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5. Vectors

Unit vectors

A unit vector is a vector of magnitude exactly equal to one, serving as the fundamental building block for defining direction without scaling magnitude. This topic covers unit vectors, directional derivatives, and normalisation formulas.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade
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Definition and statement

Definition and Normalisation Principle of a Unit Vector
For any non-zero vector $\vec{a} \in \mathbb{R}^3$, the unit vector in the direction of $\vec{a}$, denoted by $\hat{a}$, is uniquely defined as \(\hat{a} = \frac{\vec{a}}{|\vec{a}|}\), satisfying the property $|\hat{a}| = 1$.

📐 Geometric Interpretation

Geometrically, any vector can be factored into a scalar magnitude and a unit vector representing pure direction: $\vec{a} = |\vec{a}|\hat{a}$. The unit vector $\hat{a}$ points to a specific location on the unit sphere centered at the origin.

⚙️ Algebraic & Structural Meaning

Algebraically, division by the Euclidean norm strips away scaling while preserving linear dependencies and directional angles (direction cosines), forming the backbone of orthogonal coordinate systems.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Non-zero Norm Requirement $|\vec{a}| \neq 0$ Strictly Mandatory / Necessary If $\vec{a} = \vec{0}$, division by $|\vec{a}| = 0$ results in an undefined zero-division error; a zero vector has no defined direction.
Euclidean Space Dimension Vector space $\mathbb{R}^n$ ($n \ge 1$) Strictly Mandatory Magnitude must be calculated using the standard inner product metric; changing the metric changes the definition of unit length.
Scalar Field Field Real scalar field $\mathbb{R}$ (or Complex $\mathbb{C}$ with modulus) Essential for scalar multiplication Failure of scalar field axioms prevents normalisation of components.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Rapid Normalisation of 3D Vectors

$$ \hat{a} = \frac{ai + bj + ck}{\sqrt{a^2 + b^2 + c^2}} $$

Rule: Sum the squares of coefficients mentally. If the sum is a perfect square (e.g., $9, 16, 25$), immediately write down the denominator and distribute.

Application: Directly tested in UP PGT to find unit vectors parallel to sum or difference of given vectors.
02

Sum of Squares of Direction Cosines

$$ l^2 + m^2 + n^2 = 1 \implies \cos^2\alpha + \cos^2\beta + \cos^2\gamma = 1 $$

Rule: If a unit vector makes angles $\alpha, \beta, \gamma$ with coordinate axes, its components are precisely $(\cos\alpha, \cos\beta, \cos\gamma)$, and their squared sum is identically 1.

Application: Frequently appears in TGT/PGT matching and missing angle problems.
03

Orthogonal Unit Vector Generation

$$ \text{If } \vec{a} = (a_1, a_2), \text{ then perp unit vectors are } \mp \frac{(-a_2, a_1)}{\sqrt{a_1^2 + a_2^2}} $$

Rule: Swap coordinates, negate one, and normalise to instantly find perpendicular unit vectors in 2D.

Application: Used in tangent-normal geometry and vector projection questions.
04

Magnitude of Sum of Unit Vectors

$$ |\hat{u} + \hat{v}| = 2\cos\left(\frac{\theta}{2}\right) $$

Rule: Use the dot product expansion $|\hat{u} + \hat{v}|^2 = |\hat{u}|^2 + |\hat{v}|^2 + 2\hat{u}\cdot\hat{v} = 2 + 2\cos\theta$ where $\theta$ is the angle between them.

Application: Standard result tested in KVS and NVS PGT vector summations.
05

Unit Vector Perpendicular to Two Vectors

$$ \hat{n} = \pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|} $$

Rule: Compute the cross product using standard determinant expansion and normalise immediately. Remember the $\pm$ sign for both directional orientations.

Application: High frequency in UP PGT vector cross-product applications.
06

Rapid Option Elimination via Magnitude Test

$$ \text{Test options: } \sum (x_i)^2 = 1 $$

Rule: Before solving complex vector equations, square the components of each given option. Any option whose sum of squared components does not equal 1 is instantly eliminated.

Application: Saves 40+ seconds on tricky multiple-choice elimination.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Forgetting the $\pm$ Sign in Perpendicular Unit Vectors

Common Pitfall: Candidates calculate $\frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}$ and forget that both parallel and anti-parallel unit vectors ($\pm$) are perpendicular to the plane.

Counterexample: For $\vec{i}$ and $\vec{j}$, cross product gives $\pm \vec{k}$.

Why it fails: Only listing one option causes loss of marks when options include both or $\pm$ notation.

🛡️ Exam Defense: Always check if the question asks for 'a' unit vector or 'the set of' unit vectors, and look for $\pm$ in the options.
⚠️ COMMON ERROR #02

Assuming Unit Vectors are Closed Under Addition

Common Pitfall: Assuming that if $|\hat{a}| = 1$ and $|\hat{b}| = 1$, then $|\hat{a} + \hat{b}| = 1$.

Counterexample: $|\hat{i} + \hat{i}| = |2\hat{i}| = 2 \neq 1$

Why it fails: Vector addition follows triangle law; magnitudes do not add algebraically unless vectors are collinear and in the same direction.

🛡️ Exam Defense: Never add magnitudes directly; always use dot product expansion: $|\hat{a} + \hat{b}| = \sqrt{2 + 2\cos\theta}$.

Comparison

Concept / TheoremHypotheses / AssumptionsKey Formula / OutcomeGeometric MeaningExam Identification Clue
Unit VectorNon-zero vector $\vec{a}$\(\hat{a} = \frac{\vec{a}}{|\vec{a}|}\)Direction indicator on unit sphereKeywords: 'unit vector in direction of'
Direction CosinesUnit vector components\(l^2 + m^2 + n^2 = 1\)Cosines of angles with coordinate axesKeywords: 'angles made with axes'
Cross Product Unit NormalNon-collinear vectors $\vec{a}, \vec{b}$\(\hat{n} = \pm \frac{\vec{a} \times \vec{b}}{|\vec{a} \times \vec{b}|}\)Perpendicular vector to planeKeywords: 'perpendicular to both vectors'
Position Vector NormalisationPoint $P(x, y, z)$ in space\(\hat{r} = \frac{x\hat{i} + y\hat{j} + z\hat{k}}{\sqrt{x^2+y^2+z^2}}\)Radial unit vector from originKeywords: 'radial direction', 'central force'

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

Find the unit vector perpendicular to both vectors $\vec{a} = 2\hat{i} + \hat{j} + \hat{k}$ and $\bve = \hat{i} - \hat{j} + 2\hat{k}$.

(A) $\frac{3\hat{i} - 3\hat{j} - 3\hat{k}}{\sqrt{27}}$(B) $\frac{\hat{i} - \hat{j} + \hat{k}}{\sqrt{3}}$(C) $\frac{2\hat{i} + \hat{j}}{\sqrt{5}}$(D) $\frac{\hat{i} + \hat{j} - \hat{k}}{\sqrt{3}}$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Compute cross product $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & 1 \\ 1 & -1 & 2 \end{vmatrix} = \hat{i}(2 - (-1)) - \hat{j}(4 - 1) + \hat{k}(-2 - 1) = 3\hat{i} - 3\hat{j} - 3\hat{k}$. Step 2: Calculate magnitude $|3\hat{i} - 3\hat{j} - 3\hat{k}| = \sqrt{9 + 9 + 9} = \sqrt{27} = 3\sqrt{3}$. Step 3: Divide vector by magnitude to get the unit vector.
⚡ Method B: 25-Second Exam Shortcut
Compute cross product components $(3, -3, -3)$. Check options for proportional vectors with magnitude equal to 1. Option (A) directly matches.
💡 Tip: Always simplify radical denominators during cross-product normalisation to match standard answer keys.
CANONICAL PROBLEM 02 Dual-Method Breakdown

If $\hat{a}$ and $\hat{b}$ are two unit vectors such that the angle between them is $\frac{\pi}{3}$, then $|\hat{a} - \hat{b}|$ is equal to:

(A) $1$(B) $\sqrt{3}$(C) $2$(D) $\frac{1}{2}$
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Use formula $|\hat{a} - \hat{b}|^2 = |\hat{a}|^2 + |\hat{b}|^2 - 2\hat{a}\cdot\hat{b}$. Step 2: Substitute $|\hat{a}| = 1, |\hat{b}| = 1$, and $\hat{a}\cdot\hat{b} = |\hat{a}||\hat{b}|\cos(\pi/3) = 1 \cdot 1 \cdot \frac{1}{2} = \frac{1}{2}$. Step 3: $|\hat{a} - \hat{b}|^2 = 1 + 1 - 2(1/2) = 1 \implies |\hat{a} - \hat{b}| = 1$.
⚡ Method B: 25-Second Exam Shortcut
Equilateral triangle property: Two unit vectors at $\pi/3$ ($60^\circ$) form an equilateral triangle with their difference vector, hence the third side has length 1.
💡 Tip: Memorize standard results for angle $60^\circ$ (diff magnitude = 1) and angle $120^\circ$ (sum magnitude = 1).

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: If $\vec{a} = 3\hat{i} - 4\hat{j} + 12\hat{k}$, then the unit vector in the direction of $\vec{a}$ is:
A $\frac{3}{13}\hat{i} - \frac{4}{13}\hat{j} + \frac{12}{13}\hat{k}$
B $\frac{3}{5}\hat{i} - \frac{4}{5}\hat{j}$
C $\frac{3}{12}\hat{i} - \frac{4}{12}\hat{j} + \frac{12}{12}\hat{k}$
D $3\hat{i} - 4\hat{j} + 12\hat{k}$
Derivation & Analysis:

Magnitude is $|\vec{a}| = \sqrt{3^2 + (-4)^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$. Dividing each component by 13 gives Option (A).

Q2
MCQ 2: For any unit vector $\hat{a}$, the value of $(\hat{a} \times \hat{i})^2 + (\hat{a} \times \hat{j})^2 + (\hat{a} \times \hat{k})^2$ is equal to:
A $0$
B $1$
C $2$
D $3$
Derivation & Analysis:

Let $\hat{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ with $a_1^2 + a_2^2 + a_3^2 = 1$. Using Lagrange's identity or vector expansion, each cross product magnitude squared equals $1 - a_i^2$. Summing them gives $(1-a_1^2) + (1-a_2^2) + (1-a_3^2) = 3 - (a_1^2+a_2^2+a_3^2) = 3 - 1 = 2$.

Q3
MCQ 3: If vector $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$ is a unit vector, then the maximum value of $xy + yz + zx$ is:
A $1$
B $\frac{1}{2}$
C $3$
D $\frac{1}{3}$
Derivation & Analysis:

We know $(x+y+z)^2 = x^2+y^2+z^2 + 2(xy+yz+zx) = 1 + 2(xy+yz+zx)$. Also, $x^2+y^2+z^2 \ge xy+yz+zx$ (algebraic inequality). Since $x^2+y^2+z^2=1$, the maximum value of $xy+yz+zx$ is $\frac{1}{2}$.

Q4
MCQ 4: If $\vec{a}, \vec{b}, \vec{c}$ are unit vectors such that $\vec{a} + \vec{b} + \vec{c} = \vec{0}$, then the angle between $\vec{a}$ and $\vec{b}$ is:
A $\frac{\pi}{3}$
B $\frac{\pi}{2}$
C $\frac{2\pi}{3}$
D $\pi$
Derivation & Analysis:

From $\vec{a} + \vec{b} = -\vec{c}$, taking magnitude squared gives $|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b} = |-\vec{c}|^2 \implies 1 + 1 + 2\cos\theta = 1 \implies 2\cos\theta = -1 \implies \cos\theta = -\frac{1}{2} \implies \theta = \frac{2\pi}{3}$.

Q5
MCQ 5: The unit vector parallel to the xy-plane and perpendicular to the vector $4\hat{i} - 3\hat{j} + \hat{k}$ is:
A $\pm \frac{3\hat{i} + 4\hat{j}}{5}$
B $\pm \frac{4\hat{i} + 3\hat{j}}{5}$
C $\pm \frac{\hat{i} - \hat{j}}{\sqrt{2}}$
D $\pm \hat{k}$
Derivation & Analysis:

A vector parallel to the xy-plane has the form $x\hat{i} + y\hat{j}$ (i.e., $z=0$). For it to be perpendicular to $4\hat{i} - 3\hat{j} + \hat{k}$, their dot product must be zero: $4x - 3y = 0 \implies 4x = 3y$. Testing options, $\frac{3}{5}\hat{i} + \frac{4}{5}\hat{j}$ gives $4(3/5) - 3(4/5) = 0$ and has magnitude 1.

Q6
MCQ 6: If $\hat{a}$ and $\hat{b}$ are unit vectors and $\theta$ is the acute angle between them, then $\sin(\theta/2)$ is equal to:
A $\frac{1}{2}|\hat{a} - \hat{b}|$
B $\frac{1}{2}|\hat{a} + \hat{b}|$
C $|\hat{a} - \hat{b}|$
D $2|\hat{a} - \hat{b}|$
Derivation & Analysis:

$|\hat{a} - \hat{b}|^2 = |\hat{a}|^2 + |\hat{b}|^2 - 2\hat{a}\cdot\hat{b} = 2 - 2\cos\theta = 4\sin^2(\theta/2)$. Taking square roots gives $|\hat{a} - \hat{b}| = 2\sin(\theta/2) \implies \sin(\theta/2) = \frac{1}{2}|\hat{a} - \hat{b}|$.

Q7
MCQ 7: If the vectors $\alpha \hat{i} + \hat{j} + \hat{k}$, $\hat{i} + \beta \hat{j} + \hat{k}$ and $\hat{i} + \hat{j} + \gamma \hat{k}$ are mutually orthogonal unit vectors, then the value of $\alpha^2 + \beta^2 + \gamma^2$ is:
A $0$
B $1$
C $2$
D $3$
Derivation & Analysis:

Since they are unit vectors, $\alpha^2 + 1 + 1 = 1 \implies \alpha^2 = -1$. Wait, in real vector spaces $\mathbb{R}^3$, orthogonal unit vectors cannot all have 1 as other components unless they are standard basis vectors, but orthogonality requires $\alpha(1) + 1(\beta) + 1(1) = 0$. If they are standard basis vectors $\hat{i}, \hat{j}, \hat{k}$, then $\alpha=0, \beta=0, \gamma=0$, so $\alpha^2+\beta^2+\gamma^2 = 0$.

Q8
MCQ 8: If $\vec{a}$ is a non-zero vector of magnitude $a$, then $\frac{\vec{a}}{a}$ is:
A a null vector
B a unit vector in the direction of $\vec{a}$
C a unit vector in the direction opposite to $\vec{a}$
D a scalar
Derivation & Analysis:

By definition, dividing a vector by its own magnitude results in a unit vector pointing in the exact same direction: $\left|\frac{\vec{a}}{a}\right| = \frac{|\vec{a}|}{a} = \frac{a}{a} = 1$.

Q9
MCQ 9: Let $\vec{a}, \vec{b}, \vec{c}$ be three unit vectors such that $\vec{a} \cdot \vec{b} = \vec{a} \cdot \vec{c} = 0$ and the angle between $\vec{b}$ and $\vec{c}$ is $\frac{\pi}{3}$. Then $\vec{a}$ is equal to:
A $\pm 2(\vec{b} \times \vec{c})$
B $\pm \frac{2}{\sqrt{3}}(\vec{b} \times \vec{c})$
C $\pm (\vec{b} \times \vec{c})$
D $\pm \sqrt{3}(\vec{b} \times \vec{c})$
Derivation & Analysis:

Since $\vec{a}$ is orthogonal to both $\vec{b}$ and $\vec{c}$, $\vec{a}$ must be parallel to $\vec{b} \times \vec{c}$. Thus $\vec{a} = \lambda(\vec{b} \times \vec{c})$. Magnitude of cross product is $|\vec{b}||\vec{c}|\sin(\pi/3) = 1 \cdot 1 \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$. Since $|\vec{a}| = 1$, we have $|\lambda| \frac{\sqrt{3}}{2} = 1 \implies \lambda = \pm \frac{2}{\sqrt{3}}$.

Q10
MCQ 10: If $|\vec{a}| = 3, |\vec{b}| = 4$, and $|\vec{a} + \vec{b}| = 5$, then the unit vector in the direction of $\vec{a} - \vec{b}$ is:
A $\frac{3\hat{i} - 4\hat{j}}{5}$
B $\frac{4\hat{i} - 3\hat{j}}{5}$
C $\frac{3\hat{a} - 4\hat{b}}{5}$
D Cannot be determined without coordinate axes
Derivation & Analysis:

Notice that $3^2 + 4^2 = 5^2$, meaning $\vec{a}$ and $\vec{b}$ are orthogonal vectors. However, without knowing the explicit Cartesian components of $\vec{a}$ and $\vec{b}$, the specific unit vector along $\vec{a} - \vec{b}$ cannot be expressed in terms of standard basis vectors $\hat{i}, \hat{j}$, making (D) correct.

Common questions

Yes! A unit vector simply requires that the sum of the squares of its components equals 1 ($a_1^2 + a_2^2 + a_3^2 = 1$). Individual components can be negative, fractional, or zero.

Examiners test this to check if candidates remember both the magnitude normalisation and the crucial $\pm$ sign indicating both upward and downward orientations perpendicular to a plane.

Square each component in the given option and sum them up. If the sum equals exactly 1, it is a valid unit vector. This eliminates incorrect options.

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