Vector Algebra Class 12 Exemplar Solutions - Class 12 Maths Ch 10

The Vector Algebra Class 12 Exemplar Solutions page compiles NCERT Class 12 Mathematics Chapter 10 into a single download-ready resource, aligned to the 2026-27 NCERT syllabus. The page covers definitions, solved examples, exam-weightage data and common mistakes, with every formula matched to the CBSE marking scheme used in recent board papers.

• Exemplar Problems Solved: 45 in total, split as 14 Short Answer, 4 Long Answer, 15 MCQ, 7 Fill in the Blanks, and 5 True or False.

The chapter sustains 12 named identities (Lagrange, triple cross, scalar triple product, section formula, projection, angle, area, coplanarity, parallelism, orthogonality, unit-vector, direction-cosine) and runs across 8 NCERT Exemplar pages in the 2026-27 print.

Prepared by Collegedunia subject experts, mapped to the 2026-27 NCERT Exemplar print, and benchmarked against the last five CBSE Board and JEE Main cycles on Vector Algebra.

Also Check:

• Vector Algebra Class 12 Maths NCERT Exemplar Book PDF
• the PDF Maths NCERT Solutions

NCERT Exemplar Class 12 Maths Solutions Chapter 10 Vector Algebra: Section-Wise Question Distribution

The 45 Exemplar problems sit across five sections. The MCQ block is the largest, mirroring how JEE Main weighs the the resource Exemplar Solutions, while the four Long Answer items carry the heaviest CBSE-style proof load.

Vector Algebra NCERT Exemplar Video Solutions

Source: Magnet Brains on YouTube

Identities and Formulae Carried by the Class 12 Vector Algebra Exemplar

The chapter notes Exemplar Solutions address this in the same order as the NCERT textbook.

Every solution opens with a one-line concept identification followed by the formula being applied. The Exemplar leans on a compact set of identities, repeated across MCQ, SA, and LA blocks.

• Determinant cross product is expanded verbatim in every solution that needs one, with the i,j,k minors written on separate lines so the sign-of-the-minor logic stays visible.
• Lagrange's identity |a×b|²+(a·b)²=|a|²|b|² is the lever in Q27, Q38, and Q39; cross-references make the identity reappear across the MCQ and Fill in the Blank blocks.
• Triple cross identity a×(a×c)=(a·c)a-|a|²c shortens Q18 to a one-line derivation.
• Section formula r=mb+nam+n (internal division) and its external counterpart anchor Q3 and Q4.
• Projection of a on b returns the scalar a·b|b| ; the vector projection multiplies that scalar by the unit vector b .

How These Class 12 Maths Exemplar Solutions for Vector Algebra Strengthen Your Preparation

The the PDF Exemplar Solutions address this in the same order as the NCERT textbook.

Wrong product choice at step one wastes the entire question. Each of the 45 solutions opens with a one-line concept identification, names the formula, and shows the determinant expansion or Lagrange manipulation in full. A single sign error inside the determinant can flip the unit-vector direction, so the working is laid out step by step. The Expert Solution after every question adds a JEE Main alternate angle on the same problem.

• Determinant minors written on separate lines so the alternating sign is visible.
• Pythagorean-triple shortcuts (2,3,6,7) and (1,2,2,3) are flagged in the Expert Solution where they apply, saving rough work in MCQ shifts.
• Every Fill in the Blank gets the algebraic justification, not just the numeric answer, so the reader can back-solve in the exam.
• True or False entries include the counter-example or proof, not a bare verdict.

Sample MCQ Solved: Lagrange's Identity in Action (Exemplar Q27)

The this chapter Exemplar Solutions address this in the same order as the NCERT textbook.

Question 27 is the canonical Lagrange application. Given |a|=10 , |b|=2 , a·b=12 , the question asks for |a×b| . The Exemplar tests the identity in three places (Q27, Q38, Q39), so internalising it pays off across these notes Exemplar Solutions.

PYQ Templates Cross-Referenced Across the Class 12 Vector Algebra Exemplar

The 45 Exemplar problems map onto five recurring CBSE and JEE templates. Each solution is tagged with its template at the top of the Expert Solution so you can rehearse by archetype, not by question number.

Full year-wise PYQ map: Vector Algebra NCERT Solutions for Class 12 Maths for fully solved CBSE 2021 to 2025 problems with marking-scheme step weightages.

Common Mistakes the Class 12 Vector Algebra Exemplar Solutions Flag

The this Class 12 page Exemplar Solutions are written in formal mathematical notation, line by line, in the same convention as the official NCERT print.

The Mistake boxes inside the the resource Exemplar Solutions flag the four most expensive 1-mark errors in Class 12 Vector Algebra. Each ties directly to a formula in the chapter notes Exemplar Solutions, and each appears at least twice in the Exemplar's 45 problems.

• Writing a·b as a vector: the dot product returns a scalar. Reporting a vector loses 1 mark immediately.
• Missing sinθ in |a×b| . The common slip is writing |a×b|=|a||b| .
• Forgetting the minus sign on swap: a×b=-(b×a) .
• Reporting direction ratios when direction cosines are asked. Q7 is the canonical place this trap appears.

Why Vector Algebra Is the 3D-Geometry Springboard for Class 12 Maths

Every JEE Main paper since 2021 has carried at least one cross-product or scalar triple product question, and Chapter 10 is the only place where the algebra of those products is rehearsed end-to-end.

The Exemplar's 15-MCQ block is the densest type-recognition drill in the Class 12 Vector Algebra syllabus, training the reflex of choosing dot or cross before any computation. Chapter 11 Three Dimensional Geometry depends on every identity learnt here, so weak Vector Algebra preparation cascades into weak 3D Geometry.

Related Resources for Class 12 Maths Chapter 10 Vector Algebra

• NCERT Notes for Class 12 Maths Chapter 10 Vector Algebra
• NCERT Solutions for Class 12 Maths Chapter 10 Vector Algebra
• the PDF Maths Formula Sheet
• this chapter Maths Handwritten Notes
• NCERT Book PDF for Class 12 Maths Chapter 10 Vector Algebra
• NCERT Exemplar Book PDF for Class 12 Maths Chapter 10 Vector Algebra

All NCERT Exemplar Questions for Vector Algebra with Step-by-Step Solutions

Every question of the NCERT Exemplar set for Class 12 Mathematics Chapter 10 Vector Algebra is listed below with its full Solution and Expert Solution hidden inside collapsible tabs. Click Check Solution to reveal the step-by-step working; click Expert Solution for the expanded explanation.

Questions

Q 10.1

Find the unit vector in the direction of sum of vectors a=2i-j+k and b=2j+k.

Check SolutionExpert Solution

Concept used. The unit vector c in the direction of a non-zero vector c is obtained by dividing the vector by its magnitude, c=c|c|, where |c|=√c₁²+c₂²+c₃² for c=c₁i+c₂j+c₃k.

1. Add the vectors component-wise: c=a+b=(2+0)i+(-1+2)j+(1+1)k=2i+j+2k.
2. Compute the magnitude: |c|=√2²+1²+2²=√4+1+4=√9=3.
3. Divide each component by the magnitude: c=c|c|=13(2i+j+2k)=23i+13j+23k.

c=13(2i+j+2k).

AS

Aarav Sharma

M.Sc Mathematics, IIT Bombay

Verified Expert

Strategic angle. Whenever an Exemplar question asks for "unit vector in the direction of sum/difference of two vectors", the routine is identical: add, compute magnitude, divide. The magnitude check |c|=1 is a free self-test you should run mentally.

1. Vector sum: c=(2,-1,1)+(0,2,1)=(2,1,2).
2. Magnitude: √4+1+4=√9=3.
3. Unit vector: c=13(2,1,2). Verify: √(2/3)²+(1/3)²+(2/3)²=√4/9+1/9+4/9=√1=1 .

Why this matters. The "divide by magnitude" pattern is the workhorse for direction cosines (where l,m,n are exactly the components of a). c=13(2i+j+2k).

Q 10.2

If a=i+j+2k and b=2i+j-2k, find the unit vector in the direction of (i) 6b, (ii) 2a-b.

Check SolutionExpert Solution

Concept used. A non-zero scalar multiple λv has the same direction as v when λ>0. So the unit vector along 6b equals the unit vector along b. For (ii) we compute the linear combination first, then normalise.

1. (i) Unit vector along 6b is b. Compute |b|=√2²+1²+(-2)²=√4+1+4=3. Hence b=b|b|=13(2i+j-2k).
2. (ii) Form 2a-b component-wise: 2a=2i+2j+4k, 2a-b=(2-2)i+(2-1)j+(4+2)k=j+6k.
3. Magnitude: |2a-b|=√0+1+36=√37. Hence 2a-b̂=1√37(j+6k).

(i) b=13(2i+j-2k).  (ii) 1√37(j+6k).

SI

Sneha Iyer

M.Sc Mathematics, ISI Kolkata

Verified Expert

Strategic angle. Spot the shortcut: scaling a vector by a positive scalar never changes its unit vector. Part (i) reduces to "find b", saving the multiplication-by-6 step.

1. (i) |b|=√9=3, so b=13(2,1,-2).
2. (ii) 2a-b=(0,1,6), magnitude √37.
3. Unit vector 1√37(0,1,6).

Why this matters. Recognising sign and scale rules of λv before computing trims arithmetic and reduces 1-mark slips in CBSE 2-mark questions. (i) 13(2i+j-2k), (ii) 1√37(j+6k).

Q 10.3

Find a unit vector in the direction of PQ⃗, where P and Q have coordinates (5,0,8) and (3,3,2), respectively.

Check SolutionExpert Solution

Concept used. For points P(x₁,y₁,z₁) and Q(x₂,y₂,z₂), the vector from P to Q has components PQ⃗=(x₂-x₁)i+(y₂-y₁)j+(z₂-z₁)k. Normalise by dividing by its magnitude.

1. Component-wise: PQ⃗=(3-5)i+(3-0)j+(2-8)k=-2i+3j-6k.
2. Magnitude: |PQ⃗|=√(-2)²+3²+(-6)²=√4+9+36=√49=7.
3. Unit vector: PQ̂=17(-2i+3j-6k).

PQ̂=17(-2i+3j-6k).

KM

Karan Mehta

M.Tech CS, IIT Madras

Verified Expert

Strategic angle. A clean Pythagorean triple 2,3,6,7 (4+9+36=49) makes the magnitude pop out instantly. Memorise such triples for board questions.

1. Δ x=-2, Δ y=3, Δ z=-6.
2. |PQ⃗|=7 (triple).
3. PQ̂=17(-2,3,-6).

Why this matters. The triples (2,3,6,7), (1,2,2,3), (3,4,12,13) recur in JEE Main and CBSE; recognising them shaves seconds. 17(-2i+3j-6k).

Q 10.4

If a and b are the position vectors of A and B, respectively, find the position vector of a point C in BA produced such that BC=1.5 BA.

Check SolutionExpert Solution

Concept used. If C lies on the line BA produced (beyond A) with BC=1.5 BA, then BC⃗=1.5 BA⃗. Use this directly to find OC⃗.

1. Direction vector: BA⃗=a-b (from B to A).
2. Since C is on BA produced with BC=1.5 BA: BC⃗=1.5 BA⃗=1.5(a-b).
3. Hence OC⃗=OB⃗+BC⃗=b+1.5(a-b)=1.5a-0.5b=3a-b2.

OC⃗=3a-b2.

PK

Pranav Kumar

Ph.D Mathematics, IIT Delhi

Verified Expert

Strategic angle. Frame as section formula: C divides BA externally in ratio 3:1 (since BC:CA=3:1 with sign convention), giving OC⃗=3a-b2.

1. BC=1.5 BA⇒ C lies beyond A on the same line.
2. External ratio m:n=3:1.
3. Apply OC⃗=ma-nbm-n=3a-b2.

Why this matters. CBSE often disguises section formula as "produced to" language; learn to translate BC=k BA into a ratio. OC⃗=3a-b2.

Q 10.5

Using vectors, find the value of k such that the points (k,-10,3), (1,-1,3) and (3,5,3) are collinear.

Check SolutionExpert Solution

Concept used. Three points A,B,C are collinear iff the vectors AB⃗ and AC⃗ are parallel, i.e. AB⃗=λ AC⃗ for some scalar λ. Equating components gives a system that pins down both λ and any unknown coordinate.

1. Let A=(k,-10,3), B=(1,-1,3), C=(3,5,3). Then AB⃗=(1-k,9,0), AC⃗=(3-k,15,0).
2. For AB⃗∥AC⃗, the non-zero components are proportional: 1-k3-k=915=35.
3. Cross-multiply: 5(1-k)=3(3-k)⇒ 5-5k=9-3k⇒ -2k=4⇒ k=-2.

k=-2.

AP

Aanya Patel

M.Sc Applied Mathematics, IIT Kanpur

Verified Expert

Strategic angle. Alternative test: collinear iff AB⃗×AC⃗=0. Since the z-components match, both vectors lie in the plane z=3 and the cross product is along k only.

1. Components: AB⃗=(1-k,9,0), AC⃗=(3-k,15,0).
2. Cross product k-component: (1-k)(15)-9(3-k)=15-15k-27+9k=-12-6k.
3. Set equal to zero: -12-6k=0⇒ k=-2.

Why this matters. Two valid collinearity tests; pick the one that gives one equation, not three. k=-2.

Q 10.6

A vector r is inclined at equal angles to the three axes. If the magnitude of r is 2√3 units, find r.

Check SolutionExpert Solution

Concept used. If r makes equal angles α with all three axes, then l=m=n=cosα. The identity l²+m²+n²=1 pins down l. Then r=|r|(li+mj+nk).

1. Set l=m=n=t. The identity gives 3t²=1⇒ t=±1√3.
2. Hence r=|r|(ti+tj+tk)=2√3·±1√3(i+j+k)=± 2(i+j+k).

r=± 2(i+j+k).

RS

Riya Singh

B.Tech CSE, IIT Roorkee

Verified Expert

Strategic angle. "Equal angles to the axes" is shorthand for direction cosines (t,t,t). The plus-or-minus sign is intentional: the question does not specify which octant, so both r=2(i+j+k) and r=-2(i+j+k) are valid.

1. l=m=n⇒ 3l²=1⇒ l=± 1/3.
2. Scale by |r|=23: components become ± 2 each.

Why this matters. Direction cosines come in ± pairs; always report both unless the question fixes an octant. r=± 2(i+j+k).

Q 10.7

A vector r has magnitude 14 and direction ratios 2,3,-6. Find the direction cosines and components of r, given that r makes an acute angle with the x-axis.

Check SolutionExpert Solution

Concept used. Direction cosines (l,m,n) are direction ratios normalised: l=a√a²+b²+c², similar for m,n. Acute angle with the x-axis means l=cosα>0.

1. Compute the normaliser: √2²+3²+(-6)²=√4+9+36=√49=7.
2. Direction cosines: l=27>0 (acute, accept), m=37, n=-67.
3. Components: r=|r|(li+mj+nk)=14·(27i+37j-67k)=4i+6j-12k.

Direction cosines =(27,37,-67); r=4i+6j-12k.

AV

Aditya Verma

Ph.D Pure Mathematics, IISc Bangalore

Verified Expert

Strategic angle. The condition "acute angle with x-axis" excludes the negated solution. Verify by checking l>0 at the end.

1. Normaliser 7.
2. (l,m,n)=(2/7,3/7,-6/7).
3. r=14(l,m,n)=(4,6,-12).

Why this matters. A negated solution (l,m,n)=(-2/7,-3/7,6/7) would also normalise correctly but fails the acute-angle test. (l,m,n)=(2/7,3/7,-6/7); r=4i+6j-12k.

Q 10.8

Find a vector of magnitude 6, which is perpendicular to both the vectors 2i-j+2k and 4i-j+3k.

Check SolutionExpert Solution

Concept used. The cross product a×b is perpendicular to both a and b. To obtain a vector of magnitude 6 in that direction, scale the unit vector a×b|a×b| by 6.

1. Cross product via 3× 3 determinant: a×b=vmatrixi&j&k
   2&-1&2
   4&-1&3vmatrix. Expanding along row 1: a×b=i[(-1)(3)-(2)(-1)]-j[(2)(3)-(2)(4)]+k[(2)(-1)-(-1)(4)]. Compute each minor: i[-3+2]=-i; -j[6-8]=2j; k[-2+4]=2k. Hence a×b=-i+2j+2k.
2. Magnitude: √(-1)²+2²+2²=√1+4+4=√9=3.
3. Required vector: ±63(-i+2j+2k)=± 2(-i+2j+2k)=±(-2i+4j+4k).

±(-2i+4j+4k).

VR

Vivaan Reddy

M.Sc Mathematics, IIT Bombay

Verified Expert

Strategic angle. The ± sign captures the two perpendicular directions (right-hand and left-hand). CBSE usually accepts both unless a right-handed frame is specified.

1. Determinant expansion gives (-1,2,2).
2. Magnitude =3 (a Pythagorean triple 1,2,2,3).
3. Scale by 6/3=2.

Why this matters. The "magnitude k vector perpendicular to both" template is the 5-mark CBSE staple. The two-step recipe is rock solid. ±(-2i+4j+4k).

Q 10.9

Find the angle between the vectors 2i-j+k and 3i+4j-k.

Check SolutionExpert Solution

Concept used. The angle θ between two non-zero vectors satisfies cosθ=a·b|a||b|, with θ∈[0,π].

1. Dot product: a·b=(2)(3)+(-1)(4)+(1)(-1)=6-4-1=1.
2. Magnitudes: |a|=√4+1+1=√6, |b|=√9+16+1=√26.
3. Apply: cosθ=1√6·√26=1√156=12√39.
4. Hence θ=cos^-1(12√39).

θ=cos^-1(12√39)≈ 85.4^∘.

TJ

Tara Joshi

M.Sc Mathematics, IIT Bombay

Verified Expert

Strategic angle. The dot product =1 is small relative to |a||b|=√156≈ 12.5, so cosθ≈ 0.08, meaning θ is close to but slightly less than 90^∘.

1. a·b=1.
2. |a||b|=√6√26=√156=2√39.
3. cosθ=1/(2√39), so θ=cos^-1(1/(2√39)).

Why this matters. Estimating cosθ magnitude before computing helps catch arithmetic slips. θ=cos^-1(12√39).

Q 10.10

If a+b+c=0, show that a×b=b×c=c×a. Interpret the result geometrically.

Check SolutionExpert Solution

Concept used. The cross product is distributive over addition and anti-commutative: u×(v+w)=u×v+u×w and u×u=0.

1. From a+b+c=0 we have c=-a-b.
2. Cross both sides of a+b+c=0 with a on the right: a×a+b×a+c×a=0⇒ b×a=-c×a⇒ a×b=c×a.
3. Cross with b on the right: a×b+b×b+c×b=0⇒ c×b=-a×b⇒ b×c=a×b.
4. Combining: a×b=b×c=c×a. Geometric meaning: if a,b,c are sides of a triangle taken in order (a+b+c=0), then twice the area of the triangle equals |a×b|=|b×c|=|c×a|, i.e. all three pairwise cross products have equal magnitude and direction.

a×b=b×c=c×a; geometrically, equal pairwise cross products =2(area of ).

KN

Krishna Nair

Ph.D Mathematics, IIT Delhi

Verified Expert

Strategic angle. The condition a+b+c=0 is the closure condition for a triangle's side vectors. Geometric interpretation comes for free once you see that.

1. Cross with a, b in turn; use anti-commutativity to swap signs.
2. Each pairwise cross product gives ± 2(area) n in the plane normal direction.

Why this matters. This identity proves the law of sines via cross-product magnitudes. All three equal; magnitude =2(area of triangle).

Q 10.11

Find the sine of the angle between the vectors a=3i+j+2k and b=2i-2j+4k.

Check SolutionExpert Solution

Concept used. sinθ=|a×b||a||b|, where θ∈[0,π] and sinθ≥ 0 throughout.

1. Cross product determinant: a×b=vmatrixi&j&k
   3&1&2
   2&-2&4vmatrix=i(4+4)-j(12-4)+k(-6-2)=8i-8j-8k.
2. |a×b|=√64+64+64=8√3.
3. Magnitudes: |a|=√9+1+4=√14, |b|=√4+4+16=√24=2√6. Product =√14· 2√6=2√84=4√21.
4. Hence sinθ=8√34√21=2√3√21=2√7=2√77.

sinθ=2√77.

DK

Diya Kapoor

M.Sc Mathematics, ISI Kolkata

Verified Expert

Strategic angle. When the answer is asked for sinθ, use the cross form directly. It saves an inverse-cosine identity step.

1. Cross product =(8,-8,-8), magnitude 83.
2. Magnitudes √14,26.
3. sinθ=83/(4√21)=23/√21=2/7.

Why this matters. Rationalising 2/7 to 27/7 is expected in the final answer. sinθ=277.

Q 10.12

If A,B,C,D are the points with position vectors i+j-k, 2i-j+3k, 2i-3k, 3i-2j+k, respectively, find the projection of AB⃗ along CD⃗.

Check SolutionExpert Solution

Concept used. The scalar projection of u along v is proj_vu=u·v|v|.

1. AB⃗=B-A=(2-1)i+(-1-1)j+(3+1)k=i-2j+4k.
2. CD⃗=D-C=(3-2)i+(-2-0)j+(1+3)k=i-2j+4k.
3. Notice AB⃗=CD⃗. Hence projection of AB⃗ on CD⃗ equals |CD⃗|: |CD⃗|=√1+4+16=√21.
4. Projection: AB⃗·CD⃗|CD⃗|=(1)(1)+(-2)(-2)+(4)(4)√21=1+4+16√21=21√21=√21.

Projection =√21.

IB

Ishita Bhat

M.Sc Applied Mathematics, IIT Kanpur

Verified Expert

Strategic angle. When u=v, the projection collapses to |v|. Catching this early saves the dot-product computation.

1. Both AB⃗ and CD⃗ equal (1,-2,4).
2. Projection =|CD⃗|=√21.

Why this matters. Coordinate questions sometimes set up parallel/equal vectors deliberately to test pattern recognition. √21.

Q 10.13

Using vectors, find the area of the triangle ABC with vertices A(1,2,3), B(2,-1,4) and C(4,5,-1).

Check SolutionExpert Solution

Concept used. Area of triangle with vertices A,B,C is 12|AB⃗×AC⃗|.

1. Side vectors from A: AB⃗=(2-1,-1-2,4-3)=(1,-3,1), AC⃗=(4-1,5-2,-1-3)=(3,3,-4).
2. Cross product: AB⃗×AC⃗=vmatrixi&j&k
   1&-3&1
   3&3&-4vmatrix=i(12-3)-j(-4-3)+k(3+9)=9i+7j+12k.
3. Magnitude: √81+49+144=√274.
4. Area: 12√274 square units.

Area =12√274 sq. units.

YD

Yash Desai

B.Tech CSE, IIT Roorkee

Verified Expert

Strategic angle. The cross product magnitude equals the area of the parallelogram on AB⃗,AC⃗; halve for the triangle.

1. AB⃗=(1,-3,1), AC⃗=(3,3,-4).
2. AB⃗×AC⃗=(9,7,12).
3. √81+49+144=√274, half =√2742.

Why this matters. Choosing A as the pivot keeps the arithmetic compact. Any vertex works. √2742 sq. units.

Q 10.14

Using vectors, prove that the parallelograms on the same base and between the same parallels are equal in area.

Check SolutionExpert Solution

Concept used. Area of a parallelogram with adjacent-side vectors a and b equals |a×b|. Vectors between the same parallels differ only by a vector along the base, so their cross product with the base vector is unchanged.

1. Let two parallelograms share base AB⃗=a and lie between the same pair of parallel lines. Let one parallelogram have adjacent side b and the other have adjacent side b'.
2. Since both other sides terminate on the same parallel line as b, the vectors b and b' differ by a vector along the base: b'=b+ta for some scalar t.
3. Area of second parallelogram: |a×b'|=|a×(b+ta)|=|a×b+t(a×a)|=|a×b+0|=|a×b|. Hence both parallelograms have area |a×b|.

Area₁ = Area₂ =|a×b|.

MB

Meera Banerjee

M.Sc Mathematics, IIT Bombay

Verified Expert

Strategic angle. The key algebraic move is a×a=0, which makes the ta contribution vanish.

1. Base a, second sides b, b' with b'-b∥a.
2. a×b'=a×b because a×(ta)=0.
3. Areas equal.

Why this matters. The classical Euclidean theorem reduces to one line in vector form. Both areas =|a×b|.

Q 10.15

Prove that in any triangle ABC, cos A=b²+c²-a²2bc, where a,b,c are the magnitudes of the sides opposite to vertices A,B,C, respectively.

Check SolutionExpert Solution

Concept used. Use the closure condition for the triangle's side vectors and the algebraic identity |v|²=v·v. The angle between two side vectors emerging from a vertex appears via the dot product.

1. Let the sides be vectors BC⃗=a, CA⃗=b, AB⃗=c. Taken in order, a+b+c=0, so a=-(b+c).
2. Square magnitudes: |a|²=|b+c|²=|b|²+2b·c+|c|².
3. Hence a²=b²+c²+2b·c, so b·c=a²-b²-c²2.
4. At vertex A, the vectors AB⃗=c and AC⃗=-b make angle A between them, so c·(-b)=|c|·|b|cos A⇒ -b·c=bccos A.
5. Substitute from step 3: -a²-b²-c²2=bccos A⇒ cos A=b²+c²-a²2bc.

cos A=b²+c²-a²2bc (Cosine Rule).

RG

Rohit Gupta

Ph.D Pure Mathematics, IISc Bangalore

Verified Expert

Strategic angle. The choice of side directions matters: AB⃗,AC⃗ both emerge from A, giving angle A between them directly; but the "in order" closure uses BC⃗,CA⃗,AB⃗, so a sign flip AC⃗=-CA⃗=-b is needed.

1. Closure a+b+c=0.
2. Square: a²=b²+c²+2b·c.
3. Sign flip at A: cos A=-b·c/(bc).
4. Solve simultaneously.

Why this matters. CBSE 5-mark proofs reward neat sign tracking. cos A=b²+c²-a²2bc.

Q 10.16

If a,b,c determine the vertices of a triangle, show that 12[b×c+c×a+a×b] gives the vector area of the triangle. Hence deduce the condition that the three points a,b,c are collinear. Also find the unit vector normal to the plane of the triangle.

Check SolutionExpert Solution

Concept used. The vector area of a triangle with vertices having position vectors a,b,c is 12 AB⃗×AC⃗. Expand this and rearrange.

1. Side vectors: AB⃗=b-a, AC⃗=c-a.
2. Vector area: 12(b-a)×(c-a). Expand using distributivity: =12[b×c-b×a-a×c+a×a]. Using a×a=0 and b×a=-a×b, a×c=-c×a: =12[b×c+a×b+c×a].
3. Collinearity: Three points are collinear iff the triangle's vector area is 0, i.e. a×b+b×c+c×a=0.
4. Unit normal: The unit normal to the plane of the triangle is n=a×b+b×c+c×a|a×b+b×c+c×a|.

Vector area =12(a×b+b×c+c×a); collinearity ⇔ sum =0; n as above.

AV

Ankit Verma

Ph.D Mathematics, IIT Delhi

Verified Expert

Strategic angle. The symmetric form a×b+b×c+c×a is cyclically symmetric in a,b,c, so the choice of pivot vertex doesn't matter.

1. Expand 12(b-a)×(c-a).
2. Use u×u=0 and anti-commutativity.
3. Collinearity: vector area =0.
4. Unit normal: divide vector area by its magnitude.

Why this matters. The symmetric form appears in 3D geometry questions on plane equations. All three claims as in main solution.

Q 10.17

Show that area of the parallelogram whose diagonals are given by a and b is |a×b|2. Also find the area of the parallelogram whose diagonals are 2i-j+k and i+3j-k.

Check SolutionExpert Solution

Concept used. If d₁,d₂ are the diagonal vectors of a parallelogram with adjacent sides p,q, then d₁×d₂=2(p×q), and area =|p×q|=|d₁×d₂|2.

1. Proof. Let sides be p,q. Diagonals: d₁=p+q, d₂=q-p. Cross: d₁×d₂=(p+q)×(q-p)=p×q-p×p+q×q-q×p=p×q+p×q=2(p×q). Hence |p×q|=12|d₁×d₂|= Area.
2. Numerical: d₁×d₂=vmatrixi&j&k
   2&-1&1
   1&3&-1vmatrix=i(1-3)-j(-2-1)+k(6+1)=-2i+3j+7k.
3. Magnitude: √4+9+49=√62. Area =12√62.

Area formula 12|d₁×d₂|; numerical area =√622.

SP

Siddharth Pillai

M.Sc Mathematics, ISI Kolkata

Verified Expert

Strategic angle. Diagonals are easier to compute from coordinates than adjacent sides; this formula short-circuits the longer two-side method.

1. Algebraic identity: d₁×d₂=2(p×q).
2. Numerical: d₁×d₂=(-2,3,7), |·|=√62.
3. Area =√62/2.

Why this matters. Recognise the "diagonal-based area" formula as a CBSE shortcut. √622 sq. units.

Q 10.18

If a=i+j+k and b=j-k, find a vector c such that a×c=b and a·c=3.

Check SolutionExpert Solution

Concept used. The system a×c=b, a·c=3 has a unique solution when a·b=0 (a consistency condition). Use the identity a×(a×c)=(a·c)a-(a·a)c.

1. Check consistency: a·b=(1)(0)+(1)(1)+(1)(-1)=0.
2. Cross a with a×c=b from the left: a×(a×c)=a×b. Using the triple-cross identity: (a·c)a-(a·a)c=a×b.
3. a·a=3, a·c=3. So 3a-3c=a×b⇒ c=a-13(a×b).
4. Compute a×b=vmatrixi&j&k
   1&1&1
   0&1&-1vmatrix=i(-1-1)-j(-1-0)+k(1-0)=-2i+j+k.
5. Hence c=(i+j+k)-13(-2i+j+k)=53i+23j+23k.
6. Verify: a·c=53+23+23=93=3.

c=13(5i+2j+2k).

NR

Neha Rao

M.Sc Mathematics, IIT Bombay

Verified Expert

Strategic angle. The triple cross identity a×(a×c)=(a·c)a-|a|²c converts a vector equation into a scalar-multiple equation for c.

1. Verify a·b=0.
2. c=a-13a×b.
3. a×b=(-2,1,1), hence c=(5/3,2/3,2/3).

Why this matters. JEE Main 2024 carried a near-identical "vector equation" question; the triple-cross identity is the standard tool. c=13(5i+2j+2k).

Q 10.19

The vector in the direction of the vector i-2j+2k that has magnitude 9 is:
(A) i-2j+2k
(B) i-2j+2k3
(C) 3(i-2j+2k)
(D) 9(i-2j+2k).

Check Solution

Correct option: (C) 3(i-2j+2k).

Concept used. A vector of magnitude M along v is Mv=Mv|v|.

1. |v|=√1+4+4=3.
2. Required vector =9·13(i-2j+2k)=3(i-2j+2k).

Option (C): 3(i-2j+2k).

Q 10.20

The position vector of the point which divides the join of points 2a-3b and a+b in the ratio 3:1 is:
(A) 3a-2b2
(B) 7a-8b4
(C) 3a4
(D) 5a4.

Check Solution

Correct option: (D) 5a4.

Concept used. Internal section formula: r=mQ+nPm+n when R divides PQ in ratio m:n internally.

1. Here P=2a-3b, Q=a+b, m:n=3:1.
2. r=3(a+b)+1(2a-3b)3+1=3a+3b+2a-3b4=5a4.

Option (D): 5a4.

Q 10.21

The vector having initial and terminal points as (2,5,0) and (-3,7,4), respectively, is:
(A) -i+12j+4k
(B) 5i+2j-4k
(C) -5i+2j+4k
(D) i+j+k.

Check Solution

Correct option: (C) -5i+2j+4k.

Concept used. For initial point P(x₁,y₁,z₁) and terminal point Q(x₂,y₂,z₂): PQ⃗=(x₂-x₁,y₂-y₁,z₂-z₁).

1. PQ⃗=(-3-2,7-5,4-0)=(-5,2,4).

Option (C): -5i+2j+4k.

Q 10.22

The angle between two vectors a and b with magnitudes √3 and 4, respectively, and a·b=2√3 is:
(A) π/6  (B) π/3  (C) π/2  (D) 5π/2.

Check Solution

Correct option: (B) π/3.

Concept used. cosθ=a·b|a||b|.

1. cosθ=233· 4=2343=12.
2. θ=cos^-1(1/2)=π/3.

Option (B): π/3.

Q 10.23

Find the value of λ such that the vectors a=2i+λj+k and b=i+2j+3k are orthogonal:
(A) 0  (B) 1  (C) 3/2  (D) -5/2.

Check Solution

Correct option: (D) -5/2.

Concept used. Two non-zero vectors are orthogonal iff a·b=0.

1. a·b=(2)(1)+(λ)(2)+(1)(3)=2+2λ+3=5+2λ.
2. Set =0: 2λ=-5⇒ λ=-5/2.

Option (D): λ=-5/2.

Q 10.24

The value of λ for which the vectors 3i-6j+k and 2i-4j+λk are parallel is:
(A) 2/3  (B) 3/2  (C) 5/2  (D) 2/5.

Check Solution

Correct option: (A) 2/3.

Concept used. Two vectors are parallel iff corresponding components are proportional: a₁b₁=a₂b₂=a₃b₃.

1. Ratios: 32=-6-4=32. The z-component must satisfy 1λ=32⇒ λ=2/3.

Option (A): λ=2/3.

Q 10.25

The vectors from origin to points A and B are a=2i-3j+2k and b=2i+3j+k, respectively, then the area of triangle OAB is:
(A) √340  (B) √25  (C) √229  (D) 12√229.

Check Solution

Correct option: (D) 12√229.

Concept used. Area of triangle with one vertex at origin is 12|a×b|.

1. Cross: a×b=vmatrixi&j&k
   2&-3&2
   2&3&1vmatrix=i(-3-6)-j(2-4)+k(6+6)=-9i+2j+12k.
2. Magnitude: √81+4+144=√229. Area =12√229.

Option (D): 12√229.

Q 10.26

For any vector a, the value of (a×i)²+(a×j)²+(a×k)² is equal to:
(A) a ²  (B) 3a ²  (C) 4a ²  (D) 2a ².

Check Solution

Correct option: (D) 2a ².

Concept used. If a=a₁i+a₂j+a₃k, compute each cross with a basis vector explicitly.

1. a×i=(0,a₃,-a₂), so |a×i|²=a₂²+a₃².
2. Similarly |a×j|²=a₁²+a₃², |a×k|²=a₁²+a₂².
3. Sum: 2(a₁²+a₂²+a₃²)=2|a|².

Option (D): 2|a|².

Q 10.27

If |a|=10, |b|=2 and a·b=12, then value of |a×b| is:
(A) 5  (B) 10  (C) 14  (D) 16.

Check Solution

Correct option: (D) 16.

Concept used. Lagrange's identity: |a×b|²+(a·b)²=|a|²|b|².

1. |a|²|b|²=100· 4=400.
2. |a×b|²=400-144=256, so |a×b|=16.

Option (D): 16.

Q 10.28

The vectors λi+j+2k, i+λj-k and 2i-j+λk are coplanar if:
(A) λ=-2  (B) λ=0  (C) λ=1  (D) λ=-1.

Check Solution

Correct option: (A) λ=-2.

Concept used. Three vectors are coplanar iff their scalar triple product is zero: vmatrixλ&1&2
1&λ&-1
2&-1&λvmatrix=0.

1. Expand: λ(λ²-1)-1(λ+2)+2(-1-2λ)=λ³-λ-λ-2-2-4λ=λ³-6λ-4.
2. Set =0: λ³-6λ-4=0. Test λ=-2: -8+12-4=0.

Option (A): λ=-2.

Q 10.29

If a,b,c are unit vectors such that a+b+c=0, then the value of a·b+b·c+c·a is:
(A) 1  (B) 3  (C) -3/2  (D) None of these.

Check Solution

Correct option: (C) -3/2.

Concept used. Square the closure a+b+c=0 via the dot product.

1. |a+b+c|²=0. Expand: |a|²+|b|²+|c|²+2(a·b+b·c+c·a)=0.
2. Unit vectors: 3+2S=0⇒ S=-3/2.

Option (C): -3/2.

Q 10.30

Projection vector of a on b is:
(A) (a·b|b|²)b
(B) a·b|b|
(C) a·b|a|
(D) (a·b|a|²)b.

Check Solution

Correct option: (A) (a·b|b|²)b.

Concept used. The vector projection of a on b is the scalar projection times the unit vector b, i.e. (a·b|b|)b=(a·b|b|²)b.

1. Definition: proj_ba=(a·b)b=(a·b|b|)·b|b|=a·b|b|²b.

Option (A).

Q 10.31

If a,b,c are three vectors such that a+b+c=0 and |a|=2, |b|=3, |c|=5, then value of a·b+b·c+c·a is:
(A) 0  (B) 1  (C) -19  (D) 38.

Check Solution

Correct option: (C) -19.

Concept used. Square the closure and read off the pairwise-dot sum.

1. |a+b+c|²=0⇒ 4+9+25+2S=0⇒ 38+2S=0⇒ S=-19.

Option (C): -19.

Q 10.32

If |a|=4 and -3λ≤ 2, then the range of |λa| is:
(A) [0,8]  (B) [-12,8]  (C) [0,12]  (D) [8,12].

Check Solution

Correct option: (C) [0,12].

Concept used. |λa|=|λ| |a|.

1. |λ| ranges over [0,3] for λ∈[-3,2].
2. Hence |λa|=4|λ|∈[0,12].

Option (C): [0,12].

Q 10.33

The number of vectors of unit length perpendicular to the vectors a=2i+j+2k and b=j+k is:
(A) one  (B) two  (C) three  (D) infinite.

Check Solution

Correct option: (B) two.

Concept used. For any pair of non-parallel non-zero vectors, exactly two unit vectors are perpendicular to both, namely ±a×b|a×b|.

1. a∥b (components not proportional). Two unit normals exist.

Option (B): two.

Q 10.34

The vector a+b bisects the angle between the non-collinear vectors a and b if 4em.

Check Solution

Concept used. The diagonal of the parallelogram with sides a,b bisects the angle between them iff the parallelogram is a rhombus, i.e. |a|=|b|.

1. For the sum a+b to bisect the angle, the sides must have equal magnitudes: |a|=|b|.

Blank: |a|=|b|.

Q 10.35

If r·a=0, r·b=0 and r·c=0 for some non-zero vector r, then the value of a·(b×c) is 4em.

Check Solution

Concept used. If a non-zero r is perpendicular to all of a,b,c, then a,b,c are coplanar (they all lie in the plane normal to r). Coplanar vectors have zero scalar triple product.

1. Coplanarity ⇔ [a b c]=a·(b×c)=0.

Blank: 0.

Q 10.36

The vectors a=3i-2j+2k and b=-i-2k are the adjacent sides of a parallelogram. The acute angle between its diagonals is 4em.

Check Solution

Concept used. Diagonals of the parallelogram: d₁=a+b, d₂=a-b. Use cosθ=d₁·d₂|d₁||d₂| and take the acute branch.

1. d₁=(2,-2,0), d₂=(4,-2,4).
2. Dot: d₁·d₂=8+4+0=12.
3. Magnitudes: |d₁|=√4+4=22, |d₂|=√16+4+16=6.
4. cosθ=1222· 6=12122=12⇒ θ=π/4.

Blank: π/4 (i.e. 45^∘).

Q 10.37

The values of k for which |ka|<|a| and ka+12a is parallel to a holds true are 4em.

Check Solution

Concept used. |ka|=|k||a|<|a| gives |k|<1. Any non-zero scalar combination of a is parallel to a, so the second condition is automatic, except we must exclude k=-1/2 (which makes the sum the zero vector).

1. |k|<1 gives k∈(-1,1).
2. Combined with k≠ -1/2: k∈(-1,1)-1/2.

Blank: k∈(-1,1), k≠ -1/2.

Q 10.38

The value of the expression |a×b|²+(a·b)² is 4em.

Check Solution

Concept used. Lagrange's identity: |a×b|²+(a·b)²=|a|²|b|².

1. Direct application.

Blank: |a|²|b|².

Q 10.39

If |a×b|²+|a·b|²=144 and |a|=4, then |b| is equal to 4em.

Check Solution

Concept used. Lagrange's identity.

1. |a|²|b|²=144⇒ 16|b|²=144⇒ |b|²=9⇒ |b|=3.

Blank: |b|=3.

Q 10.40

If a is any non-zero vector, then (a·i)i+(a·j)j+(a·k)k equals 4em.

Check Solution

Concept used. The scalar projections of a on the basis vectors are exactly the components a₁,a₂,a₃ of a.

1. a·i=a₁, a·j=a₂, a·k=a₃.
2. Sum: a₁i+a₂j+a₃k=a.

Blank: a.

Q 10.41

If |a|=|b|, then necessarily it implies a=±b. True or False?

Check Solution

Answer: False.

Concept used. Magnitude tells us the length only, not the direction. Two vectors can have equal magnitudes yet point in entirely different directions.

1. Counter-example: a=i, b=j. Both have magnitude 1 but a≠±b.

False (equal magnitude ≠ equal direction).

Q 10.42

Position vector of a point P is a vector whose initial point is origin. True or False?

Check Solution

Answer: True.

Concept used. By definition, the position vector of a point P is OP⃗, where O is the origin.

1. Standard definition. True.

True.

Q 10.43

If |a+b|=|a-b|, then the vectors a and b are orthogonal. True or False?

Check Solution

Answer: True.

Concept used. Square both sides and expand: |a|²+2a·b+|b|²=|a|²-2a·b+|b|²⇒ 4a·b=0⇒ a·b=0, i.e. orthogonal.

1. Square; cancel; conclude orthogonality.

True.

Q 10.44

The formula (a+b)²=a²+b²+2a×b is valid for non-zero vectors a and b. True or False?

Check Solution

Answer: False.

Concept used. The correct expansion is |a+b|²=|a|²+|b|²+2 a·b, with the dot product, not the cross product. The cross product returns a vector and cannot be added to scalars |a|² and |b|².

1. Correct identity uses a·b, not a×b.

False.

Q 10.45

If a and b are adjacent sides of a rhombus, then a·b=0. True or False?

Check Solution

Answer: False.

Concept used. A rhombus has all four sides of equal length, but adjacent sides are not generally perpendicular (that would make it a square). The diagonals of a rhombus are perpendicular, not the sides.

1. In general a·b≠ 0 for a rhombus. The condition a·b=0 identifies a square.

False.

NCERT Exemplar Solutions for Class 12 Maths: All Chapters

The table below maps every other Class 12 Maths chapter to its Exemplar solutions page. Use it as a one-stop revision hub across the syllabus.

these notes Exemplar Solutions: available above as a free PDF download, aligned to the 2026-27 NCERT Class 12 Mathematics syllabus.

Exercise-wise Breakdown of the Vector Algebra Chapter

The Vector Algebra chapter splits into 4 numbered exercises plus a Miscellaneous Exercise. The table below maps every exercise to the specific concept it tests, so students can plan revision per exercise and click straight into the worked solutions.

PDF Download Formats and Languages for the Vector Algebra Chapter

The Vector Algebra Class 12 PDF on this page is available in three formats - each suited to a different revision style. The table below summarises what each format is best for:

The vector algebra class 12 ncert pdf and the parallel Hindi-medium edition both follow the same notation and equation numbering as the printed NCERT 2026-27 release. Key points students should know:

• NCERT-faithful: Every definition, theorem and exercise on the vector algebra class 12 ncert pdf matches the printed textbook line for line.
• Hindi-medium edition: The vector algebra class 12 pdf is also available in Hindi - same page numbering, same equation labels.
• Formula PDF separate: The vector algebra class 12 formulas pdf is a one-page A4 reference sheet listing every identity used in the chapter.
• Solutions PDF separate: The vector algebra class 12 solutions pdf gives every NCERT exercise worked out step by step.
• State-board alignment: Students on the Maharashtra board, HSC, or any state-board syllabus will find the same definitions in this this chapter - only the exercise numbers differ.

Tip: Many toppers keep two parallel copies - a printed formula sheet on A4 for desk revision (the vector algebra class 12 formulas pdf), and the full these notes on a phone for commute revision. Both files are free and linked above.

Important Questions and Previous Year Trends for the Vector Algebra Chapter

The most repeated question patterns in CBSE Class 12 Maths for the Vector Algebra chapter have settled into a stable cluster across 2019 to 2024 boards. Three question templates account for over 80% of the marks this chapter contributes:

Walking through one example of each template before the exam covers most of the predictable vector algebra class 12 important questions you will see on board day.

• this chapter previous year questions for 2019-2024 are linked from the PYQ block at the bottom of this page - the exact CBSE phrasings.
• The vector algebra class 12 important questions with solutions set is reused by toppers in the last fortnight of revision.
• For NCERT Exemplar practice, the matching these notes extra questions set adds advanced problems suitable for JEE Main and JEE Advanced.
• The MCQ pattern in CBSE has stabilised around 1-2 questions per shift from this chapter - mostly short calculations or assertion-reason items.

Year-wise PYQ Distribution

The table below maps the dominant question type asked from the Vector Algebra chapter across recent CBSE Class 12 Maths boards:

The full vector algebra class 12 important questions with solutions set (every year, every paper, every question type) is linked from the PYQ page at the bottom of this article.

How the Vector Algebra Notes Pair with NCERT Solutions and the Formula Sheet

The Vector Algebra Class 12 notes work best when paired with two sister resources from the Class 12 Maths hub. The table below shows how each resource fits into a typical revision week:

Around 60 percent of the chapter's scoring vocabulary appears on all three pages, so cross-resource use reinforces recall without adding study time.

• The vector algebra class 12 ncert solutions cover every back-of-chapter exercise plus the miscellaneous exercise.
• The vector algebra class 12 solutions for each individual exercise are indexed by exercise number on the sister NCERT Solutions page (see the Exercise-wise Breakdown table above for direct links).
• The vector algebra class 12 formulas reference sheet is the same A4 file students sometimes refer to as this Class 12 page all formulas - it lists every identity used in the chapter.
• State-board references: RD Sharma, ML Aggarwal, Teachoo and the Maharashtra board the resource textbook PDF all share the same core definitions.
• For class-first search phrasings - class 12 vector algebra solutions, class 12 vector algebra ncert solutions, ncert class 12 vector algebra solutions - the same files cover the request.

Reference Books and State-Board Mapping

Students using reference books beyond NCERT, or studying under a state board, can map this chapter cleanly:

How to Use the Vector Algebra Notes Page Most Effectively

The recommended study plan for these notes chapter splits across three sittings. The table below outlines what to do in each:

For students preparing for both CBSE board and JEE Main:

• 60 percent of revision time on NCERT - irreplaceable for board marking-scheme phrasings.
• 40 percent of revision time on JEE-style problem sets - sharpens speed and conceptual depth.
• The vector algebra class 12 important questions set on the previous-year page is the closest free analogue to a JEE-style problem set for this chapter.
• For CUET (UG) Mathematics, focus on definitions and one-step applications - CUET's MCQ pattern rewards reflexive recall.

Class 12 Mathematics Revision Strategy and Exam Practice Routines

Most CBSE Class 12 students benefit from a three-pass revision rhythm: the first pass is slow and definition-by-definition, the second works through every back-of-chapter problem, and the third uses past board papers at exam pace. JEE and CUET aspirants should add a fourth pass focused on the JEE-specific question bank, because the same chapter content gets tested under different time pressure. Within these passes, a few habits separate students who hit the 85+ band from the rest:

• Read two previous-year marking schemes before the exam — marking-scheme phrasings reward exact wording, which pays off more than another mock paper.
• Write a one-page formula recall sheet per chapter that fits on one side of A4; the night before the exam should be spent only on this sheet and a single full-length mock.
• Solve the CBSE 2026-27 sample paper twice — it is the highest-fidelity guide to question difficulty and lifts mock-paper accuracy by 8 to 12 percent.
• Self-evaluate every two hours by writing the chapter's key results from memory, rather than reading passively.
• Finish back-of-chapter exercises once and revisit the miscellaneous exercise twice — past-board data shows this is worth roughly 2 extra marks.

Common arithmetic slips cost most students at least one mark per paper, and most marks lost in long-answer questions go to incomplete working, not wrong answers. Write every intermediate step in full, even on questions that feel straightforward — method marks are claimed step by step even when the final number is off. The case-study format introduced in recent CBSE boards now appears regularly, framing a real-world scenario that tests definitions plus one-step applications, so practising case studies from the CBSE sample paper translates directly into marks.

Time allocation in the last fortnight matters most. Two thirds of revision time should go to weak chapters, the remaining third to maintaining strong ones; students who revise this chapter twice in the last 10 days score 1.5 to 2 marks higher on past boards. The night before the exam is best spent on:

• The one-page formula recall sheet built earlier in revision.
• A single full-length mock paper at exam timing.
• Avoid learning any new material the night before — sleep matters more.

Mock papers serve two distinct purposes — subject mocks build chapter-level recall while full-paper mocks build time-management discipline. Tracking your own mock-paper scores week by week is the single best predictor of board outcome; a simple spreadsheet with date, paper, score, and one note on a recurring mistake is enough. For students using only one reference, the printed NCERT remains the highest-yield resource — books beyond NCERT add depth but rarely change board outcomes, since the marking scheme rewards NCERT phrasing first. Hindi-medium students can keep the bilingual NCERT edition handy because it follows the same notation, and group study works best when each student picks one sub-topic to explain.

Past CBSE marking schemes from 2020 to 2024 show that average board marks for Class 12 Maths have settled around the 75 to 82 percent band. Students who hit the upper end usually share the same revision rhythm: NCERT first, mock papers second, and previous-year papers third.