MathsISC Class 12 · Chapter 10

Vector Algebra ISC Class 12: Notes, Products and Solved Questions

Updated 14 October 2026 · Fact-checked

Vector Algebra is Chapter 10 of ISC Class 12 Mathematics. It covers vectors and scalars, magnitude and direction, direction cosines and ratios, types of vectors, position vectors and the section formula, the scalar (dot) product with projections and angles, and the vector (cross) product with areas of triangles and parallelograms and collinearity.

What Vector Algebra is about

A vector has magnitude and direction; a scalar has magnitude only. In ISC 2027 Vector Algebra is a compulsory unit worth 5 of the 80 marks (it was optional before), and it underpins Three-dimensional Geometry.

Vectors are written in terms of the unit vectors $\hat{i}$ , $\hat{j}$ , $\hat{k}$ along the axes: $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ , with magnitude $∣ a ∣ = a_{1}^{2} + a_{2}^{2} + a_{3}^{2}$ and direction cosines $\frac{a _{1}}{∣ a ∣}, \frac{a _{2}}{∣ a ∣}, \frac{a _{3}}{∣ a ∣}$ (any multiples are direction ratios). You need the types (equal, unit, zero, parallel/collinear), the negative of a vector, addition (triangle and parallelogram laws), scalar multiplication, the position vector of a point and the vector joining two points ( $A B = O B - O A$ ), and the section formula for internal and external division.

The scalar (dot) product $a \cdot b = ∣ a ∣∣ b ∣ cos θ = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$ gives angles, tests perpendicularity ( $a \cdot b = 0$ ) and the projection of one vector on another, $\frac{a \cdot b}{∣ a ∣}$ (scalar) or that times $\overset{a}{^}$ (vector). The vector (cross) product $a \times b$ has magnitude $∣ a ∣∣ b ∣ sin θ$ and is perpendicular to both; its magnitude is the area of the parallelogram on $a$ and $b$ , so a triangle's area is half of it, and $a \times b = 0$ means the vectors are parallel. Proofs of geometrical theorems by vectors are excluded.

Weightage: Vector Algebra is part of the unit Vector Algebra, which carries 5 of the 80 marks in the ISC Mathematics. Questions on it can appear in any section of the paper.Source: CISCE ISC Mathematics syllabus, examination year 2027, distribution of marks

Key concepts

Magnitude, direction cosines and unit vectors: For $a = a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ : $∣ a ∣ = a_{1}^{2} + a_{2}^{2} + a_{3}^{2}$ . The direction cosines l, m, n are the cosines of the angles the vector makes with the axes: $l = \frac{a _{1}}{∣ a ∣}$ etc., with $l^{2} + m^{2} + n^{2} = 1$ . Any numbers proportional to them (such as $a_{1}, a_{2}, a_{3}$ ) are direction ratios. The unit vector in the direction of $a$ is $\overset{a}{^} = \frac{a}{∣ a ∣}$ ; a vector of magnitude k in that direction is $k \overset{a}{^}$ .; Worked example
For $a = 2 \hat{i} - 3 \hat{j} + 6 \hat{k}$ , find $∣ a ∣$ , its direction cosines, and a vector of magnitude 14 in its direction.
Try it first, then show the solutionSolution
$∣ a ∣ = 4 + 9 + 36 = 7$
Direction cosines $\frac{2}{7}, - \frac{3}{7}, \frac{6}{7}$
$14 \overset{a}{^} = 2 a = 4 \hat{i} - 6 \hat{j} + 12 \hat{k}$
Answer: 7; (2/7, −3/7, 6/7); $4 \hat{i} - 6 \hat{j} + 12 \hat{k}$
Types of vectors and collinearity: Zero vector: magnitude 0, any direction. Unit vector: magnitude 1. Equal vectors: same magnitude and direction (equal components). Negative of $a$ : same magnitude, opposite direction. Collinear (parallel) vectors: $b = λ a$ for some scalar λ, i.e. their components are proportional. Three points A, B, C are collinear if $A B$ and $A C$ are parallel.; Worked example
Show that $a = 2 \hat{i} + 3 \hat{j} - \hat{k}$ and $b = - 4 \hat{i} - 6 \hat{j} + 2 \hat{k}$ are collinear. Are they in the same direction?
Try it first, then show the solutionSolution
$b = - 2 a$ : components proportional (−4/2 = −6/3 = 2/−1 = −2).
λ = −2 < 0, so they point in opposite directions.
Answer: Collinear, opposite directions
Position vectors and the section formula: The position vector of P(x, y, z) is $O P = x \hat{i} + y \hat{j} + z \hat{k}$ . The vector from A to B is $A B = O B - O A$ (head minus tail). Section formula: the point R dividing AB internally in the ratio m : n has $r = \frac{m b + n a}{m + n}$ ; externally, $r = \frac{m b - n a}{m - n}$ . The mid-point is $\frac{a + b}{2}$ .; Worked example
P and Q have position vectors $\hat{i} + 2 \hat{j} + 3 \hat{k}$ and $4 \hat{i} - \hat{j} + 6 \hat{k}$ . Find the point dividing PQ in the ratio 2 : 1 (a) internally, (b) externally.
Try it first, then show the solutionSolution
(a) $\frac{2 ( 4 , - 1 , 6 ) + 1 ( 1 , 2 , 3 )}{3} = \frac{( 9 , 0 , 15 )}{3} = (3, 0, 5)$
(b) $\frac{2 ( 4 , - 1 , 6 ) - 1 ( 1 , 2 , 3 )}{2 - 1} = (7, - 4, 9)$
Answer: (a) $3 \hat{i} + 5 \hat{k}$ (b) $7 \hat{i} - 4 \hat{j} + 9 \hat{k}$
Scalar (dot) product and angles: $a \cdot b = ∣ a ∣∣ b ∣ cos θ = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$ , a scalar. So $cos θ = \frac{a \cdot b}{∣ a ∣∣ b ∣}$ . Properties: commutative; distributive over addition; $a \cdot a = ∣ a ∣^{2}$ ; $\hat{i} \cdot \hat{i} = 1$ , $\hat{i} \cdot \hat{j} = 0$ . Non-zero vectors are perpendicular if and only if $a \cdot b = 0$ . Useful identity: $(a + b) \cdot (a - b) = ∣ a ∣^{2} - ∣ b ∣^{2}$ .; Worked example
Find the angle between $a = \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $b = 2 \hat{i} - \hat{j} + 2 \hat{k}$ .
Try it first, then show the solutionSolution
$a \cdot b = 2 - 2 + 4 = 4$ ; $∣ a ∣ = ∣ b ∣ = 3$
$cos θ = \frac{4}{9}$
Answer: $θ = cos^{- 1} \frac{4}{9}$
Projections: The scalar projection of $b$ on $a$ is $\frac{a \cdot b}{∣ a ∣} = ∣ b ∣ cos θ$ ; it is negative if θ > 90° (do not take its absolute value). The vector projection is $(\frac{a \cdot b}{∣ a ∣ ^{2}}) a$ . Geometrically, it is the length of the 'shadow' of $b$ on the line of $a$ . Note the order: projection of $b$ on $a$ divides by $∣ a ∣$ .; Worked example
Find the scalar and vector projections of $b = 2 \hat{i} - \hat{j} + 2 \hat{k}$ on $a = \hat{i} + 2 \hat{j} + 2 \hat{k}$ .
Try it first, then show the solutionSolution
Scalar: $\frac{a \cdot b}{∣ a ∣} = \frac{4}{3}$
Vector: $\frac{4}{9} (\hat{i} + 2 \hat{j} + 2 \hat{k})$
Answer: 4/3; $\frac{4}{9} (\hat{i} + 2 \hat{j} + 2 \hat{k})$
Vector (cross) product and areas: $a \times b = ∣ a ∣∣ b ∣ sin θ \overset{n}{^}$ , where $\overset{n}{^}$ is perpendicular to both (right-hand rule). In components: $a \times b = \hat{i} a_{1} b_{1} \hat{j} a_{2} b_{2} \hat{k} a_{3} b_{3}$ . Properties: $a \times b = - b \times a$ ; $a \times a = 0$ ; $\hat{i} \times \hat{j} = \hat{k}$ , $\hat{j} \times \hat{k} = \hat{i}$ , $\hat{k} \times \hat{i} = \hat{j}$ ; distributive. Geometry: $∣ a \times b ∣$ = area of the parallelogram with adjacent sides $a$ , $b$ ; triangle area = ½ of it; parallelogram with diagonals $d_{1}$ , $d_{2}$ has area $\frac{1}{2} ∣ d_{1} \times d_{2} ∣$ . $a \times b = 0$ (non-zero vectors) means they are parallel.; Worked example
For $a = \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $b = 2 \hat{i} - \hat{j} + 2 \hat{k}$ , find $a \times b$ , the area of the parallelogram on them and a unit vector perpendicular to both.
Try it first, then show the solutionSolution
$a \times b = \hat{i} (4 + 2) - \hat{j} (2 - 4) + \hat{k} (- 1 - 4) = 6 \hat{i} + 2 \hat{j} - 5 \hat{k}$
Area $= 36 + 4 + 25 = 65$
Unit perpendicular: $\pm \frac{1}{65} (6 \hat{i} + 2 \hat{j} - 5 \hat{k})$
Answer: $6 \hat{i} + 2 \hat{j} - 5 \hat{k}$ ; √65; $\pm \frac{6 i ^ + 2 j ^ - 5 k ^}{65}$

Maths in real life: Vector Algebra

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Work done as a dot product

Suppose a trolley is pushed by a constant force $F = 3 \hat{i} + 2 \hat{j} + \hat{k}$ newtons while it moves from the point (1, 1, 1) to (4, 3, 2), in metres.

Find the work done by the force.

Try it first, then show the solutionSolution

Displacement $d = (4 - 1) \hat{i} + (3 - 1) \hat{j} + (2 - 1) \hat{k} = 3 \hat{i} + 2 \hat{j} + \hat{k}$
$W = F \cdot d = 9 + 4 + 1 = 14$

Answer: 14 J

Memory hook: Work is force dot displacement.

Moment of a force as a cross product

Imagine tightening a nut with a spanner: the hand is at $r = 0.2 \hat{i}$ m from the nut and pushes with $F = 10 \hat{j}$ N.

Find the moment (turning effect) about the nut.

Try it first, then show the solutionSolution

$τ = r \times F = 0.2 \times 10 (\hat{i} \times \hat{j})$
$= 2 \hat{k}$

Answer: 2 N m about the z-axis (anticlockwise seen from above)

Memory hook: Turning effect is r cross F.

Magnitude of a displacement

Suppose a drone flies 3 km east, 4 km north and climbs 0.12 km.

How far is it from its starting point, and what is the cosine of the angle its path makes with the vertical?

Try it first, then show the solutionSolution

$d = 3 \hat{i} + 4 \hat{j} + 0.12 \hat{k}$ , $∣ d ∣ = 9 + 16 + 0.0144 \approx 5.0014$ km
Direction cosine with the vertical: $\frac{0.12}{5.0014} \approx 0.024$

Answer: About 5.0 km; cos ≈ 0.024 (almost horizontal)

Memory hook: Magnitude is Pythagoras in three directions.

Topics in this chapter

Vector Algebra formula sheet

Magnitude and unit vector: $∣ a ∣ = a_{1}^{2} + a_{2}^{2} + a_{3}^{2}, \overset{a}{^} = \frac{a}{∣ a ∣}$
Direction cosines: $l = \frac{a _{1}}{∣ a ∣}, m = \frac{a _{2}}{∣ a ∣}, n = \frac{a _{3}}{∣ a ∣}; l^{2} + m^{2} + n^{2} = 1$
Vector joining two points: $A B = O B - O A$; Head minus tail.
Section formula: $r = \frac{m b + n a}{m + n} (internal); r = \frac{m b - n a}{m - n} (external)$; Mid-point: (a + b)/2.
Dot product: $a \cdot b = ∣ a ∣∣ b ∣ cos θ = a_{1} b_{1} + a_{2} b_{2} + a_{3} b_{3}$; Zero for perpendicular vectors.
Projections: $scalar = \frac{a \cdot b}{∣ a ∣}; vector = \frac{a \cdot b}{∣ a ∣ ^{2}} a$; Projection of b on a.
Cross product: $a \times b = \hat{i} a_{1} b_{1} \hat{j} a_{2} b_{2} \hat{k} a_{3} b_{3}, ∣ a \times b ∣ = ∣ a ∣∣ b ∣ sin θ$; Perpendicular to both.
Areas: $parallelogram = ∣ a \times b ∣; triangle = \frac{1}{2} ∣ A B \times A C ∣; from diagonals = \frac{1}{2} ∣ d_{1} \times d_{2} ∣$
Identity: $∣ a \times b ∣^{2} + (a \cdot b)^{2} = ∣ a ∣^{2} ∣ b ∣^{2}$; Lagrange's identity.

How the board asks Vector Algebra

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Section A one-markers	Magnitude of a combination like $∣ a - 2 b ∣$ ; product of cross products; unit vectors; projections.	Combine components first, then take magnitudes.
Basic vectors (2 marks)	Vector of given magnitude along a direction; λ for perpendicular or parallel vectors; section formula.	Unit vector times magnitude; dot = 0 for ⊥; proportional components for ∥.
Angles and projections	Angle between vectors; projection of $A B$ on $B C$ .	Get directions right ( $A B = b - a$ ); keep the sign of the projection.
Cross product and area (3 marks)	Area of triangle ABC from position vectors; unit vector perpendicular to two vectors; parallelogram from diagonals.	Determinant expansion with signs; magnitude; halve for a triangle.
Real-life vectors	Displacement, work done by a force, moment of a force.	Work = $F \cdot d$ ; moment = $r \times F$ .
Short proofs	Show $∣ a + b ∣ = ∣ a - b ∣ \Rightarrow a ⊥ b$ ; find θ when $∣ a \times b ∣ = k (a \cdot b)$ .	Square magnitudes using dot products; divide sin by cos.

Tricks to remember Vector Algebra

Short memory aids. Say each one aloud once, then use it on a question the same day.

Head minus tail
$A B$ = position vector of B minus that of A.
Use it when: Avoiding the reversed subtraction examiners report.
Dot gives cos, cross gives sin
a·b = ab cos θ (a number); |a × b| = ab sin θ (a vector's size).
Use it when: Angles and areas.
i-j-k cycle
Go round the circle i → j → k → i: forward gives +, backward gives −.
Use it when: Products of unit vectors.
Determinant for cross
Put i, j, k in the top row and expand: remember the minus on the j term.
Use it when: Computing $a \times b$ .
Half for triangles
Parallelogram = |cross|; triangle = half; diagonals form = half too.
Use it when: Area questions.

Vector Algebra: board-level solved questions

Ordered from board level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Magnitude of a combination · 1 mark · MCQBoard level

If $a = \hat{i} + 2 \hat{j} - \hat{k}$ and $b = 2 \hat{i} + \hat{j} + \hat{k}$ , then $∣ a - 2 b ∣$ is:
(a) $14$
(b) $32$
(c) 6
(d) $10$

Try it first, then show the solutionSolution

$a - 2 b = - 3 \hat{i} + 0 \hat{j} - 3 \hat{k}$ ; magnitude $18 = 32$ .

Answer: (b)

Example 2: A dot-product identity · 1 mark · Assertion-reasonBoard level

Assertion: For any vectors $a$ and $b$ , $(a + b) \cdot (a - b) = ∣ a ∣^{2} - ∣ b ∣^{2}$ .
Reason: The scalar product is commutative.
(a) Both Assertion and Reason are true, and Reason is the correct explanation of Assertion.
(b) Both Assertion and Reason are true, but Reason is not the correct explanation of Assertion.
(c) Assertion is true and Reason is false.
(d) Assertion is false and Reason is true.

Try it first, then show the solutionSolution

Expanding gives $a \cdot a - a \cdot b + b \cdot a - b \cdot b$ ; the middle terms cancel because $a \cdot b = b \cdot a$ .

Answer: (a)

Example 3: A vector of given length · 2 marks · Short answerHarder board level

Find a vector of magnitude 14 units in the direction of $2 \hat{i} - 3 \hat{j} + 6 \hat{k}$ . [2]

Try it first, then show the solutionSolution

Magnitude of the given vector: 7; unit vector $\frac{1}{7} (2 \hat{i} - 3 \hat{j} + 6 \hat{k})$ .
Required vector $= 14 \times$ unit vector $= 4 \hat{i} - 6 \hat{j} + 12 \hat{k}$ .

Answer: $4 \hat{i} - 6 \hat{j} + 12 \hat{k}$

Example 4: Internal and external division · 3 marks · Short answerHarder board level

Find the position vectors of the points dividing the join of P( $\hat{i} + 2 \hat{j} + 3 \hat{k}$ ) and Q( $4 \hat{i} - \hat{j} + 6 \hat{k}$ ) in the ratio 2 : 1 internally and externally. [3]

Try it first, then show the solutionSolution

Internal: $\frac{2 q + p}{3} = \frac{9 i ^ + 0 j ^ + 15 k ^}{3} = 3 \hat{i} + 5 \hat{k}$ .
External: $\frac{2 q - p}{2 - 1} = 7 \hat{i} - 4 \hat{j} + 9 \hat{k}$ .
Check: the internal point lies between P and Q; the external one lies beyond Q.

Answer: $3 \hat{i} + 5 \hat{k}$ ; $7 \hat{i} - 4 \hat{j} + 9 \hat{k}$

Example 5: Angle and projection · 3 marks · Short answerHarder board level

For $a = \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $b = 2 \hat{i} - \hat{j} + 2 \hat{k}$ , find the angle between them and the projection of $b$ on $a$ . [3]

Try it first, then show the solutionSolution

$a \cdot b = 4$ , $∣ a ∣ = ∣ b ∣ = 3$ .
$cos θ = \frac{4}{9}$ , $θ = cos^{- 1} \frac{4}{9}$ .
Projection of $b$ on $a$ : $\frac{4}{3}$ .

Answer: $cos^{- 1} \frac{4}{9}$ ; 4/3

Example 6: Area of a triangle from position vectors · 5 marks · Long answerHarder board level

The points A, B, C have position vectors $\hat{i} + \hat{j} + \hat{k}$ , $2 \hat{i} + 3 \hat{j} + \hat{k}$ and $3 \hat{i} + \hat{j} + 2 \hat{k}$ .
(a) Find $A B$ and $A C$ . [1]
(b) Find $A B \times A C$ and the area of triangle ABC. [3]
(c) Find the angle A. [1]

Try it first, then show the solutionSolution

(a) $A B = \hat{i} + 2 \hat{j}$ , $A C = 2 \hat{i} + \hat{k}$ .
(b) $A B \times A C = \hat{i} 12 \hat{j} 20 \hat{k} 01 = 2 \hat{i} - \hat{j} - 4 \hat{k}$ .
$∣ A B \times A C ∣ = 4 + 1 + 16 = 21$ .
Area $= \frac{21}{2} \approx 2.29$ square units.
(c) $cos A = \frac{A B \cdot A C}{∣ A B ∣∣ A C ∣} = \frac{2}{5 5} = \frac{2}{5}$ , so A = $cos^{- 1} \frac{2}{5}$ .

Answer: (b) $\frac{21}{2}$ (c) $cos^{- 1} \frac{2}{5}$

Example 7: Angle from a cross-dot relation · 3 marks · Short answerHOTS / competency

Non-zero vectors $a$ and $b$ satisfy $∣ a \times b ∣ = 3 (a \cdot b)$ . Find the angle between them. If $∣ a ∣ = 2$ and $∣ b ∣ = 3$ , find $a \cdot b$ . [3]

Try it first, then show the solutionSolution

$∣ a ∣∣ b ∣ sin θ = 3 ∣ a ∣∣ b ∣ cos θ \Rightarrow tan θ = 3$ .
θ = π/3 (θ in [0, π], and a·b must be ≥ 0 since the left side is non-negative).
$a \cdot b = 2 \times 3 \times \frac{1}{2} = 3$ .

Answer: π/3; 3

Example 8: Collinear points · 2 marks · Short answerHOTS / competency

Show that the points A(1, 2, 3), B(2, 3, 4) and C(4, 5, 6) are collinear and find the ratio in which B divides AC. [2]

Try it first, then show the solutionSolution

$A B = \hat{i} + \hat{j} + \hat{k}$ , $A C = 3 \hat{i} + 3 \hat{j} + 3 \hat{k} = 3 A B$ : parallel with a common point A, so collinear.
AB : BC = 1 : 2 (since AC = 3 AB).

Answer: Collinear; B divides AC in the ratio 1 : 2

Case-study practice (Section D style)

Read the situation, then answer every part. In the board paper this is a 5-mark question split into 1 + 1 + 2 + 1 marks.

Measuring a triangular sail

Suppose the corners of a triangular sail are at A(1, 1, 1), B(2, 3, 1) and C(3, 1, 2), in metres, relative to the base of a mast.

(i) · 1 mark
Find $A B$ and $A C$ .
Answer
Answer: $\hat{i} + 2 \hat{j}$ and $2 \hat{i} + \hat{k}$
Head minus tail.
(ii) · 1 mark
Find the projection of $A B$ on $A C$ .
Answer
Answer: $\frac{2}{5} \approx 0.89$ m
$\frac{A B \cdot A C}{∣ A C ∣} = \frac{2}{5}$
(iii) · 2 marks
Find the area of the sail.
Answer
Answer: $\frac{21}{2} \approx 2.29$ m²
½|2i − j − 4k| = ½√21.
(iv) · 1 mark
Find a unit vector perpendicular to the sail.
Answer
Answer: $\pm \frac{1}{21} (2 \hat{i} - \hat{j} - 4 \hat{k})$
Normalise the cross product.

Forces on a trolley

Suppose a trolley is pushed with a force $F = 3 \hat{i} + 2 \hat{j} + \hat{k}$ N while it moves from (1, 1, 1) to (4, 3, 2), in metres, and the handle is at $r = 0.5 \hat{i} + \hat{k}$ m from the front axle.

(i) · 1 mark
Find the displacement vector.
Answer
Answer: $3 \hat{i} + 2 \hat{j} + \hat{k}$
(4 − 1, 3 − 1, 2 − 1).
(ii) · 1 mark
Find the work done.
Answer
Answer: 14 J
$F \cdot d = 9 + 4 + 1$
(iii) · 1 mark
Find the angle between the force and the displacement.
Answer
Answer: 0°
They are the same vector, so cos θ = 1: the force acts along the motion.
(iv) · 2 marks
Find the moment of the force about the front axle, $r \times F$ .
Answer
Answer: $- 2 \hat{i} + 2.5 \hat{j} + \hat{k}$ N m
$\hat{i} 0.5 3 \hat{j} 02 \hat{k} 11 = \hat{i} (0 - 2) - \hat{j} (0.5 - 3) + \hat{k} (1 - 0) = - 2 \hat{i} + 2.5 \hat{j} + \hat{k}$

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Find λ so that $2 \hat{i} + λ \hat{j} + \hat{k}$ is perpendicular to $\hat{i} - 2 \hat{j} + 3 \hat{k}$ . (2 marks)

Step	Marks
Sets the dot product to zero: 2 − 2λ + 3 = 0	1
λ = 5/2	1

Find the angle between $\hat{i} + 2 \hat{j} + 2 \hat{k}$ and $2 \hat{i} - \hat{j} + 2 \hat{k}$ , and the projection of the second on the first. (3 marks)

Step	Marks
Dot product 4 and magnitudes 3, 3	1
θ = cos⁻¹(4/9)	1
Projection 4/3	1

Find the area of triangle ABC with position vectors $\hat{i} + \hat{j} + \hat{k}$ , $2 \hat{i} + 3 \hat{j} + \hat{k}$ , $3 \hat{i} + \hat{j} + 2 \hat{k}$ , and a unit vector perpendicular to its plane. (5 marks)

Step	Marks
$A B$ and $A C$ correct (head − tail)	1
Cross product $2 \hat{i} - \hat{j} - 4 \hat{k}$	1
Magnitude √21	1
Area √21/2	1
Unit normal ±(2i − j − 4k)/√21	1

Where students lose marks in Vector Algebra

The slips examiners see most often, and what to write instead.

Mistake: Subtracting position vectors in the wrong order
CISCE examiners' comments, 2025: candidates calculated $A B$ as $O A - O B$ and made arithmetic errors in $A B$ and $B C$ .
Wrong working
$A B = O A - O B$
Right working
$A B = O B - O A$ (head minus tail).
Do this: Write 'head − tail' above the subtraction.
Usually costs: 1-2 marks (errors carry forward).
Mistake: Wrong projection formula or taking its absolute value
CISCE examiners' comments, 2024 and 2025: candidates used an incorrect formula for the projection of $b$ on $a$ , and some took the absolute value of a negative scalar projection.
Wrong working
Projection of $b$ on $a$ = $\frac{a \cdot b}{∣ b ∣}$ , or |−2/3| = 2/3.
Right working
$\frac{a \cdot b}{∣ a ∣}$ , keeping its sign; negative when the angle exceeds 90°.
Do this: Divide by the magnitude of the vector you project on.
Usually costs: 1 mark.
Mistake: Not using the distributive law with a zero cross product
CISCE examiners' comments, 2024: proving that $a \times b = a \times c$ implies $b = c$ or $a ∥ (b - c)$ , candidates did not write $a \times (b - c) = 0$ .
Wrong working
Cancels $a$ to get $b = c$ directly.
Right working
$a \times (b - c) = 0$ , so either $b - c = 0$ or it is parallel to $a$ .
Do this: Vectors cannot be 'cancelled'; bring terms to one side.
Usually costs: 1-2 marks.
Mistake: Sign error on the ĵ term of a cross product
A frequent arithmetic slip in determinant expansion.
Wrong working
$\hat{i} (a_{2} b_{3} - a_{3} b_{2}) + \hat{j} (a_{1} b_{3} - a_{3} b_{1}) + \dots$
Right working
$\hat{i} (a_{2} b_{3} - a_{3} b_{2}) - \hat{j} (a_{1} b_{3} - a_{3} b_{1}) + \hat{k} (a_{1} b_{2} - a_{2} b_{1})$
Do this: Check: the answer's dot product with each vector must be 0.
Usually costs: 1-2 marks.

Practice questions on Vector Algebra

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The unit vector in the direction of $3 \hat{i} - 4 \hat{j}$ is:
(a) $3 \hat{i} - 4 \hat{j}$
(b) $\frac{1}{5} (3 \hat{i} - 4 \hat{j})$
(c) $\frac{1}{7} (3 \hat{i} - 4 \hat{j})$
(d) $\frac{1}{25} (3 \hat{i} - 4 \hat{j})$
Hint
Magnitude 5.
Answer
(b)
Q2 · 1 mark · MCQBoard level
$\hat{j} \times \hat{i}$ equals:
(a) $\hat{k}$
(b) $- \hat{k}$
(c) 0
(d) 1
Hint
Backward round the cycle.
Answer
(b)
Q3 · 1 mark · MCQBoard level
If $a \cdot b = 0$ for non-zero vectors, then the vectors are:
(a) parallel
(b) perpendicular
(c) equal
(d) collinear
Hint
cos θ = 0.
Answer
(b)
Q4 · 1 mark · Very short answerBoard level
Write the direction cosines of $2 \hat{i} - 3 \hat{j} + 6 \hat{k}$ .
Hint
Divide by 7.
Answer
2/7, −3/7, 6/7
Q5 · 2 marks · Short answerBoard level
Find λ if $2 \hat{i} + λ \hat{j} + \hat{k}$ and $\hat{i} - 2 \hat{j} + 3 \hat{k}$ are perpendicular.
Hint
Dot product zero.
Answer
λ = 5/2
Q6 · 2 marks · Short answerHarder board level
Find the projection of $b = 2 \hat{i} - \hat{j} + 2 \hat{k}$ on $a = \hat{i} + 2 \hat{j} + 2 \hat{k}$ , and the vector projection.
Hint
a·b/|a|.
Answer
4/3; $\frac{4}{9} (\hat{i} + 2 \hat{j} + 2 \hat{k})$
Q7 · 2 marks · Short answerHarder board level
Find the area of the parallelogram whose diagonals are $3 \hat{i} + \hat{j} - 2 \hat{k}$ and $\hat{i} - 3 \hat{j} + 4 \hat{k}$ .
Hint
Half the cross product of diagonals.
Answer
$d_{1} \times d_{2} = - 2 \hat{i} - 14 \hat{j} - 10 \hat{k}$ , magnitude 10√3; area 5√3.
Q8 · 2 marks · Short answerHOTS / competency
If $∣ a + b ∣ = ∣ a - b ∣$ , show that $a$ and $b$ are perpendicular.
Hint
Square both sides.
Answer
$∣ a ∣^{2} + 2 a \cdot b + ∣ b ∣^{2} = ∣ a ∣^{2} - 2 a \cdot b + ∣ b ∣^{2} \Rightarrow 4 a \cdot b = 0$ , so they are perpendicular.
Q9 · 3 marks · Short answerHarder board level
For $a = \hat{i} + 2 \hat{j} + 2 \hat{k}$ and $b = 2 \hat{i} - \hat{j} + 2 \hat{k}$ , find $a \times b$ and verify that it is perpendicular to both.
Hint
Dot with each vector.
Answer
$6 \hat{i} + 2 \hat{j} - 5 \hat{k}$ ; dot with a: 6 + 4 − 10 = 0; with b: 12 − 2 − 10 = 0.
Q10 · 3 marks · Short answerHarder board level
A force $3 \hat{i} + 2 \hat{j} + \hat{k}$ N moves a particle from (1, 1, 1) to (4, 3, 2) m. Find the work done. Also find the moment about the origin of the force $10 \hat{j}$ N acting at $0.2 \hat{i}$ m.
Hint
F·d and r × F.
Answer
Work 14 J; moment $2 \hat{k}$ N m.
Q11 · 3 marks · Short answerHarder board level
Show that A(1, 2, 3), B(2, 3, 4) and C(4, 5, 6) are collinear, and find the position vector of the point dividing AC in the ratio 1 : 2.
Hint
AC = 3AB.
Answer
$A C = 3 A B$ , so collinear. Point = $\frac{1 \cdot c + 2 a}{3} = (2, 3, 4)$ , which is B.
Q12 · 5 marks · Long answerHarder board level
(a) Find the area of the triangle with vertices A(1, 1, 1), B(2, 3, 1), C(3, 1, 2). [3]
(b) Find the angle at A. [2]
Hint
AB × AC.
Answer
(a) $\frac{21}{2}$ (b) $cos^{- 1} \frac{2}{5}$
Q13 · 2 marks · Short answerHarder board level
If $∣ a ∣ = 3$ , $∣ b ∣ = 4$ and $a \cdot b = 6$ , find the angle between them and $∣ a \times b ∣$ .
Hint
cos θ = 1/2.
Answer
θ = π/3; $∣ a \times b ∣ = 12 \times \frac{3}{2} = 63$

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Magnitude of $a_{1} \hat{i} + a_{2} \hat{j} + a_{3} \hat{k}$ ?Show answer
$a_{1}^{2} + a_{2}^{2} + a_{3}^{2}$ .
2.Sum of squares of direction cosines?Show answer
1.
3.Section formula (internal, m : n)?Show answer
$\frac{m b + n a}{m + n}$ .
4.Condition for perpendicular vectors?Show answer
$a \cdot b = 0$ .
5.Condition for parallel vectors?Show answer
$a \times b = 0$ (components proportional).
6.Projection of $b$ on $a$ ?Show answer
$\frac{a \cdot b}{∣ a ∣}$ .
7.Area of triangle ABC by vectors?Show answer
$\frac{1}{2} ∣ A B \times A C ∣$ .
8. $\hat{k} \times \hat{j}$ = ?Show answer
$- \hat{i}$ .

Best order to study it

Learn components, magnitude, unit vectors and direction cosines.
Learn types of vectors, collinearity and position vectors.
Practise the section formula (internal and external).
Master the dot product: angles, perpendicularity and projections.
Master the cross product: determinant expansion, areas and unit normals.
Finish with real-life uses (work, moment) and short proofs.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you find a vector of given magnitude along a direction?
Can you write direction cosines and ratios?
Can you use the section formula both ways?
Can you find the angle between two vectors?
Can you find scalar and vector projections with the right sign?
Can you expand a cross product without sign errors?
Can you find the area of a triangle or parallelogram?
Can you test collinearity of three points?

Vector Algebra: frequently asked questions

Is Vector Algebra compulsory in ISC 2027?

Yes. It is now Unit 4 (5 marks) of the compulsory paper; earlier it was part of the optional Section B.

Are proofs of geometrical theorems by vectors required?

No. The syllabus excludes proofs of geometrical theorems using vector algebra.

Is the scalar triple product in the syllabus?

It is not listed in the ISC 2027 Vector Algebra content, which covers the scalar and vector products with their geometrical significance.

What is the difference between scalar and vector projection?

The scalar projection of \vec{b} on \vec{a} is the signed length \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|}; the vector projection is that length times the unit vector \hat{a}.

Why is the cross product anticommutative?

Reversing the order reverses the direction given by the right-hand rule, so \vec{b} \times \vec{a} = -\vec{a} \times \vec{b}.

How many marks does this chapter carry?

Vector Algebra carries 5 of the 80 marks, and its methods are also used in Three-dimensional Geometry (6 marks).

Based on: CISCE ISC 2027 Mathematics syllabus; CISCE ISC 2027 Mathematics specimen question paper and answer key; CISCE Analysis of Pupil Performance, ISC Mathematics, 2025; CISCE Analysis of Pupil Performance, ISC Mathematics, 2024