Vector Algebra Class 12 Notes: Types, Formulas, and Worked Examples

These vector algebra class 12 notes condense the entire chapter into revision order: definitions, formulas, laws, and solved methods with fresh numbers. You get five skill blocks — foundations (scalars, vectors, position vector), direction cosines, vector operations (addition, scalar multiplication, components), the section formula, and the dot product with projection.

Each idea maps to its NCERT page so you can cross-check fast. Work the examples yourself before peeking at the steps, then check your arithmetic. If you are short on time, start with the formula table and the ten-minute recap at the end.

Vector Algebra Class 12 Notes: What the Chapter Builds

The chapter builds ideas in a strict order — each block uses the one before it. This is the order a teacher follows, and the order these notes follow:

1. Scalar vs vector — what makes a quantity a vector (magnitude + direction).
2. Position vector — writing a point as \( \overrightarrow{OP} = x\hat i + y\hat j + z\hat k \) and its magnitude.
3. Direction cosines — the angles a vector makes with the three axes.
4. Types of vectors — zero, unit, coinitial, collinear, equal, negative, free.
5. Addition — triangle law and parallelogram law.
6. Scalar multiplication — scales length, flips or keeps direction, builds the unit vector.
7. Component form — addition, subtraction, equality done component-wise.
8. Section formula — splitting a segment internally or externally (NCERT, p. 15).

The chapter then introduces two products: the scalar (dot) product and the vector (cross) product (NCERT, p. 18). The exercise coverage in this revision scope (10.1–10.3) works only with the scalar product, which is the heart of the chapter’s second half. For the wider picture of where vectors feed in, see the class 12 mathematics notes hub, or the class 12 notes index.

Scalars, Vectors, and the Position Vector: The Foundations

Definition 1 (NCERT, p. 1): A quantity that has magnitude as well as direction is called a vector.

A scalar has only a magnitude — a single real number. Your height (1.6 m), length, mass, time, speed, area, volume, temperature, work, money, voltage, density and resistance are all scalars. A vector has magnitude and direction — displacement, velocity, acceleration, force, weight, momentum and electric field intensity are vectors.

The grounding example that separates them: “What is your height?” gets a one-number answer (a scalar). “How should a football player hit the ball to pass to a teammate?” needs force — a magnitude plus a direction — so it is a vector.

The directed line segment

A line can carry two directions (shown by arrowheads). Add an arrow and the line becomes a directed line. Restrict it to a segment and you get a directed line segment — which has both magnitude and direction, so it is a vector, written \( \overrightarrow{AB} \) or \( \vec a \) (NCERT, p. 1).

• Initial point — where the vector starts (A in \( \overrightarrow{AB} \)).
• Terminal point — where it ends (B).
• Magnitude or length — the distance between initial and terminal points, written \( |\overrightarrow{AB}| \), \( |\vec a| \), or \( a \).

The arrow shows direction, and because a length is never negative, the notation \( |\vec a| \lt 0 \) has no meaning (NCERT, p. 1). Hold on to this — it is the first trap students hit.

Position vector and its magnitude

For a point \( P(x, y, z) \) in space with origin \( O(0, 0, 0) \), the vector \( \overrightarrow{OP} \) is the position vector of P with respect to O (NCERT, p. 2). Its magnitude comes straight from the distance formula:

\[ |\overrightarrow{OP}| = \sqrt{x^2 + y^2 + z^2} \]

The vectors \( \overrightarrow{OA}, \overrightarrow{OB}, \overrightarrow{OC} \) to fixed points are often written \( \vec a, \vec b, \vec c \). This position-vector machinery is exactly what carries into three-dimensional geometry class 12 notes.

Direction Cosines: The Angles a Vector Makes With the Axes

For the position vector \( \vec r \) of \( P(x, y, z) \), let the angles it makes with the positive x-, y- and z-axes be \( \alpha, \beta, \gamma \) — these are the direction angles (NCERT, p. 3). The cosines of these angles are the direction cosines, written \( l, m, n \):

\[ l = \cos\alpha = \frac{x}{r}, \quad m = \cos\beta = \frac{y}{r}, \quad n = \cos\gamma = \frac{z}{r} \]

Why the fractions? Drop a perpendicular from P to the x-axis, forming the right triangle OAP. In that triangle, \( \cos\alpha = \text{adjacent}/\text{hypotenuse} = x/r \). The same construction works for the y- and z-axes (NCERT, p. 3). This lets you write the point’s coordinates as \( (lr, mr, nr) \).

Mammoth exam identity: since the components describe one fixed vector, its direction cosines always satisfy \[ l^2 + m^2 + n^2 = 1 \]

Direction ratios \( a, b, c \) are numbers proportional to the direction cosines — for a vector \( x\hat i + y\hat j + z\hat k \) they are just the components \( x, y, z \) (NCERT, p. 12). The crucial contrast: \( l^2 + m^2 + n^2 = 1 \) always, but \( a^2 + b^2 + c^2 \neq 1 \) in general (NCERT, p. 3).

Types of Vectors in One Table

NCERT lists eight vector types you must be able to name and tell apart (NCERT, pp. 2, 4). Learn them from one comparison table:

Two remarks from the grounding are exam-gold. First, the zero vector cannot be assigned a definite direction because it has zero magnitude (NCERT, p. 4). Second, all vectors in this chapter are free vectors — you may displace them parallel without changing them, which is precisely what makes the triangle law possible (NCERT, p. 4).

Adding and Subtracting Vectors: Triangle Law and Parallelogram Law

Triangle law of vector addition (NCERT, p. 6): if a girl walks from A to B and then B to C, her net displacement is \( \overrightarrow{AC} \), and \[ \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} \]

In general, to add two vectors \( \vec a \) and \( \vec b \), slide one so its initial point meets the other’s terminal point; the third side of the triangle is the sum. Because \( \overrightarrow{AC} = -\overrightarrow{CA} \), walking the three sides of a triangle in order returns you to the start:

\[ \overrightarrow{AB} + \overrightarrow{BC} + \overrightarrow{CA} = \overrightarrow{AA} = \vec 0 \]

Subtraction is just addition of the negative: construct \( \overrightarrow{BC’} = -\overrightarrow{BC} \), then \( \overrightarrow{AC’} = \vec a – \vec b \) (NCERT, p. 6).

Parallelogram law (NCERT, p. 6): when two vectors are drawn as adjacent sides of a parallelogram from a common point, their sum is the diagonal through that point. The two laws agree — the diagonal of the parallelogram is the third side of the triangle it splits into.

Real-life use: the boat crossing a river

The textbook’s application is a boat steering perpendicular to a river’s flow. Two velocities act on it at once — the engine’s velocity and the river current’s velocity. The boat actually travels along neither; its resultant velocity is the parallelogram diagonal of the two velocity vectors (NCERT, p. 6).

This is why the parallelogram law exists: to combine two simultaneous pushes into one effective motion.

Properties of addition

• Commutative: \( \vec a + \vec b = \vec b + \vec a \) (NCERT, p. 7).
• Associative: \( (\vec a + \vec b) + \vec c = \vec a + (\vec b + \vec c) \) — brackets become unnecessary (NCERT, p. 7).
• Additive identity: \( \vec a + \vec 0 = \vec a \); the zero vector is the identity (NCERT, p. 7).

Multiplying a Vector by a Scalar: Direction and Magnitude

For a vector \( \vec a \) and a scalar \( \lambda \), the product \( \lambda\vec a \) is a vector collinear to \( \vec a \) (NCERT, p. 9). The rule:

• Same direction as \( \vec a \) if \( \lambda \gt 0 \); opposite direction if \( \lambda \lt 0 \).
• Magnitude scales by \( |\lambda| \): \( |\lambda\vec a| = |\lambda|\,|\vec a| \).
• With \( \lambda = -1 \), \( \lambda\vec a = -\vec a \) is the additive inverse — \( \vec a + (-\vec a) = \vec 0 \) (NCERT, p. 9).
• For any scalar \( k \), \( k\vec 0 = \vec 0 \) (NCERT, p. 9).

The unit-vector recipe (the most-used formula of the chapter): if \( \vec a \neq \vec 0 \), take \( \lambda = 1/|\vec a| \). Then \( |\lambda\vec a| = 1 \), so \( \lambda\vec a \) is the unit vector in the direction of \( \vec a \):

\[ \hat a = \frac{1}{|\vec a|} \, \vec a \]

Distributive laws (NCERT, p. 10)

• \( k\vec a + m\vec a = (k + m)\vec a \)
• \( k(m\vec a) = (km)\vec a \)
• \( k(\vec a + \vec b) = k\vec a + k\vec b \)

The collinearity test

Since \( \lambda\vec a \) is always collinear to \( \vec a \), two vectors are collinear iff there is a nonzero scalar \( \lambda \) with \( \vec b = \lambda\vec a \) (NCERT, p. 12). In components this reads \[ \frac{b_1}{a_1} = \frac{b_2}{a_2} = \frac{b_3}{a_3} = \lambda \]

Component Form: Writing Every Vector as x i + y j + z k

The unit vectors along the axes are \( \hat i \) (x-axis), \( \hat j \) (y-axis), \( \hat k \) (z-axis), each with magnitude 1 (NCERT, p. 10). Any position vector breaks into these:

\[ \vec r = \overrightarrow{OP} = x\hat i + y\hat j + z\hat k \]

• Scalar components: \( x, y, z \) — also called rectangular components.
• Vector components: \( x\hat i, y\hat j, z\hat k \) — the vectors along each axis.

Why the magnitude formula works — Pythagoras twice (NCERT, p. 10): first in the xy-plane, \( |\overrightarrow{OP_1}| = \sqrt{x^2 + y^2} \), then lift up the z-axis, \( |\vec r| = \sqrt{(x^2 + y^2) + z^2} \). One right triangle in the plane, a second one in the vertical direction.

With \( \vec a = a_1\hat i + a_2\hat j + a_3\hat k \) and \( \vec b = b_1\hat i + b_2\hat j + b_3\hat k \), everything is component-wise (NCERT, p. 10):

• Sum: \( \vec a + \vec b = (a_1 + b_1)\hat i + (a_2 + b_2)\hat j + (a_3 + b_3)\hat k \)
• Difference: \( \vec a – \vec b = (a_1 – b_1)\hat i + (a_2 – b_2)\hat j + (a_3 – b_3)\hat k \)
• Equality: \( \vec a = \vec b \) iff \( a_1 = b_1,\; a_2 = b_2,\; a_3 = b_3 \) — all three components match.
• Scalar multiple: \( \lambda\vec a = (\lambda a_1)\hat i + (\lambda a_2)\hat j + (\lambda a_3)\hat k \)

This is the workhorse section — most board problems reduce to matching components. Example from NCERT: the vectors \( x\hat i + 2\hat j + z\hat k \) and \( 2\hat i + y\hat j + \hat k \) are equal only when \( x = 2, y = 2, z = 1 \) (NCERT, p. 12).

Vector Joining Two Points and the Section Formula

The vector from \( P_1(x_1, y_1, z_1) \) to \( P_2(x_2, y_2, z_2) \) — terminal minus initial, component by component (NCERT, p. 14):

\[ \overrightarrow{P_1P_2} = (x_2 – x_1)\hat i + (y_2 – y_1)\hat j + (z_2 – z_1)\hat k \]

\[ |\overrightarrow{P_1P_2}| = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2 + (z_2 – z_1)^2} \]

This follows from the triangle law on triangle \( OP_1P_2 \): \( \overrightarrow{OP_1} + \overrightarrow{P_1P_2} = \overrightarrow{OP_2} \), so the joining vector is the difference of the two position vectors.

Section formula (NCERT, p. 15)

Let R divide the segment PQ, where P and Q have position vectors \( \vec a \) and \( \vec b \), in the ratio \( m : n \).

Case 1 — internal division (R lies between P and Q):

\[ \overrightarrow{OR} = \frac{m\vec b + n\vec a}{m + n} \]

Case 2 — external division (R lies on the line beyond P or Q):

\[ \overrightarrow{OR} = \frac{m\vec b – n\vec a}{m – n} \]

Midpoint is internal division with \( m = n \):

\[ \overrightarrow{OR} = \frac{\vec a + \vec b}{2} \]

Classic slip to avoid: the coefficient \( m \) sits on \( \vec b \) and \( n \) on \( \vec a \) — it is \( (m\vec b + n\vec a) \), not \( (m\vec a + n\vec b) \). The point that carries the \( m \)-part of the ratio multiplies the opposite-end position vector.

Every Formula You Need in One Place

This is the quick-scan box. Each entry names every symbol so you never substitute blind.

Remember \( \theta \) is the angle between the two vectors, always taken in \( 0 \le \theta \le \pi \). Magnitudes carry length units; the dot product carries the product of the two quantities’ units.

Dot Product of Two Vectors: Definition, Laws, and Tests

Definition 2 (NCERT, p. 18): the scalar product of two nonzero vectors is \[ \vec a \cdot \vec b = |\vec a||\vec b|\cos\theta \]

where \( \theta \) is the angle between them. The result is a real number (a scalar) — that is why it is called the scalar product. If either vector is \( \vec 0 \), \( \theta \) is not defined and \( \vec a\cdot\vec b = 0 \) by definition.

The seven observations that run the chapter (NCERT, p. 19)

1. \( \vec a\cdot\vec b \) is a real number.
2. Perpendicularity test: for nonzero vectors, \( \vec a\cdot\vec b = 0 \Leftrightarrow \vec a \perp \vec b \).
3. If \( \theta = 0 \), \( \vec a\cdot\vec b = |\vec a||\vec b| \); in particular \( \vec a\cdot\vec a = |\vec a|^2 \).
4. If \( \theta = \pi \), \( \vec a\cdot\vec b = -|\vec a||\vec b| \).
5. For the axis unit vectors: \( \hat i\cdot\hat i = \hat j\cdot\hat j = \hat k\cdot\hat k = 1 \) and \( \hat i\cdot\hat j = \hat j\cdot\hat k = \hat k\cdot\hat i = 0 \).
6. Angle formula: \( \cos\theta = \dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|} \), i.e. \( \theta = \cos^{-1}\left(\dfrac{\vec a\cdot\vec b}{|\vec a||\vec b|}\right) \).
7. Commutativity: \( \vec a\cdot\vec b = \vec b\cdot\vec a \).

Memory device for observation 5: “Match = 1, mismatch = 0.” Compare the letters — the same axis unit vector dotted with itself gives 1 (match), any two different axes give 0 (mismatch).

This is exactly why \( \vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3 \): expanding the product, every mismatched pair like \( a_1b_2(\hat i\cdot\hat j) \) dies to zero, leaving only the same-index terms (NCERT, p. 19).

Component form: \( \vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3 \). This is the formula you substitute into.

Laws of the dot product (NCERT, p. 19)

• Distributivity over addition: \( \vec a\cdot(\vec b + \vec c) = \vec a\cdot\vec b + \vec a\cdot\vec c \).
• Scalar factor moves: \( (\lambda\vec a)\cdot\vec b = \lambda(\vec a\cdot\vec b) = \vec a\cdot(\lambda\vec b) \).

Edge case: a·b = 0 does not mean one vector is zero

Students read \( \vec a\cdot\vec b = 0 \) as “one of them must be zero.” False. If \( \vec a = \vec 0 \) or \( \vec b = \vec 0 \) then indeed the dot product is 0, but the converse fails (Exercise 10.3, Q14).

Counterexample: take \( \vec a = \hat i \) and \( \vec b = \hat j \), two nonzero unit vectors. Then \( \vec a\cdot\vec b = \hat i\cdot\hat j = 0 \) — neither vector is zero; they are perpendicular. So for nonzero vectors, a zero dot product is the definition of perpendicularity, not the sign of a zero vector (NCERT, p. 19).

Projection of a Vector on a Line

If \( \overrightarrow{AB} \) makes angle \( \theta \) with a directed line \( l \), its projection vector on \( l \) has magnitude \( |\overrightarrow{AB}|\,|\cos\theta| \). Its direction matches the line when \( \cos\theta \gt 0 \) and flips when \( \cos\theta \lt 0 \) (NCERT, p. 20).

Projection of \( \vec a \) on \( \vec b \), all three equivalent forms (NCERT, p. 21):

\[ \vec a\cdot\hat b = \vec a\cdot\frac{\vec b}{|\vec b|} = \frac{\vec a\cdot\vec b}{|\vec b|} \]

Special angles (NCERT, p. 21):

• \( \theta = 0 \): projection is \( \overrightarrow{AB} \) itself.
• \( \theta = \pi \): projection is \( \overrightarrow{BA} \) (reversed).
• \( \theta = \pi/2 \) or \( 3\pi/2 \): projection is the zero vector.

Why this matters for components: the scalar components \( a_1, a_2, a_3 \) of \( \vec a \) are precisely its projections along the x-, y- and z-axes (NCERT, p. 21). So projection is not an isolated idea — it is the machinery that builds component form in the first place.

If \( \vec a \) is a unit vector, it can be written in terms of its direction cosines as \[ \vec a = \cos\alpha\,\hat i + \cos\beta\,\hat j + \cos\gamma\,\hat k \]

Worked Examples Solved Step by Step

Each example names its method first, then shows every substitution with units. Work them fully before reading the steps.

Example 1: Unit vector in the direction of a = 3i − 4j + 12k

Method: unit-vector recipe — divide the vector by its magnitude, \( \hat a = \vec a/|\vec a| \).

Example 2: Vector joining P(1, 2, −3) to Q(4, −1, 2), with its magnitude

Method: terminal minus initial, component by component.

Example 3: Section formula, internal and external, ratio 2 : 3

Method: section formula. Let P and Q have position vectors \( \vec a \) and \( \vec b \), and let R divide PQ in the ratio \( m : n = 2 : 3 \).

Notice the coefficient \( m = 2 \) multiplies \( \vec b \), the position vector of Q — the far end of the division. Getting this backwards is the chapter’s most common algebraic slip.

Example 4: Angle between a = i + j and b = j + k

Method: angle formula \( \cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|} \).

Example 5: Projection of a = 2i + 3j − k on b = i − 2j

Method: projection formula — projection of \( \vec a \) on \( \vec b \) is \( \frac{\vec a\cdot\vec b}{|\vec b|} \).

Common Mistakes Students Make in Vector Algebra

These are the exact slips and how to catch yourself.

Exam Notes: How to Earn the Mark in Vector Problems

These are observed mark-earning habits — write them into your working, not just your head.

• Write the formula before substituting. For magnitude, \( |\vec a| = \sqrt{x^2+y^2+z^2} \) in full; for the dot product, \( \vec a\cdot\vec b = a_1b_1+a_2b_2+a_3b_3 \) before plugging in. The formula line earns the method mark even if arithmetic slips.
• Prove collinearity by ratios. Show \( \frac{b_1}{a_1} = \frac{b_2}{a_2} = \frac{b_3}{a_3} = \lambda \) explicitly; each equal ratio is a scored line.
• For a right-angled triangle, compute all three squared lengths and identify which is the sum of the other two — the largest square is the hypotenuse (NCERT, p. 15).
• State internal or external before writing a section formula. The two differ only by the minus sign and denominator, and examiners check that you chose correctly.
• Test perpendicularity via the dot product: write “two nonzero vectors are perpendicular iff their scalar product is 0” before computing it (NCERT, p. 19).

Mapping the exercises: Exercise 10.1 (p. 5) is concept classification — scalars vs vectors, and true/false on collinear and equal vectors. Exercise 10.2 (p. 17) is the working core: magnitudes, unit vectors, direction cosines, section formula, and right-angled triangles. Exercise 10.3 (p. 24) is the scalar-product block: angles, projections, and perpendicularity proofs.

For full-marks answers, every numerical must show the formula, the substitution, and the final statement of the result with its unit.

You can verify any formula against the official text at NCERT Mathematics Part II (lemh2) — chapter 10 opens directly there. For the chapters that follow this one, see three-dimensional geometry class 12 notes and the rest of the differential equations class 12 notes collection.

Rapid Revision Recap: The Whole Chapter in One Screen

Ten minutes of reading. Each line is a complete statement.

Definitions (one line each)

• Scalar: quantity with magnitude only. Vector: quantity with magnitude and direction (p. 1).
• Position vector of \( P(x,y,z) \): \( \overrightarrow{OP} = x\hat i + y\hat j + z\hat k \), magnitude \( \sqrt{x^2+y^2+z^2} \) (p. 2).
• Unit vector: magnitude 1; \( \hat a = \vec a/|\vec a| \) (p. 9).
• Zero vector: initial = terminal point, magnitude 0 (p. 4).
• Equal vectors: same magnitude and direction (p. 4). Collinear: parallel to one line (p. 4).
• Negative: \( \overrightarrow{BA} = -\overrightarrow{AB} \) (p. 4).

Laws

• Triangle law: \( \overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC} \); closed triangle sums to \( \vec 0 \) (p. 6).
• Parallelogram law: sum of two adjacent sides = the diagonal through the common point (p. 6).
• Commutative and associative addition; zero vector is the additive identity (p. 7).

Formulas

• Direction cosines: \( l = x/r, m = y/r, n = z/r \), with \( l^2+m^2+n^2 = 1 \) (p. 3).
• Vector joining two points: \( (x_2-x_1)\hat i + (y_2-y_1)\hat j + (z_2-z_1)\hat k \) (p. 14).
• Section formula: internal \( \frac{m\vec b+n\vec a}{m+n} \), external \( \frac{m\vec b-n\vec a}{m-n} \), midpoint \( \frac{\vec a+\vec b}{2} \) (p. 15).
• Dot product: \( |\vec a||\vec b|\cos\theta = a_1b_1+a_2b_2+a_3b_3 \); \( \vec a\cdot\vec a = |\vec a|^2 \) (pp. 18–19).
• Projection of \( \vec a \) on \( \vec b \): \( \frac{\vec a\cdot\vec b}{|\vec b|} \) (p. 21).

Tests

• Perpendicular: \( \vec a\cdot\vec b = 0 \Leftrightarrow \vec a \perp \vec b \), nonzero vectors (p. 19).
• Collinear: \( \vec b = \lambda\vec a \Leftrightarrow b_1/a_1 = b_2/a_2 = b_3/a_3 = \lambda \) (p. 12).
• Inequalities: \( |\vec a\cdot\vec b| \le |\vec a||\vec b| \) and \( |\vec a+\vec b| \le |\vec a| + |\vec b| \) (p. 20).

For a final sweep across the subject, return to the CBSE notes index.

Frequently Asked Questions About Vector Algebra

Why can the magnitude of a vector never be negative?

Magnitude is a length — the distance between the initial and terminal points of the vector (NCERT, p. 1), and a distance is always \( \ge 0 \). The formula \( |\vec r| = \sqrt{x^2+y^2+z^2} \) is a sum of squares inside a square root, which can never come out negative, so the notation \( |\vec a| \lt 0 \) has no meaning.

What is the difference between direction cosines and direction ratios of a vector?

Direction cosines \( l, m, n \) are the cosines of the angles the vector makes with the axes, so they always satisfy \( l^2 + m^2 + n^2 = 1 \) (NCERT, p. 3). Direction ratios \( a, b, c \) are numbers proportional to the cosines — for \( x\hat i + y\hat j + z\hat k \) they are just \( x, y, z \) — and \( a^2 + b^2 + c^2 \neq 1 \) in general.

Cosines are normalised; ratios are not.

When should I use the internal section formula and when the external section formula?

Use the internal formula \( \frac{m\vec b+n\vec a}{m+n} \) when the dividing point R lies between P and Q, and the external formula \( \frac{m\vec b-n\vec a}{m-n} \) when R lies on the line beyond one of them (NCERT, p. 15).

The ratio \( m:n \) always lists the \( m \)-part first; that coefficient multiplies \( \vec b \), the position vector of Q.

If the dot product of two vectors is zero, are the vectors always perpendicular?

Only when both vectors are nonzero. For nonzero vectors, \( \vec a\cdot\vec b = 0 \Leftrightarrow \vec a \perp \vec b \) (NCERT, p. 19). If either vector is \( \vec 0 \), the dot product is 0 by definition without any perpendicularity.

Counterexample: \( \vec a = \hat i \), \( \vec b = \hat j \) — both unit vectors, \( \vec a\cdot\vec b = \hat i\cdot\hat j = 0 \), neither is zero, and they are perpendicular.

How do I find the projection of vector a on vector b without drawing a diagram?

Use the formula directly: the projection of \( \vec a \) on \( \vec b \) is \( \frac{\vec a\cdot\vec b}{|\vec b|} = \vec a\cdot\hat b \) (NCERT, p. 21). Compute the dot product in components and divide by the magnitude of \( \vec b \).

For example, projection of \( \vec a = 2\hat i + 3\hat j – \hat k \) on \( \vec b = \hat i – 2\hat j \) is \( \frac{2-6+0}{\sqrt{5}} = \frac{-4}{\sqrt{5}} \) — the negative sign tells you the projection points opposite to \( \vec b \).

Reference: NCERT Class 12 Mathematics Part II textbook, chapter Vector Algebra.