\vec{a} \cdot \vec{b} = (2)(1) + (3)(-1) + (1)(1) = 2 - 3 + 1 = 0$$

Yes, they are perpendicular.

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5. Cross Product (Vector Product)

5.1 Definition

The cross product of two vectors $\vec{a}$ and $\vec{b}$ is:

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

where:

• $\theta$ is angle between vectors

• $\hat{n}$ is unit vector perpendicular to both $\vec{a}$ and $\vec{b}$

• Direction determined by right-hand rule

5.2 Right-Hand Rule

Point fingers in direction of $\vec{a}$, curl toward $\vec{b}$, thumb points in direction of $\vec{a} \times \vec{b}$.

5.3 Component Form

If $\vec{a} = \langle a_1, a_2, a_3 \rangle$ and $\vec{b} = \langle b_1, b_2, b_3 \rangle$, then:

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$$

Expanding the determinant:

$$\vec{a} \times \vec{b} = (a_2b_3 - a_3b_2)\hat{i} - (a_1b_3 - a_3b_1)\hat{j} + (a_1b_2 - a_2b_1)\hat{k}$$

5.4 Geometric Interpretation

1. Magnitude: $|\vec{a} \times \vec{b}| = |\vec{a}| |\vec{b}| \sin\theta$

2. Area of parallelogram with sides $\vec{a}$ and $\vec{b}$ = $|\vec{a} \times \vec{b}|$

3. Area of triangle with sides $\vec{a}$ and $\vec{b}$ = $\frac{1}{2}|\vec{a} \times \vec{b}|$

4. Parallel vectors: $\vec{a} \times \vec{b} = \vec{0} \iff \vec{a} \parallel \vec{b}$

5.5 Properties of Cross Product

1. Anti-commutative: $\vec{a} \times \vec{b} = -\vec{b} \times \vec{a}$

2. Distributive: $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$

3. Scalar multiplication: $(k\vec{a}) \times \vec{b} = k(\vec{a} \times \vec{b}) = \vec{a} \times (k\vec{b})$

4. Self product: $\vec{a} \times \vec{a} = \vec{0}$

5. Cross product with zero: $\vec{a} \times \vec{0} = \vec{0} \times \vec{a} = \vec{0}$

5.6 Applications

a) Finding Area

Example: Find area of parallelogram with adjacent sides $\vec{a} = 2\hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = \hat{i} + 3\hat{j} - 2\hat{k}$

Solution:

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -1 \\ 1 & 3 & -2 \end{vmatrix}$$

$$= \hat{i}[(1)(-2) - (-1)(3)] - \hat{j}[(2)(-2) - (-1)(1)] + \hat{k}[(2)(3) - (1)(1)]$$

$$= \hat{i}[-2 + 3] - \hat{j}[-4 + 1] + \hat{k}[6 - 1]$$

$$= \hat{i}(1) - \hat{j}(-3) + \hat{k}(5) = \hat{i} + 3\hat{j} + 5\hat{k}$$

Area = $|\vec{a} \times \vec{b}| = \sqrt{1^2 + 3^2 + 5^2} = \sqrt{1 + 9 + 25} = \sqrt{35}$

b) Finding Unit Perpendicular Vector

Example: Find unit vector perpendicular to both $\vec{a} = \hat{i} + \hat{j} + \hat{k}$ and $\vec{b} = 2\hat{i} + \hat{j} - \hat{k}$

Solution:

First find $\vec{a} \times \vec{b}$:

$$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 1 & -1 \end{vmatrix}$$

$$= \hat{i}[(1)(-1) - (1)(1)] - \hat{j}[(1)(-1) - (1)(2)] + \hat{k}[(1)(1) - (1)(2)]$$

$$= \hat{i}[-1 - 1] - \hat{j}[-1 - 2] + \hat{k}[1 - 2]$$

$$= \hat{i}(-2) - \hat{j}(-3) + \hat{k}(-1) = -2\hat{i} + 3\hat{j} - \hat{k}$$

Magnitude = $\sqrt{(-2)^2 + 3^2 + (-1)^2} = \sqrt{4 + 9 + 1} = \sqrt{14}$

Unit vector = $\frac{-2\hat{i} + 3\hat{j} - \hat{k}}{\sqrt{14}}$

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6. Scalar Triple Product

6.1 Definition

The scalar triple product of three vectors $\vec{a}, \vec{b}, \vec{c}$ is:

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

6.2 Determinant Form

If $\vec{a} = \langle a_1, a_2, a_3 \rangle$, $\vec{b} = \langle b_1, b_2, b_3 \rangle$, $\vec{c} = \langle c_1, c_2, c_3 \rangle$, then:

$$[\vec{a} \ \vec{b} \ \vec{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$$

6.3 Geometric Interpretation

1. Volume of parallelepiped with edges $\vec{a}, \vec{b}, \vec{c}$ = $|[\vec{a} \ \vec{b} \ \vec{c}]|$

2. Volume of tetrahedron with edges $\vec{a}, \vec{b}, \vec{c}$ = $\frac{1}{6}|[\vec{a} \ \vec{b} \ \vec{c}]|$

3. Coplanarity test: $[\vec{a} \ \vec{b} \ \vec{c}] = 0 \iff \vec{a}, \vec{b}, \vec{c}$ are coplanar

6.4 Properties

1. Cyclic property: $[\vec{a} \ \vec{b} \ \vec{c}] = [\vec{b} \ \vec{c} \ \vec{a}] = [\vec{c} \ \vec{a} \ \vec{b}]$

2. Anti-cyclic property: $[\vec{a} \ \vec{b} \ \vec{c}] = -[\vec{a} \ \vec{c} \ \vec{b}]$

3. Linear in each argument: $[k\vec{a} \ \vec{b} \ \vec{c}] = k[\vec{a} \ \vec{b} \ \vec{c}]$

4. If two vectors equal: $[\vec{a} \ \vec{a} \ \vec{c}] = 0$

6.5 Applications

a) Checking Coplanarity

Example: Check if vectors $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = 4\hat{i} + 5\hat{j} + 6\hat{k}$, $\vec{c} = 7\hat{i} + 8\hat{j} + 9\hat{k}$ are coplanar.

Solution:

Compute scalar triple product:

$$[\vec{a} \ \vec{b} \ \vec{c}] = \begin{vmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{vmatrix}$$

Using Sarrus' rule:

• Left-to-right: $(1)(5)(9) + (2)(6)(7) + (3)(4)(8) = 45 + 84 + 96 = 225$

• Right-to-left: $(3)(5)(7) + (1)(6)(8) + (2)(4)(9) = 105 + 48 + 72 = 225$

Determinant = $225 - 225 = 0$

So vectors are coplanar.

b) Finding Volume

Example: Find volume of parallelepiped with edges $\vec{a} = \hat{i} + \hat{j} + \hat{k}$, $\vec{b} = 2\hat{i} + 3\hat{j} - \hat{k}$, $\vec{c} = \hat{i} - \hat{j} + \hat{k}$

Solution:

$$[\vec{a} \ \vec{b} \ \vec{c}] = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 3 & -1 \\ 1 & -1 & 1 \end{vmatrix}$$

Expand along first row:

$$= 1\begin{vmatrix}3 & -1 \\ -1 & 1\end{vmatrix} - 1\begin{vmatrix}2 & -1 \\ 1 & 1\end{vmatrix} + 1\begin{vmatrix}2 & 3 \\ 1 & -1\end{vmatrix}$$

$$= 1(3\cdot1 - (-1)(-1)) - 1(2\cdot1 - (-1)(1)) + 1(2(-1) - 3\cdot1)$$

$$= 1(3 - 1) - 1(2 + 1) + 1(-2 - 3)$$

$$= 2 - 3 - 5 = -6$$

Volume = $|[\vec{a} \ \vec{b} \ \vec{c}]| = |-6| = 6$

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7. Vector Triple Product

7.1 Definition

The vector triple product of three vectors $\vec{a}, \vec{b}, \vec{c}$ is:

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

7.2 Important Identity (BAC-CAB Rule)

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

Mnemonic: "BAC minus CAB"

7.3 Properties

1. Not associative: $\vec{a} \times (\vec{b} \times \vec{c}) \neq (\vec{a} \times \vec{b}) \times \vec{c}$

2. Jacobi identity: $\vec{a} \times (\vec{b} \times \vec{c}) + \vec{b} \times (\vec{c} \times \vec{a}) + \vec{c} \times (\vec{a} \times \vec{b}) = \vec{0}$

3. Result lies in plane of $\vec{b}$ and $\vec{c}$

7.4 Application

Example: Simplify $\vec{a} \times (\vec{b} \times \vec{c}) + \vec{b} \times (\vec{c} \times \vec{a}) + \vec{c} \times (\vec{a} \times \vec{b})$

Solution:

Using BAC-CAB rule:

First term: $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$

Second term: $(\vec{b} \cdot \vec{a})\vec{c} - (\vec{b} \cdot \vec{c})\vec{a}$

Third term: $(\vec{c} \cdot \vec{b})\vec{a} - (\vec{c} \cdot \vec{a})\vec{b}$

Add all terms:

From first: $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$

From second: $(\vec{a} \cdot \vec{b})\vec{c} - (\vec{b} \cdot \vec{c})\vec{a}$

From third: $(\vec{b} \cdot \vec{c})\vec{a} - (\vec{a} \cdot \vec{c})\vec{b}$

Sum = $\vec{0}$

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8. Projection of Vectors

8.1 Scalar Projection

Scalar projection of $\vec{b}$ onto $\vec{a}$:

$$\text{comp}_{\vec{a}}\vec{b} = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}$$

This is the signed length of projection.

8.2 Vector Projection

Vector projection of $\vec{b}$ onto $\vec{a}$:

$$\text{proj}_{\vec{a}}\vec{b} = \left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right)\vec{a}$$

This is the vector component of $\vec{b}$ in direction of $\vec{a}$.

8.3 Orthogonal Component

Component of $\vec{b}$ perpendicular to $\vec{a}$:

$$\vec{b}_{\perp} = \vec{b} - \text{proj}_{\vec{a}}\vec{b}$$

Example: Find projection of $\vec{b} = 4\hat{i} + 5\hat{j} + 6\hat{k}$ onto $\vec{a} = \hat{i} + 2\hat{j} + 2\hat{k}$

Solution:

$$\vec{a} \cdot \vec{b} = (1)(4) + (2)(5) + (2)(6) = 4 + 10 + 12 = 26$$

$$|\vec{a}|^2 = 1^2 + 2^2 + 2^2 = 1 + 4 + 4 = 9$$

$$\text{proj}_{\vec{a}}\vec{b} = \frac{26}{9}\vec{a} = \frac{26}{9}(\hat{i} + 2\hat{j} + 2\hat{k}) = \frac{26}{9}\hat{i} + \frac{52}{9}\hat{j} + \frac{52}{9}\hat{k}$$

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9. Lines in Space

9.1 Vector Equation of a Line

Through point with position vector $\vec{a}$ and parallel to $\vec{b}$:

$$\vec{r} = \vec{a} + \lambda\vec{b}, \quad \lambda \in \mathbb{R}$$

where $\vec{r}$ is position vector of any point on line.

9.2 Parametric Form

If $\vec{a} = \langle x_1, y_1, z_1 \rangle$ and $\vec{b} = \langle a, b, c \rangle$, then:

$$x = x_1 + \lambda a, \quad y = y_1 + \lambda b, \quad z = z_1 + \lambda c$$

9.3 Symmetric Form

From parametric form (eliminating $\lambda$):

$$\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$$

9.4 Line Through Two Points

Through points with position vectors $\vec{a}$ and $\vec{b}$:

$$\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$$

or

$$\frac{x - x_1}{x_2 - x_1} = \frac{y - y_1}{y_2 - y_1} = \frac{z - z_1}{z_2 - z_1}$$

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10. Planes in Space

10.1 Vector Equation of a Plane

a) Through point $\vec{a}$ and perpendicular to $\vec{n}$:

$$(\vec{r} - \vec{a}) \cdot \vec{n} = 0$$

or

$$\vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$$

b) Through three points $\vec{a}, \vec{b}, \vec{c}$:

$$(\vec{r} - \vec{a}) \cdot [(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})] = 0$$

or

$$[\vec{r} - \vec{a} \ \vec{b} - \vec{a} \ \vec{c} - \vec{a}] = 0$$

10.2 Cartesian Equation

If $\vec{n} = \langle A, B, C \rangle$ and point is $(x_1, y_1, z_1)$, then:

$$A(x - x_1) + B(y - y_1) + C(z - z_1) = 0$$

or

$$Ax + By + Cz + D = 0$$

10.3 Intercept Form

If plane cuts x, y, z axes at $a, b, c$ respectively:

$$\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$$

10.4 Distance from Point to Plane

Point $P(x_0, y_0, z_0)$ to plane $Ax + By + Cz + D = 0$:

$$d = \frac{|Ax_0 + By_0 + Cz_0 + D|}{\sqrt{A^2 + B^2 + C^2}}$$

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11. Important Formulas Summary

11.1 Basic Operations

• Addition: $\vec{a} + \vec{b} = \langle a_1+b_1, a_2+b_2, a_3+b_3 \rangle$

• Subtraction: $\vec{a} - \vec{b} = \langle a_1-b_1, a_2-b_2, a_3-b_3 \rangle$

• Scalar multiplication: $k\vec{a} = \langle ka_1, ka_2, ka_3 \rangle$

• Magnitude: $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$

11.2 Dot Product

• $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 = |\vec{a}||\vec{b}|\cos\theta$

• Angle: $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$

• Orthogonal: $\vec{a} \cdot \vec{b} = 0 \iff \vec{a} \perp \vec{b}$

11.3 Cross Product

\hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}$$

• Magnitude: $|\vec{a} \times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta$

• Parallel: $\vec{a} \times \vec{b} = \vec{0} \iff \vec{a} \parallel \vec{b}$

11.4 Scalar Triple Product

a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{vmatrix}$$

• Coplanar: $[\vec{a} \ \vec{b} \ \vec{c}] = 0 \iff \vec{a}, \vec{b}, \vec{c}$ coplanar

11.5 Projection

• Scalar projection of $\vec{b}$ on $\vec{a}$: $\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}$

• Vector projection of $\vec{b}$ on $\vec{a}$: $\left(\frac{\vec{a} \cdot \vec{b}}{|\vec{a}|^2}\right)\vec{a}$

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12. Solved Examples

Example 1: Finding Unit Vector

Find unit vector in direction of $\vec{a} = 2\hat{i} - 3\hat{j} + 6\hat{k}$

Solution:

$$|\vec{a}| = \sqrt{2^2 + (-3)^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$$

Unit vector = $\frac{\vec{a}}{|\vec{a}|} = \frac{2\hat{i} - 3\hat{j} + 6\hat{k}}{7} = \frac{2}{7}\hat{i} - \frac{3}{7}\hat{j} + \frac{6}{7}\hat{k}$

Example 2: Angle Between Vectors

Find angle between $\vec{a} = \hat{i} + \hat{j} - \hat{k}$ and $\vec{b} = 2\hat{i} - \hat{j} + \hat{k}$

Solution:

$$\vec{a} \cdot \vec{b} = (1)(2) + (1)(-1) + (-1)(1) = 2 - 1 - 1 = 0$$

Since dot product = 0, vectors are perpendicular. Angle = $90^\circ$

Example 3: Area of Triangle

Find area of triangle with vertices $A(1,1,1)$, $B(2,3,5)$, $C(-1,0,2)$

Solution:

Vectors:

$$\vec{AB} = (2-1)\hat{i} + (3-1)\hat{j} + (5-1)\hat{k} = \hat{i} + 2\hat{j} + 4\hat{k}$$

$$\vec{AC} = (-1-1)\hat{i} + (0-1)\hat{j} + (2-1)\hat{k} = -2\hat{i} - \hat{j} + \hat{k}$$

Area = $\frac{1}{2}|\vec{AB} \times \vec{AC}|$

$$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 4 \\ -2 & -1 & 1 \end{vmatrix}$$

$$= \hat{i}[(2)(1) - (4)(-1)] - \hat{j}[(1)(1) - (4)(-2)] + \hat{k}[(1)(-1) - (2)(-2)]$$

$$= \hat{i}[2 + 4] - \hat{j}[1 + 8] + \hat{k}[-1 + 4]$$

$$= 6\hat{i} - 9\hat{j} + 3\hat{k}$$

Magnitude = $\sqrt{6^2 + (-9)^2 + 3^2} = \sqrt{36 + 81 + 9} = \sqrt{126}$

Area = $\frac{1}{2}\sqrt{126} = \frac{3\sqrt{14}}{2}$

Example 4: Line Equation

Find vector equation of line through $A(1,2,3)$ and $B(4,5,6)$

Solution:

Position vectors: $\vec{a} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\vec{b} = 4\hat{i} + 5\hat{j} + 6\hat{k}$

Direction vector: $\vec{b} - \vec{a} = 3\hat{i} + 3\hat{j} + 3\hat{k}$

Equation: $\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$

$$\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(3\hat{i} + 3\hat{j} + 3\hat{k})$$

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13. Exam Tips and Common Mistakes

13.1 Common Mistakes

1. Confusing dot and cross products: Dot gives scalar, cross gives vector

2. Direction of cross product: Use right-hand rule correctly

3. Unit vector calculation: Forgetting to divide by magnitude

4. Coplanarity test: Remember $[\vec{a} \ \vec{b} \ \vec{c}] = 0$ for coplanar vectors

5. Parallel vectors: $\vec{a} \times \vec{b} = \vec{0}$ for parallel vectors

13.2 Problem-Solving Strategy

1. Identify what's given and what's asked

2. Choose appropriate formula (dot, cross, triple product, etc.)

3. Compute step by step

4. Check answer for reasonableness

5. Include units if applicable

13.3 Quick Checks

1. Dot product result is always scalar

2. Cross product result is always vector perpendicular to both

3. Scalar triple product is scalar (volume)

4. Vector triple product is vector in plane of last two vectors

5. Unit vector has magnitude 1

This comprehensive theory covers all aspects of vector algebra with detailed explanations and examples, providing complete preparation for the entrance examination.