Vector Algebra
Vector Algebra for JEE Main & Advanced
Foundations of Vectors and Algebraic Structure
Scalars, Vectors, and Their ClassificationTopic 1
A physical quantity having only magnitude (size or number) is termed a scalar. Examples include mass, time, temperature, and arc length. In contrast, a quantity possessing both magnitude and a specific direction which obeys the parallelogram law of addition is termed a vector. Examples include displacement, velocity, acceleration, force, and torque.
A vector is typically denoted by a bold lowercase letter $\mathbf{a}$ or by directed segment notation $\overrightarrow{AB}$, where $A$ is the initial point (tail) and $B$ is the terminal point (head). The magnitude of vector $\mathbf{a}$ is denoted $|\mathbf{a}|$ and represents the length of the directed segment.
Classification of Vectors:- Zero (Null) Vector $\mathbf{0}$: A vector with zero magnitude and an indeterminate direction. The starting and terminal points coincide.
- Unit Vector $\hat{\mathbf{a}}$: A vector of magnitude unity in the direction of $\mathbf{a}$, defined as $\hat{\mathbf{a}} = \dfrac{\mathbf{a}}{|\mathbf{a}|}$, valid only when $|\mathbf{a}| \neq 0$.
- Equal Vectors: Two vectors are equal if they have the same magnitude and the same direction, regardless of initial point positions.
- Negative Vector: The vector $-\mathbf{a}$ has the same magnitude as $\mathbf{a}$ but opposite direction.
- Collinear (Parallel) Vectors: Vectors lying along the same line or parallel lines, irrespective of magnitude and sense. They satisfy $\mathbf{a} = \lambda \mathbf{b}$ for some scalar $\lambda \in \mathbb{R}$.
- Co-initial Vectors: Vectors sharing the same initial point.
- Coplanar Vectors: Three or more vectors lying in or parallel to the same plane.
- Free Vector: A vector whose position in space is arbitrary; only magnitude and direction matter.
- Localised Vector: A vector with a fixed initial point (used in mechanics for forces applied at specific points).
The position vector of a point $P(x, y, z)$ relative to the origin $O$ is $\overrightarrow{OP} = x\hat{i} + y\hat{j} + z\hat{k}$, with magnitude $|\overrightarrow{OP}| = \sqrt{x^2 + y^2 + z^2}$. Here, $\hat{i}$, $\hat{j}$, $\hat{k}$ are mutually perpendicular unit vectors along the positive $x$, $y$, $z$ axes respectively.
Find a unit vector in the direction of $\mathbf{a} = 2\hat{i} - 3\hat{j} + 6\hat{k}$, and also find a vector of magnitude $14$ in the same direction.
Show solution
First, compute the magnitude of the given vector: \[ |\mathbf{a}| = \sqrt{(2)^2 + (-3)^2 + (6)^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7 \] The unit vector in the direction of $\mathbf{a}$ is: \[ \hat{\mathbf{a}} = \frac{\mathbf{a}}{|\mathbf{a}|} = \frac{2\hat{i} - 3\hat{j} + 6\hat{k}}{7} = \frac{2}{7}\hat{i} - \frac{3}{7}\hat{j} + \frac{6}{7}\hat{k} \] A vector of magnitude $14$ in the same direction equals $14$ times the unit vector: \[ \mathbf{v} = 14 \hat{\mathbf{a}} = 14 \cdot \frac{1}{7}(2\hat{i} - 3\hat{j} + 6\hat{k}) = 2(2\hat{i} - 3\hat{j} + 6\hat{k}) = 4\hat{i} - 6\hat{j} + 12\hat{k} \] Final Answer: $\hat{\mathbf{a}} = \dfrac{1}{7}(2\hat{i} - 3\hat{j} + 6\hat{k})$ and $\mathbf{v} = 4\hat{i} - 6\hat{j} + 12\hat{k}$.
Vector Addition, Section Formula, and Linear CombinationsTopic 2
- Triangle Law: If the tail of $\mathbf{b}$ is placed at the head of $\mathbf{a}$, then $\mathbf{a} + \mathbf{b}$ is the vector from the tail of $\mathbf{a}$ to the head of $\mathbf{b}$.
- Parallelogram Law: If $\mathbf{a}$ and $\mathbf{b}$ are co-initial, then their sum is the diagonal of the parallelogram formed by $\mathbf{a}$ and $\mathbf{b}$.
- Commutativity: $\mathbf{a} + \mathbf{b} = \mathbf{b} + \mathbf{a}$
- Associativity: $(\mathbf{a} + \mathbf{b}) + \mathbf{c} = \mathbf{a} + (\mathbf{b} + \mathbf{c})$
- Existence of identity: $\mathbf{a} + \mathbf{0} = \mathbf{a}$
- Existence of inverse: $\mathbf{a} + (-\mathbf{a}) = \mathbf{0}$
Section Formula: If point $P$ divides the line segment joining points $A$ (position vector $\mathbf{a}$) and $B$ (position vector $\mathbf{b}$) in the ratio $m : n$, then the position vector of $P$ is: \[ \mathbf{r}_P = \frac{m\mathbf{b} + n\mathbf{a}}{m + n} \quad \text{(internal division)} \] \[ \mathbf{r}_P = \frac{m\mathbf{b} - n\mathbf{a}}{m - n} \quad \text{(external division)} \] The midpoint of $AB$ has position vector $\mathbf{r}_M = \dfrac{\mathbf{a} + \mathbf{b}}{2}$.
Linear Combination and Linear Independence: A vector $\mathbf{r}$ is a linear combination of vectors $\mathbf{a}_1, \mathbf{a}_2, \ldots, \mathbf{a}_n$ if there exist scalars $\lambda_1, \lambda_2, \ldots, \lambda_n$ such that $\mathbf{r} = \lambda_1 \mathbf{a}_1 + \lambda_2 \mathbf{a}_2 + \cdots + \lambda_n \mathbf{a}_n$.
Collinearity & Coplanarity Conditions:- Three points $A$, $B$, $C$ with position vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are collinear iff $\overrightarrow{AB} = \lambda \overrightarrow{AC}$, equivalently $\exists$ scalars $x, y, z$ (not all zero) such that $x\mathbf{a} + y\mathbf{b} + z\mathbf{c} = \mathbf{0}$ with $x + y + z = 0$.
- Four points $A, B, C, D$ are coplanar iff $\exists$ scalars $x, y, z, w$ (not all zero) such that $x\mathbf{a} + y\mathbf{b} + z\mathbf{c} + w\mathbf{d} = \mathbf{0}$ and $x + y + z + w = 0$.
The position vectors of points $A$, $B$, $C$ are $\mathbf{a} = \hat{i} + 2\hat{j} + 3\hat{k}$, $\mathbf{b} = -\hat{i} + \hat{j} + 8\hat{k}$, $\mathbf{c} = -4\hat{i} + 4\hat{j} + 6\hat{k}$. Find the position vector of point $D$ such that $ABCD$ forms a parallelogram (with $AB \parallel DC$).
Show solution
In a parallelogram $ABCD$, the diagonals $AC$ and $BD$ bisect each other. The midpoint of $AC$ equals the midpoint of $BD$. \[ \text{Midpoint of } AC = \frac{\mathbf{a} + \mathbf{c}}{2} = \frac{(\hat{i} + 2\hat{j} + 3\hat{k}) + (-4\hat{i} + 4\hat{j} + 6\hat{k})}{2} = \frac{-3\hat{i} + 6\hat{j} + 9\hat{k}}{2} \] Let position vector of $D$ be $\mathbf{d}$. Midpoint of $BD$: \[ \frac{\mathbf{b} + \mathbf{d}}{2} = \frac{-3\hat{i} + 6\hat{j} + 9\hat{k}}{2} \] \[ \implies \mathbf{d} = -3\hat{i} + 6\hat{j} + 9\hat{k} - \mathbf{b} = -3\hat{i} + 6\hat{j} + 9\hat{k} - (-\hat{i} + \hat{j} + 8\hat{k}) \] \[ = -3\hat{i} + 6\hat{j} + 9\hat{k} + \hat{i} - \hat{j} - 8\hat{k} = -2\hat{i} + 5\hat{j} + \hat{k} \] Final Answer: $\mathbf{d} = -2\hat{i} + 5\hat{j} + \hat{k}$.
Scalar (Dot) Product and Geometric Applications
Dot Product Definition, Properties and ProjectionsTopic 1
The scalar product (dot product) of two non-zero vectors $\mathbf{a}$ and $\mathbf{b}$, inclined at angle $\theta$ ($0 \leq \theta \leq \pi$), is the scalar: \[ \mathbf{a} \cdot \mathbf{b} = |\mathbf{a}||\mathbf{b}|\cos\theta \] If $\mathbf{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\mathbf{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, then in component form: \[ \mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3 \]
Key Properties:- Commutativity: $\mathbf{a} \cdot \mathbf{b} = \mathbf{b} \cdot \mathbf{a}$
- Distributivity: $\mathbf{a} \cdot (\mathbf{b} + \mathbf{c}) = \mathbf{a} \cdot \mathbf{b} + \mathbf{a} \cdot \mathbf{c}$
- Scalar multiplication: $(\lambda \mathbf{a}) \cdot \mathbf{b} = \lambda(\mathbf{a} \cdot \mathbf{b})$
- $\mathbf{a} \cdot \mathbf{a} = |\mathbf{a}|^2$
- Perpendicularity: $\mathbf{a} \perp \mathbf{b} \iff \mathbf{a} \cdot \mathbf{b} = 0$ (for non-zero vectors)
- For standard basis: $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$; $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$
Angle Between Two Vectors: \[ \cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} \]
Projection (Component) of $\mathbf{a}$ along $\mathbf{b}$: The scalar projection (signed length) of $\mathbf{a}$ on $\mathbf{b}$ is: \[ \text{Scalar projection of } \mathbf{a} \text{ on } \mathbf{b} = |\mathbf{a}|\cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|} \] The vector projection of $\mathbf{a}$ on $\mathbf{b}$ is: \[ \text{proj}_\mathbf{b}\mathbf{a} = \left(\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2}\right)\mathbf{b} = \left(\frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|}\right)\hat{\mathbf{b}} \]
Direction Cosines: If $\mathbf{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, then the direction cosines (with respect to coordinate axes) are: \[ \ell = \frac{a_1}{|\mathbf{a}|}, \quad m = \frac{a_2}{|\mathbf{a}|}, \quad n = \frac{a_3}{|\mathbf{a}|}, \quad \text{and} \quad \ell^2 + m^2 + n^2 = 1 \]
Useful Identity (Polarization): \[ |\mathbf{a} + \mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 + 2(\mathbf{a} \cdot \mathbf{b}) \] \[ |\mathbf{a} - \mathbf{b}|^2 = |\mathbf{a}|^2 + |\mathbf{b}|^2 - 2(\mathbf{a} \cdot \mathbf{b}) \] \[ |\mathbf{a} + \mathbf{b}|^2 + |\mathbf{a} - \mathbf{b}|^2 = 2(|\mathbf{a}|^2 + |\mathbf{b}|^2) \quad \text{(Parallelogram identity)} \]
If $\mathbf{a} = 2\hat{i} + 2\hat{j} - \hat{k}$ and $\mathbf{b} = 6\hat{i} - 3\hat{j} + 2\hat{k}$, find: (i) the angle between $\mathbf{a}$ and $\mathbf{b}$, (ii) the vector projection of $\mathbf{a}$ on $\mathbf{b}$.
Show solution
Compute $|\mathbf{a}|$, $|\mathbf{b}|$, and $\mathbf{a} \cdot \mathbf{b}$: \[ |\mathbf{a}| = \sqrt{4 + 4 + 1} = 3, \qquad |\mathbf{b}| = \sqrt{36 + 9 + 4} = 7 \] \[ \mathbf{a} \cdot \mathbf{b} = (2)(6) + (2)(-3) + (-1)(2) = 12 - 6 - 2 = 4 \] (i) The angle between $\mathbf{a}$ and $\mathbf{b}$: \[ \cos\theta = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{a}||\mathbf{b}|} = \frac{4}{3 \times 7} = \frac{4}{21} \implies \theta = \cos^{-1}\left(\frac{4}{21}\right) \] (ii) Vector projection of $\mathbf{a}$ on $\mathbf{b}$: \[ \text{proj}_\mathbf{b}\mathbf{a} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2}\mathbf{b} = \frac{4}{49}(6\hat{i} - 3\hat{j} + 2\hat{k}) = \frac{24}{49}\hat{i} - \frac{12}{49}\hat{j} + \frac{8}{49}\hat{k} \] Final Answer: $\theta = \cos^{-1}\left(\dfrac{4}{21}\right)$ and $\text{proj}_\mathbf{b}\mathbf{a} = \dfrac{4}{49}(6\hat{i} - 3\hat{j} + 2\hat{k})$.
Geometric Applications of the Dot ProductTopic 2
The dot product offers an elegant route to several classical results in geometry:
Cosine Rule (Vector Proof): In triangle $ABC$ with $\overrightarrow{BC} = \mathbf{a}$, $\overrightarrow{CA} = \mathbf{b}$, $\overrightarrow{AB} = \mathbf{c}$, the relation $\mathbf{a} + \mathbf{b} + \mathbf{c} = \mathbf{0}$ yields: \[ a^2 = b^2 + c^2 - 2bc \cos A \]
Diagonals of a Parallelogram: For a parallelogram with adjacent sides $\mathbf{a}$ and $\mathbf{b}$, the diagonals are $\mathbf{d}_1 = \mathbf{a} + \mathbf{b}$ and $\mathbf{d}_2 = \mathbf{a} - \mathbf{b}$. The diagonals are:- Equal in length iff $|\mathbf{a} + \mathbf{b}| = |\mathbf{a} - \mathbf{b}|$ iff $\mathbf{a} \cdot \mathbf{b} = 0$ (rectangle)
- Perpendicular iff $|\mathbf{a}| = |\mathbf{b}|$ (rhombus)
Equation of a Line: The vector equation of a line passing through point with position vector $\mathbf{a}$ and parallel to vector $\mathbf{b}$ is: \[ \mathbf{r} = \mathbf{a} + \lambda \mathbf{b}, \quad \lambda \in \mathbb{R} \] The line through two points $\mathbf{a}$ and $\mathbf{b}$ has equation: \[ \mathbf{r} = \mathbf{a} + \lambda(\mathbf{b} - \mathbf{a}) = (1 - \lambda)\mathbf{a} + \lambda \mathbf{b} \]
Component Resolution: Any vector $\mathbf{a}$ can be decomposed uniquely into components parallel and perpendicular to a unit vector $\hat{\mathbf{b}}$: \[ \mathbf{a} = (\mathbf{a} \cdot \hat{\mathbf{b}})\hat{\mathbf{b}} + \left[\mathbf{a} - (\mathbf{a} \cdot \hat{\mathbf{b}})\hat{\mathbf{b}}\right] \] The first term is parallel to $\mathbf{b}$, the bracket is perpendicular to $\mathbf{b}$.
Work Done by a Force: If a constant force $\mathbf{F}$ produces displacement $\mathbf{d}$, the work done is the scalar: \[ W = \mathbf{F} \cdot \mathbf{d} = |\mathbf{F}||\mathbf{d}|\cos\theta \]
Resolve the vector $\mathbf{a} = 2\hat{i} + \hat{j} - 3\hat{k}$ into components parallel and perpendicular to $\mathbf{b} = \hat{i} - 2\hat{j} + 2\hat{k}$.
Show solution
Compute the necessary quantities:
\[ \mathbf{a} \cdot \mathbf{b} = (2)(1) + (1)(-2) + (-3)(2) = 2 - 2 - 6 = -6 \]
\[ |\mathbf{b}|^2 = 1 + 4 + 4 = 9 \]
Component parallel to $\mathbf{b}$:
\[ \mathbf{a}_{\parallel} = \frac{\mathbf{a} \cdot \mathbf{b}}{|\mathbf{b}|^2}\mathbf{b} = \frac{-6}{9}(\hat{i} - 2\hat{j} + 2\hat{k}) = -\frac{2}{3}\hat{i} + \frac{4}{3}\hat{j} - \frac{4}{3}\hat{k} \]
Component perpendicular to $\mathbf{b}$:
\[ \mathbf{a}_{\perp} = \mathbf{a} - \mathbf{a}_{\parallel} = (2\hat{i} + \hat{j} - 3\hat{k}) - \left(-\frac{2}{3}\hat{i} + \frac{4}{3}\hat{j} - \frac{4}{3}\hat{k}\right) \]
\[ = \frac{8}{3}\hat{i} - \frac{1}{3}\hat{j} - \frac{5}{3}\hat{k} \]
Verification: $\mathbf{a}_{\perp} \cdot \mathbf{b} = \dfrac{8}{3} - \dfrac{-2}{3}(-1)\cdot\ldots = \dfrac{8}{3}(1) + \left(-\dfrac{1}{3}\right)(-2) + \left(-\dfrac{5}{3}\right)(2) = \dfrac{8 + 2 - 10}{3} = 0$. \checkmark
Final Answer: $\mathbf{a}_{\parallel} = -\dfrac{2}{3}\hat{i} + \dfrac{4}{3}\hat{j} - \dfrac{4}{3}\hat{k}$, $\mathbf{a}_{\perp} = \dfrac{8}{3}\hat{i} - \dfrac{1}{3}\hat{j} - \dfrac{5}{3}\hat{k}$.
Vector (Cross) Product and Area Applications
Cross Product Definition and PropertiesTopic 1
The vector product (cross product) of two non-zero vectors $\mathbf{a}$ and $\mathbf{b}$ at angle $\theta$ ($0 \leq \theta \leq \pi$) is defined as the vector: \[ \mathbf{a} \times \mathbf{b} = |\mathbf{a}||\mathbf{b}|\sin\theta \, \hat{\mathbf{n}} \] where $\hat{\mathbf{n}}$ is the unit vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$ such that $\{\mathbf{a}, \mathbf{b}, \hat{\mathbf{n}}\}$ forms a right-handed system.
Determinant Form (Coordinate Definition): If $\mathbf{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\mathbf{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, then: \[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix} = (a_2 b_3 - a_3 b_2)\hat{i} - (a_1 b_3 - a_3 b_1)\hat{j} + (a_1 b_2 - a_2 b_1)\hat{k} \]
Key Properties:- Anti-commutativity: $\mathbf{a} \times \mathbf{b} = -(\mathbf{b} \times \mathbf{a})$
- Distributivity: $\mathbf{a} \times (\mathbf{b} + \mathbf{c}) = \mathbf{a} \times \mathbf{b} + \mathbf{a} \times \mathbf{c}$
- $\mathbf{a} \times \mathbf{a} = \mathbf{0}$
- Parallelism: $\mathbf{a} \parallel \mathbf{b} \iff \mathbf{a} \times \mathbf{b} = \mathbf{0}$
- Standard basis: $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{i} = \hat{j}$ (cyclic)
- Lagrange's Identity: $|\mathbf{a} \times \mathbf{b}|^2 = |\mathbf{a}|^2|\mathbf{b}|^2 - (\mathbf{a} \cdot \mathbf{b})^2$
- $|\mathbf{a} \times \mathbf{b}|$ = area of parallelogram with adjacent sides $\mathbf{a}$ and $\mathbf{b}$
- Area of triangle with adjacent sides $\mathbf{a}, \mathbf{b}$ = $\dfrac{1}{2}|\mathbf{a} \times \mathbf{b}|$
- Area of triangle with vertices $A, B, C$ = $\dfrac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|$
- Unit vector perpendicular to both $\mathbf{a}$ and $\mathbf{b}$: $\hat{\mathbf{n}} = \pm \dfrac{\mathbf{a} \times \mathbf{b}}{|\mathbf{a} \times \mathbf{b}|}$
- Moment of force $\mathbf{F}$ about point $O$ with position vector $\mathbf{r}$: $\boldsymbol{\tau} = \mathbf{r} \times \mathbf{F}$
- Velocity of a point in rotation: $\mathbf{v} = \boldsymbol{\omega} \times \mathbf{r}$
Find the area of the triangle with vertices $A(1, -1, 2)$, $B(2, 1, -1)$, $C(3, -1, 2)$.
Show solution
Compute the vectors $\overrightarrow{AB}$ and $\overrightarrow{AC}$: \[ \overrightarrow{AB} = (2-1)\hat{i} + (1-(-1))\hat{j} + (-1-2)\hat{k} = \hat{i} + 2\hat{j} - 3\hat{k} \] \[ \overrightarrow{AC} = (3-1)\hat{i} + (-1-(-1))\hat{j} + (2-2)\hat{k} = 2\hat{i} + 0\hat{j} + 0\hat{k} = 2\hat{i} \] Compute the cross product: \[ \overrightarrow{AB} \times \overrightarrow{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & -3 \\ 2 & 0 & 0 \end{vmatrix} = \hat{i}(2 \cdot 0 - (-3) \cdot 0) - \hat{j}(1 \cdot 0 - (-3) \cdot 2) + \hat{k}(1 \cdot 0 - 2 \cdot 2) \] \[ = \hat{i}(0) - \hat{j}(6) + \hat{k}(-4) = -6\hat{j} - 4\hat{k} \] The magnitude is: \[ |\overrightarrow{AB} \times \overrightarrow{AC}| = \sqrt{0 + 36 + 16} = \sqrt{52} = 2\sqrt{13} \] The area of the triangle is: \[ \text{Area} = \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}| = \frac{1}{2}(2\sqrt{13}) = \sqrt{13} \] Final Answer: Area $= \sqrt{13}$ square units.
Scalar Triple Product (Box Product) and VolumeTopic 2
The scalar triple product (STP) of three vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$ is the scalar: \[ [\mathbf{a}\,\mathbf{b}\,\mathbf{c}] = \mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \times \mathbf{b}) \cdot \mathbf{c} \] In component form, if $\mathbf{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, etc.: \[ [\mathbf{a}\,\mathbf{b}\,\mathbf{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix} \]
Properties of STP:- Cyclic property: $[\mathbf{a}\,\mathbf{b}\,\mathbf{c}] = [\mathbf{b}\,\mathbf{c}\,\mathbf{a}] = [\mathbf{c}\,\mathbf{a}\,\mathbf{b}]$
- Sign change with interchange: $[\mathbf{a}\,\mathbf{b}\,\mathbf{c}] = -[\mathbf{b}\,\mathbf{a}\,\mathbf{c}]$
- Coplanarity test: $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are coplanar iff $[\mathbf{a}\,\mathbf{b}\,\mathbf{c}] = 0$
- If any two of $\mathbf{a}, \mathbf{b}, \mathbf{c}$ are equal or parallel, then $[\mathbf{a}\,\mathbf{b}\,\mathbf{c}] = 0$
- Scalar multiplication: $[\lambda \mathbf{a}\,\mathbf{b}\,\mathbf{c}] = \lambda [\mathbf{a}\,\mathbf{b}\,\mathbf{c}]$
- $[\mathbf{a} + \mathbf{d}\,\mathbf{b}\,\mathbf{c}] = [\mathbf{a}\,\mathbf{b}\,\mathbf{c}] + [\mathbf{d}\,\mathbf{b}\,\mathbf{c}]$
- Volume of tetrahedron with concurrent edges $\mathbf{a}, \mathbf{b}, \mathbf{c}$ from one vertex: $V = \dfrac{1}{6}|[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]|$
- Four points $A, B, C, D$ with position vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}, \mathbf{d}$ are coplanar iff $[\overrightarrow{AB}\,\overrightarrow{AC}\,\overrightarrow{AD}] = 0$
Vector Triple Product (BAC--CAB Rule): \[ \mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c} \] \[ (\mathbf{a} \times \mathbf{b}) \times \mathbf{c} = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{b} \cdot \mathbf{c})\mathbf{a} \] Note that the vector triple product is not associative; in general, $\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) \neq (\mathbf{a} \times \mathbf{b}) \times \mathbf{c}$.
Lagrange's Identity for STP: \[ [\mathbf{a} \times \mathbf{b}\,\,\mathbf{c} \times \mathbf{d}\,\,\mathbf{e} \times \mathbf{f}] \text{ (advanced expansion exists)} \] A more frequently used identity: \[ (\mathbf{a} \times \mathbf{b}) \cdot (\mathbf{c} \times \mathbf{d}) = \begin{vmatrix} \mathbf{a}\cdot\mathbf{c} & \mathbf{a}\cdot\mathbf{d} \\ \mathbf{b}\cdot\mathbf{c} & \mathbf{b}\cdot\mathbf{d} \end{vmatrix} \]
Show that the points with position vectors $\mathbf{a} = \hat{i} - \hat{j} + 3\hat{k}$, $\mathbf{b} = 2\hat{i} + 3\hat{j} - 4\hat{k}$, $\mathbf{c} = -3\hat{i} + 4\hat{j} - 7\hat{k}$, and $\mathbf{d} = 4\hat{i} + \hat{j} + 0\hat{k}$ are coplanar.
Show solution
Compute vectors from $A$: \[ \overrightarrow{AB} = \mathbf{b} - \mathbf{a} = (2-1)\hat{i} + (3-(-1))\hat{j} + (-4-3)\hat{k} = \hat{i} + 4\hat{j} - 7\hat{k} \] \[ \overrightarrow{AC} = \mathbf{c} - \mathbf{a} = -4\hat{i} + 5\hat{j} - 10\hat{k} \] \[ \overrightarrow{AD} = \mathbf{d} - \mathbf{a} = 3\hat{i} + 2\hat{j} - 3\hat{k} \] Compute the STP: \[ [\overrightarrow{AB}\,\overrightarrow{AC}\,\overrightarrow{AD}] = \begin{vmatrix} 1 & 4 & -7 \\ -4 & 5 & -10 \\ 3 & 2 & -3 \end{vmatrix} \] Expand along the first row: \[ = 1\begin{vmatrix} 5 & -10 \\ 2 & -3 \end{vmatrix} - 4\begin{vmatrix} -4 & -10 \\ 3 & -3 \end{vmatrix} + (-7)\begin{vmatrix} -4 & 5 \\ 3 & 2 \end{vmatrix} \] \[ = 1(-15 - (-20)) - 4(12 - (-30)) - 7(-8 - 15) \] \[ = 1(5) - 4(42) - 7(-23) = 5 - 168 + 161 = -2 \]
Since the STP $\neq 0$, the four points are not coplanar. (This illustrates the need to verify coplanarity through computation rather than assertion.)
Final Answer: The given points are not coplanar; $[\overrightarrow{AB}\,\overrightarrow{AC}\,\overrightarrow{AD}] = -2$.
Vector Equations of Lines, Planes & Reciprocal Systems
Vector Equations of Lines and Shortest DistanceTopic 1
A line in space is uniquely determined by a point and a direction vector. The vector equation of a line through a point with position vector $\mathbf{a}$ and parallel to $\mathbf{b}$ is: \[ \mathbf{r} = \mathbf{a} + \lambda \mathbf{b}, \quad \lambda \in \mathbb{R} \]
Distance from a Point to a Line: The perpendicular distance $d$ from a point $P$ with position vector $\mathbf{p}$ to the line $\mathbf{r} = \mathbf{a} + \lambda\mathbf{b}$ is: \[ d = \frac{|(\mathbf{p} - \mathbf{a}) \times \mathbf{b}|}{|\mathbf{b}|} \]
Shortest Distance between Two Skew Lines: Consider two lines: \[ L_1: \mathbf{r} = \mathbf{a}_1 + \lambda \mathbf{b}_1, \qquad L_2: \mathbf{r} = \mathbf{a}_2 + \mu \mathbf{b}_2 \] These lines are skew (non-intersecting and non-parallel) when $\mathbf{b}_1 \times \mathbf{b}_2 \neq \mathbf{0}$ and the common perpendicular has non-zero length. The shortest distance between them is: \[ d = \frac{|[\mathbf{a}_2 - \mathbf{a}_1\,\,\mathbf{b}_1\,\,\mathbf{b}_2]|}{|\mathbf{b}_1 \times \mathbf{b}_2|} = \frac{|(\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2)|}{|\mathbf{b}_1 \times \mathbf{b}_2|} \]
If $\mathbf{b}_1 \times \mathbf{b}_2 = \mathbf{0}$ (the lines are parallel), the shortest distance is computed using the point-to-line formula.
Condition for Coplanarity of Two Lines: Two lines $L_1$ and $L_2$ are coplanar iff: \[ [\mathbf{a}_2 - \mathbf{a}_1\,\,\mathbf{b}_1\,\,\mathbf{b}_2] = 0 \]
Vector Equation of a Plane:- Plane through point $\mathbf{a}$ with normal $\mathbf{n}$: $\quad (\mathbf{r} - \mathbf{a}) \cdot \mathbf{n} = 0$, equivalently $\mathbf{r} \cdot \mathbf{n} = \mathbf{a} \cdot \mathbf{n} = d$
- Plane through three non-collinear points $\mathbf{a}, \mathbf{b}, \mathbf{c}$: $\quad (\mathbf{r} - \mathbf{a}) \cdot [(\mathbf{b} - \mathbf{a}) \times (\mathbf{c} - \mathbf{a})] = 0$
- Plane through point $\mathbf{a}$ parallel to vectors $\mathbf{b}$ and $\mathbf{c}$: $\quad \mathbf{r} = \mathbf{a} + \lambda \mathbf{b} + \mu \mathbf{c}$
Distance from Point to Plane: The perpendicular distance from point $\mathbf{p}$ to the plane $\mathbf{r} \cdot \hat{\mathbf{n}} = d$ is: \[ \text{Distance} = |\mathbf{p} \cdot \hat{\mathbf{n}} - d| \]
Find the shortest distance between the lines: \[ L_1: \mathbf{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 4\hat{k}) \] \[ L_2: \mathbf{r} = (2\hat{i} + 4\hat{j} + 5\hat{k}) + \mu(3\hat{i} + 4\hat{j} + 5\hat{k}) \]
Show solution
Identify: $\mathbf{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$, $\mathbf{b}_1 = 2\hat{i} + 3\hat{j} + 4\hat{k}$, $\mathbf{a}_2 = 2\hat{i} + 4\hat{j} + 5\hat{k}$, $\mathbf{b}_2 = 3\hat{i} + 4\hat{j} + 5\hat{k}$.
Compute $\mathbf{a}_2 - \mathbf{a}_1 = \hat{i} + 2\hat{j} + 2\hat{k}$.
Compute $\mathbf{b}_1 \times \mathbf{b}_2$: \[ \mathbf{b}_1 \times \mathbf{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 3 & 4 & 5 \end{vmatrix} = \hat{i}(15 - 16) - \hat{j}(10 - 12) + \hat{k}(8 - 9) = -\hat{i} + 2\hat{j} - \hat{k} \] \[ |\mathbf{b}_1 \times \mathbf{b}_2| = \sqrt{1 + 4 + 1} = \sqrt{6} \] Compute the numerator: \[ (\mathbf{a}_2 - \mathbf{a}_1) \cdot (\mathbf{b}_1 \times \mathbf{b}_2) = (1)(-1) + (2)(2) + (2)(-1) = -1 + 4 - 2 = 1 \] Therefore: \[ d = \frac{|1|}{\sqrt{6}} = \frac{1}{\sqrt{6}} \] Final Answer: Shortest distance $= \dfrac{1}{\sqrt{6}}$ units.
Reciprocal System of Vectors and DifferentiationTopic 2
Reciprocal System of Vectors (JEE Advanced): Given three non-coplanar vectors $\mathbf{a}, \mathbf{b}, \mathbf{c}$, their reciprocal system $\mathbf{a}', \mathbf{b}', \mathbf{c}'$ is defined by: \[ \mathbf{a}' = \frac{\mathbf{b} \times \mathbf{c}}{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}, \quad \mathbf{b}' = \frac{\mathbf{c} \times \mathbf{a}}{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}, \quad \mathbf{c}' = \frac{\mathbf{a} \times \mathbf{b}}{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]} \]
Properties of Reciprocal Systems:- Biorthogonality: $\mathbf{a} \cdot \mathbf{a}' = \mathbf{b} \cdot \mathbf{b}' = \mathbf{c} \cdot \mathbf{c}' = 1$, and all other cross dot products vanish.
- Reciprocal STP: $[\mathbf{a}'\,\mathbf{b}'\,\mathbf{c}'] = \dfrac{1}{[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]}$
- The reciprocal of a reciprocal system gives the original system.
- Any vector $\mathbf{r}$ can be expanded as: \[ \mathbf{r} = (\mathbf{r} \cdot \mathbf{a}')\mathbf{a} + (\mathbf{r} \cdot \mathbf{b}')\mathbf{b} + (\mathbf{r} \cdot \mathbf{c}')\mathbf{c} \]
The reciprocal system is also called the dual basis and is fundamental in crystallography (used to describe reciprocal lattices).
Differentiation of Vector Functions: A vector-valued function $\mathbf{a}(t) = a_1(t)\hat{i} + a_2(t)\hat{j} + a_3(t)\hat{k}$ is differentiable if each component is differentiable. Then: \[ \frac{d\mathbf{a}}{dt} = \frac{da_1}{dt}\hat{i} + \frac{da_2}{dt}\hat{j} + \frac{da_3}{dt}\hat{k} \]
Rules of Vector Differentiation:- Linearity: $\dfrac{d}{dt}[\mathbf{a} + \mathbf{b}] = \dfrac{d\mathbf{a}}{dt} + \dfrac{d\mathbf{b}}{dt}$
- Scalar product rule: $\dfrac{d}{dt}[\phi(t)\mathbf{a}] = \phi'(t)\mathbf{a} + \phi(t)\dfrac{d\mathbf{a}}{dt}$
- Dot product rule: $\dfrac{d}{dt}[\mathbf{a} \cdot \mathbf{b}] = \dfrac{d\mathbf{a}}{dt} \cdot \mathbf{b} + \mathbf{a} \cdot \dfrac{d\mathbf{b}}{dt}$
- Cross product rule: $\dfrac{d}{dt}[\mathbf{a} \times \mathbf{b}] = \dfrac{d\mathbf{a}}{dt} \times \mathbf{b} + \mathbf{a} \times \dfrac{d\mathbf{b}}{dt}$ (order matters!)
- STP rule: $\dfrac{d}{dt}[\mathbf{a}\,\mathbf{b}\,\mathbf{c}] = [\mathbf{a}'\,\mathbf{b}\,\mathbf{c}] + [\mathbf{a}\,\mathbf{b}'\,\mathbf{c}] + [\mathbf{a}\,\mathbf{b}\,\mathbf{c}']$
Important Theorem: If $|\mathbf{a}(t)| = $ constant, then $\mathbf{a}(t) \cdot \dfrac{d\mathbf{a}}{dt} = 0$. Hence, a vector of constant magnitude is perpendicular to its derivative. This explains why position vectors on a sphere always have radial-tangential decomposition.
Find the reciprocal system to the vectors $\mathbf{a} = 2\hat{i} + 3\hat{j} - \hat{k}$, $\mathbf{b} = \hat{i} - \hat{j} - 2\hat{k}$, $\mathbf{c} = -\hat{i} + 2\hat{j} + 2\hat{k}$.
Show solution
First compute $[\mathbf{a}\,\mathbf{b}\,\mathbf{c}]$: \[ [\mathbf{a}\,\mathbf{b}\,\mathbf{c}] = \begin{vmatrix} 2 & 3 & -1 \\ 1 & -1 & -2 \\ -1 & 2 & 2 \end{vmatrix} = 2(-2+4) - 3(2-2) + (-1)(2-1) = 4 - 0 - 1 = 3 \] Compute the cross products: \[ \mathbf{b} \times \mathbf{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & -1 & -2 \\ -1 & 2 & 2 \end{vmatrix} = \hat{i}(-2 + 4) - \hat{j}(2 - 2) + \hat{k}(2 - 1) = 2\hat{i} + \hat{k} \] \[ \mathbf{c} \times \mathbf{a} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 2 \\ 2 & 3 & -1 \end{vmatrix} = \hat{i}(-2 - 6) - \hat{j}(1 - 4) + \hat{k}(-3 - 4) = -8\hat{i} + 3\hat{j} - 7\hat{k} \] \[ \mathbf{a} \times \mathbf{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & -1 \\ 1 & -1 & -2 \end{vmatrix} = \hat{i}(-6 - 1) - \hat{j}(-4 + 1) + \hat{k}(-2 - 3) = -7\hat{i} + 3\hat{j} - 5\hat{k} \] Therefore: \[ \mathbf{a}' = \frac{1}{3}(2\hat{i} + \hat{k}), \quad \mathbf{b}' = \frac{1}{3}(-8\hat{i} + 3\hat{j} - 7\hat{k}), \quad \mathbf{c}' = \frac{1}{3}(-7\hat{i} + 3\hat{j} - 5\hat{k}) \] Final Answer: As stated above; one can verify $\mathbf{a} \cdot \mathbf{a}' = 1$ as required.
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