Class 12 Maths · Chapter 10
Short answer:
Is chapter me NCERT ke teeno exercises (10.1, 10.2, 10.3) aur Miscellaneous Exercise milake tests-worthy 20+ questions cover kiye gaye hain — direction cosines/ratios nikalna, vector addition-subtraction aur section formula, unit vector aur scalar multiplication, dot product se angle aur projection nikalna, aur cross product se area of triangle/parallelogram nikalna. Har question ka poora step-by-step working diya hai taaki exam me speed aur accuracy dono bane.
Vector Algebra wahi chapter hai jahan physics aur maths ekdum mil jaate hain — force, velocity, displacement sab vectors hi hain. Yahan hum seekhte hain vector ko represent kaise karte hain (magnitude + direction), unko add/subtract/scale kaise karte hain, aur do vectors ko multiply karne ke do alag tareeke — dot product (jo scalar deta hai) aur cross product (jo vector deta hai). Dono ka apna real-life matlab hai: dot product work-done jaisa concept hai, cross product torque jaisa. Is chapter ki examples 3D coordinate geometry (Chapter 11) ki foundation banati hain, isliye concept clear hona zaroori hai, sirf formula ratna kaafi nahi.
Chapter 10 Summary — 5 Minute Revision
1. Basic Concepts
Scalar sirf magnitude rakhta hai (jaise mass, temperature, speed). Vector magnitude aur direction dono rakhta hai (jaise displacement, velocity, force). Vector ko hum bold letter ya arrow se likhte hain — is spec me hum "vector a" ya a likhenge kyunki HTML me arrow-over-letter theek se nahi dikhta.
Ek point P(x, y, z) ka position vector origin O se hota hai: OP = x î + y ĵ + z k̂, aur iski magnitude |OP| = √(x² + y² + z²).
Direction cosines (l, m, n) us vector ke x, y, z axes ke saath bane angles ke cosine hote hain. Agar vector a = x î + y ĵ + z k̂ hai aur |a| = r, toh:
l = x/r, m = y/r, n = z/r, aur hamesha l² + m² + n² = 1
Direction ratios (a, b, c) kisi bhi numbers hote hain jo direction cosines ke proportional hon (jaise khud x, y, z). Direction ratios se direction cosines nikalne ke liye normalize karo: l = a/√(a²+b²+c²), aur similarly m, n.
2. Types of Vectors
- Zero vector (null vector): magnitude zero, direction undefined — likhte hain 0
- Unit vector: magnitude exactly 1, direction diya hua — â se denote karte hain, â = a/|a|
- Coinitial vectors: jinka starting point same ho
- Collinear vectors: jo same line ke parallel hon (ek dusre ka scalar multiple), chahe direction same ho ya opposite
- Equal vectors: same magnitude aur same direction (starting point alag ho sakta hai)
- Negative of a vector: same magnitude, opposite direction — likhte hain −a
3. Addition of Vectors
Triangle law: agar vector AB aur BC diye hain toh AB + BC = AC.
Parallelogram law: agar do vectors ek point se coinitial hain aur parallelogram ke adjacent sides bante hain, toh unka sum diagonal hota hai.
Component form me: agar a = a1 î + a2 ĵ + a3 k̂ aur b = b1 î + b2 ĵ + b3 k̂, toh a + b = (a1+b1) î + (a2+b2) ĵ + (a3+b3) k̂. Subtraction bhi component-wise hi hota hai: a − b = a + (−b).
Vector addition commutative (a + b = b + a) aur associative [(a+b)+c = a+(b+c)] dono hoti hai.
4. Multiplication of a Vector by a Scalar
Agar λ ek scalar hai aur a ek vector, toh λa ek naya vector hai jiski magnitude |λ||a| hai. Agar λ > 0, direction same rehti hai; agar λ < 0, direction opposite ho jaati hai.
Component form: λa = λa1 î + λa2 ĵ + λa3 k̂
5. Section Formula for Vectors
Agar point P, line segment AB ko m:n ratio me divide karta hai (position vectors a aur b diye hain), toh:
Internal division: OP = (m·b + n·a)/(m+n)
External division: OP = (m·b − n·a)/(m−n)
Midpoint ka special case (m=n=1): OP = (a+b)/2
6. Product of Two Vectors — Scalar (Dot) Product
Dot product ek scalar deta hai: a · b = |a| |b| cos θ, jahan θ dono vectors ke beech ka angle hai.
Component form: agar a = a1î+a2ĵ+a3k̂, b = b1î+b2ĵ+b3k̂, toh a · b = a1b1 + a2b2 + a3b3
Angle nikalne ke liye: cos θ = (a · b)/(|a||b|)
Special cases: a · a = |a|², î·î = ĵ·ĵ = k̂·k̂ = 1, î·ĵ = ĵ·k̂ = k̂·î = 0. Agar a · b = 0 (aur dono non-zero) toh a perpendicular hai b se.
Projection of a on b = (a · b)/|b| — yeh ek scalar hai jo batata hai vector a, vector b ki direction me kitna "फैला" hua hai.
7. Product of Two Vectors — Vector (Cross) Product
Cross product ek vector deta hai: a × b = |a||b| sin θ n̂, jahan n̂ unit vector hai us plane ke perpendicular (right-hand rule se direction).
Component form (determinant se):
| a × b = determinant of row1: î, ĵ, k̂ row2: a1, a2, a3 row3: b1, b2, b3 |
↔ Table ko side me swipe karein
= (a2b3 − a3b2) î − (a1b3 − a3b1) ĵ + (a1b2 − a2b1) k̂
Cross product anti-commutative hai: a × b = −(b × a). Agar a × b = 0 (dono non-zero) toh a, b collinear/parallel hain.
Area of triangle (do adjacent sides a aur b) = ½ |a × b|
Area of parallelogram (adjacent sides a aur b) = |a × b|
8. Scalar Triple Product (brief)
Teen vectors a, b, c ka scalar triple product likhte hain [a b c] = a · (b × c). Yeh determinant se bhi nikal sakte hain (a1,a2,a3 / b1,b2,b3 / c1,c2,c3 rows). Agar [a b c] = 0 toh teeno vectors coplanar hain — yeh volume of parallelepiped bhi deta hai (magnitude me).
Dot vs Cross — yaad rakhne ka tarika
Dot product ek scalar deta hai, cross product ek vector — yaad rakhne ka tarika: dot 'D' se 'digit' (number/scalar), cross 'C' se... bas practice se yaad ho jaata hai kaunsa kya deta hai.
Poore Class 12 Maths ke handwritten colour notes
IITian & district toppers ke banaye short notes — revision-ready, diagram ke saath. Board se pehle poora syllabus 3 din me revise.
Exercise Questions — Solutions (Q1–Q22)
Q1. Represent graphically a displacement of 40 km, 30° east of north. Find its magnitude and direction (in terms of unit vector components if origin is placed at start).
Yeh vector origin se 30° east of north banata hai. Iski magnitude 40 km hai.
Agar north ko ĵ aur east ko î maanein, toh angle north se 30° east ki taraf hai.
x-component (east) = 40 sin 30° = 40 × 0.5 = 20 km
y-component (north) = 40 cos 30° = 40 × (√3/2) = 20√3 ≈ 34.64 km
Vector = 20 î + 20√3 ĵ, magnitude = √(20² + (20√3)²) = √(400+1200) = √1600 = 40 km ✓
Q2. Classify the following as scalar and vector quantities: time, distance, force, velocity, work done, speed, acceleration, density, number of moles.
Scalar: time, distance, work done, speed, density, number of moles.
Vector: force, velocity, acceleration.
Rule: agar direction ka matlab hai (jaise force kis taraf lag raha hai) toh vector; sirf 'kitna' matlab hai toh scalar.
Q3. In the given figure, identify the coinitial vectors, collinear vectors, and equal vectors among a, b, c, d, e, f (where a, d are equal; b, d, f are collinear; a, b, c are coinitial).
Typical NCERT figure ke basis pe:
Coinitial vectors: a, b, c (sab same starting point se)
Collinear vectors: b, d, f (sab same line ke parallel)
Equal vectors: a and d (same magnitude, same direction)
Q4. Find the values of x, y, z so that vectors a = x î + 2 ĵ + z k̂ and b = 2 î + y ĵ + k̂ are equal.
Do vectors equal hote hain jab unke corresponding components equal hon.
x = 2
2 = y ⟹ y = 2
z = 1
Answer: x = 2, y = 2, z = 1
Q5. Find the scalar and vector components of the vector with initial point P(2, 1) and terminal point Q(−5, 7).
Vector PQ = position vector of Q − position vector of P
PQ = (−5 − 2) î + (7 − 1) ĵ = −7 î + 6 ĵ
Scalar components: −7 and 6
Vector components: −7 î and 6 ĵ
Q6. Find the sum of vectors a = î − 2ĵ + k̂, b = −2î + 4ĵ + 5k̂, c = î − 6ĵ − 7k̂. Also find the unit vector in the direction of the sum.
Add component-wise:
a + b + c = (1−2+1) î + (−2+4−6) ĵ + (1+5−7) k̂ = 0î − 4ĵ − 1k̂ = −4ĵ − k̂
Magnitude = √(0² + (−4)² + (−1)²) = √(16+1) = √17
Unit vector = (−4ĵ − k̂)/√17 = (−4/√17) ĵ + (−1/√17) k̂
Q7. Find the unit vector in the direction of vector a = 2î + 3ĵ + k̂.
|a| = √(2² + 3² + 1²) = √(4+9+1) = √14
â = a/|a| = (2î + 3ĵ + k̂)/√14 = (2/√14) î + (3/√14) ĵ + (1/√14) k̂
Q8. Find a vector in the direction of vector a = î − 2ĵ that has magnitude 7 units.
|a| = √(1² + (−2)²) = √5
Unit vector â = (î − 2ĵ)/√5
Required vector = 7â = 7(î − 2ĵ)/√5 = (7/√5) î − (14/√5) ĵ
Q9. Show that the points A(1, 2, 7), B(2, 6, 3), C(3, 10, −1) are collinear.
Vector AB = (2−1)î + (6−2)ĵ + (3−7)k̂ = î + 4ĵ − 4k̂
Vector BC = (3−2)î + (10−6)ĵ + (−1−3)k̂ = î + 4ĵ − 4k̂
AB = BC, aur B dono me common point hai, isliye A, B, C collinear hain (in fact AB = BC).
Q10. Find the position vector of a point R which divides the line joining points P(2î + 3ĵ − 4k̂) and Q(4î + ĵ + 2k̂) in the ratio 2:1 (i) internally (ii) externally.
(i) Internal division (m:n = 2:1):
OR = (m·b + n·a)/(m+n) = [2(4î+ĵ+2k̂) + 1(2î+3ĵ−4k̂)] / 3
= [8î+2ĵ+4k̂ + 2î+3ĵ−4k̂] / 3 = (10î + 5ĵ + 0k̂)/3 = (10/3)î + (5/3)ĵ
(ii) External division:
OR = (m·b − n·a)/(m−n) = [2(4î+ĵ+2k̂) − 1(2î+3ĵ−4k̂)] / 1
= [8î+2ĵ+4k̂ − 2î−3ĵ+4k̂] = 6î − ĵ + 8k̂
Q11. Find the position vector of the midpoint of the vector joining points P(5, 4, 2) and Q(1, −2, 4).
Midpoint position vector = (a+b)/2 = [(5+1)î + (4−2)ĵ + (2+4)k̂]/2
= (6î + 2ĵ + 6k̂)/2 = 3î + ĵ + 3k̂
Q12. Find the magnitude of two vectors a and b having the same magnitude and such that the angle between them is 60° and their scalar product is 1/2.
Given |a| = |b|, θ = 60°, a·b = 1/2
a·b = |a||b| cos θ ⟹ 1/2 = |a|² × cos 60° = |a|² × (1/2)
|a|² = 1 ⟹ |a| = 1, so |b| = 1
Q13. Find the angle between two vectors a and b with magnitudes √3 and 2 respectively, such that a·b = √6.
cos θ = (a·b)/(|a||b|) = √6/(√3 × 2) = √6/(2√3)
= √(6/3)/2 = √2/2 = 1/√2
θ = cos⁻¹(1/√2) = 45°
Q14. Find |a| and |b| if (a+b)·(a−b) = 8 and |a| = 8|b|.
(a+b)·(a−b) = |a|² − |b|² = 8
Given |a| = 8|b|, so |a|² = 64|b|²
64|b|² − |b|² = 8 ⟹ 63|b|² = 8 ⟹ |b|² = 8/63
|b| = √(8/63) = 2√2/(3√7), aur |a| = 8|b| = 16√2/(3√7)
Q15. Show that the vectors 2î − ĵ + k̂, î − 3ĵ − 5k̂, and 3î − 4ĵ − 4k̂ form the vertices of a right-angled triangle.
Maano A=2î−ĵ+k̂, B=î−3ĵ−5k̂, C=3î−4ĵ−4k̂
AB = B−A = −î−2ĵ−6k̂, |AB|² = 1+4+36 = 41
BC = C−B = 2î−ĵ+k̂, |BC|² = 4+1+1 = 6
CA = A−C = −î+3ĵ+5k̂, |CA|² = 1+9+25 = 35
Check: |BC|² + |CA|² = 6+35 = 41 = |AB|² ⟹ right angle at C. Triangle inequality bhi satisfy hoti hai, so it's a valid right-angled triangle.
Q16. Find |a × b| if a = î − 7ĵ + 7k̂ and b = 3î − 2ĵ + 2k̂.
a × b = determinant [î,ĵ,k̂ / 1,−7,7 / 3,−2,2]
î component: (−7)(2) − (7)(−2) = −14+14 = 0
ĵ component: −[(1)(2) − (7)(3)] = −[2−21] = 19
k̂ component: (1)(−2) − (−7)(3) = −2+21 = 19
a × b = 0î + 19ĵ + 19k̂
|a × b| = √(0+361+361) = √722 = 19√2
Q17. Find a unit vector perpendicular to each of the vectors a+b and a−b, where a = î+ĵ+k̂ and b = î+2ĵ+3k̂.
a+b = 2î+3ĵ+4k̂, a−b = 0î−ĵ−2k̂
(a+b)×(a−b) = determinant [î,ĵ,k̂ / 2,3,4 / 0,−1,−2]
î: (3)(−2)−(4)(−1) = −6+4 = −2
ĵ: −[(2)(−2)−(4)(0)] = −[−4−0] = 4
k̂: (2)(−1)−(3)(0) = −2
Result = −2î + 4ĵ − 2k̂, magnitude = √(4+16+4) = √24 = 2√6
Unit vector = (−2î+4ĵ−2k̂)/(2√6) = (−1/√6) î + (2/√6) ĵ + (−1/√6) k̂
Q18. If a is a unit vector and (x−a)·(x+a) = 8, find |x|.
(x−a)·(x+a) = |x|² − |a|² = 8
Given |a| = 1 (unit vector), so |x|² − 1 = 8 ⟹ |x|² = 9
|x| = 3
Q19. Find λ and μ if (2î + 6ĵ + 27k̂) × (î + λĵ + μk̂) = 0.
Cross product zero matlab dono vectors parallel/collinear hain — components proportional honge.
2/1 = 6/λ = 27/μ
6/λ = 2 ⟹ λ = 3
27/μ = 2 ⟹ μ = 27/2
Q20. Show that (a−b) × (a+b) = 2(a × b).
(a−b) × (a+b) = a×a + a×b − b×a − b×b
a×a = 0 and b×b = 0 (any vector cross itself is zero)
= a×b − b×a = a×b − (−a×b) [since b×a = −a×b]
= a×b + a×b = 2(a×b) ✓ proved
Q21. Find the area of a triangle having vertices A(1,1,1), B(1,2,3), C(2,3,1) using vectors.
AB = B−A = 0î+1ĵ+2k̂, AC = C−A = 1î+2ĵ+0k̂
AB × AC = determinant [î,ĵ,k̂ / 0,1,2 / 1,2,0]
î: (1)(0)−(2)(2) = −4
ĵ: −[(0)(0)−(2)(1)] = −[−2] = 2
k̂: (0)(2)−(1)(1) = −1
AB × AC = −4î + 2ĵ − k̂, |AB×AC| = √(16+4+1) = √21
Area = ½ × √21 = √21/2 sq. units
Q22. Find the area of the parallelogram whose adjacent sides are given by vectors a = 3î + ĵ + 4k̂ and b = î − ĵ + k̂.
a × b = determinant [î,ĵ,k̂ / 3,1,4 / 1,−1,1]
î: (1)(1)−(4)(−1) = 1+4 = 5
ĵ: −[(3)(1)−(4)(1)] = −[3−4] = 1
k̂: (3)(−1)−(1)(1) = −3−1 = −4
a × b = 5î + ĵ − 4k̂, |a×b| = √(25+1+16) = √42
Area of parallelogram = |a×b| = √42 sq. units
Important Equations — Ek Nazar Me
| Concept | Formula |
|---|---|
| Magnitude of vector a = xî+yĵ+zk̂ | |a| = √(x² + y² + z²) |
| Direction cosines | l = x/r, m = y/r, n = z/r (r = |a|); l²+m²+n² = 1 |
| Unit vector | â = a/|a| |
| Vector addition (component form) | a+b = (a1+b1)î + (a2+b2)ĵ + (a3+b3)k̂ |
| Scalar multiplication | λa = λa1î + λa2ĵ + λa3k̂ |
| Section formula (internal, ratio m:n) | OP = (m·b + n·a)/(m+n) |
| Section formula (external, ratio m:n) | OP = (m·b − n·a)/(m−n) |
| Midpoint formula | OP = (a+b)/2 |
| Dot (scalar) product | a·b = |a||b| cos θ = a1b1 + a2b2 + a3b3 |
| Angle between two vectors | cos θ = (a·b)/(|a||b|) |
| Projection of a on b | (a·b)/|b| |
| Cross (vector) product | a×b = |a||b| sin θ n̂ = determinant[î,ĵ,k̂ / a1,a2,a3 / b1,b2,b3] |
| Area of triangle (sides a, b) | ½ |a × b| |
| Area of parallelogram (sides a, b) | |a × b| |
| Scalar triple product | [a b c] = a·(b×c); = 0 means coplanar |
↔ Table ko side me swipe karein
Common Mistakes — Yahan Marks Kat te Hain
- Dot product ka result vector likh dena. a·b hamesha ek scalar (sirf number) hota hai, vector nahi — ismein î, ĵ, k̂ nahi aata. Cross product hi vector deta hai.
- Cross product ke determinant expansion me sign galat lagana. ĵ component ke saamne hamesha minus sign lagta hai: a×b = (a2b3−a3b2)î − (a1b3−a3b1)ĵ + (a1b2−a2b1)k̂ — beech wale term ka sign bhoolna sabse common mistake hai.
- Cross product ko commutative maan lena. a×b ≠ b×a — yeh anti-commutative hai: a×b = −(b×a). Order badalte hi answer ka sign ulat jaata hai.
- Angle nikalne ke liye cross product formula use karna. Angle between vectors hamesha dot product se aata hai: cos θ = (a·b)/(|a||b|). Cross product se sin θ milta hai, jo directly angle ke liye kam use hota hai (aur ambiguous ho sakta hai 0-180° range me).
- Unit vector nikalte waqt normalize karna bhool jaana. Sirf direction batane se unit vector nahi banta — vector ko uski apni magnitude se divide karna zaroori hai: â = a/|a|. Bina divide kiye likha vector unit vector nahi hai.
- Position vector aur displacement vector me confuse ho jaana. Vector PQ (P se Q tak) = (position vector of Q) − (position vector of P), na ki Q ka position vector khud PQ ho jaaye — order aur subtraction dono zaroori hain.
Board-Style Important Questions
- 1 mark: Find the magnitude of vector a = 2î − 3ĵ + 6k̂.
- 1 mark: If a and b are unit vectors and θ is the angle between them, write the condition for a+b to also be a unit vector.
- 2 marks: Find a unit vector in the direction of vector PQ, where P and Q have coordinates (1, 3, 0) and (4, 5, 6) respectively.
- 3 marks: Find the position vector of a point which divides the join of points with position vectors a−2b and 2a+b externally in the ratio 2:1.
- 4 marks: Show that the vectors 2î−ĵ+k̂, î−3ĵ−5k̂ and 3î−4ĵ−4k̂ form the vertices of a right-angled triangle.
- 5 marks: Find the area of a parallelogram whose diagonals are given by vectors a = 3î+ĵ−2k̂ and b = î−3ĵ+4k̂.
Aksar Poochhe Jaane Wale Sawaal
Dot product aur cross product me main difference kya hai?
Dot product ek scalar (number) deta hai aur formula hai |a||b|cos θ — angle nikalne aur projection ke liye use hota hai. Cross product ek vector deta hai jo dono original vectors ke perpendicular hota hai, formula hai |a||b|sin θ n̂ — area aur perpendicular direction nikalne ke liye use hota hai.
Agar do vectors ka dot product zero ho toh iska kya matlab hai?
Iska matlab hai dono vectors perpendicular (90°) hain, bashart dono non-zero vectors hon. Yeh check karne ka fastest tarika hai ki do lines/vectors perpendicular hain ya nahi.
Agar do vectors ka cross product zero vector ho toh kya matlab hai?
Iska matlab hai dono vectors parallel ya collinear hain (angle 0° ya 180°), bashart dono non-zero hon. Isse hum collinearity prove karte hain.
Direction cosines aur direction ratios me kya farak hai?
Direction ratios (a,b,c) kisi bhi proportional numbers ho sakte hain jo vector ki direction batate hain. Direction cosines (l,m,n) unhi ka normalized version hain jahan l²+m²+n²=1 hamesha satisfy hota hai — direction ratios ko unki magnitude se divide karke direction cosines milte hain.
Section formula me internal aur external division ka farak kya hai?
Internal division me point P line segment AB ke beech me hota hai (formula me plus sign), external division me P line ko extend karke bahar hota hai (formula me minus sign). Dono formulas alag denominators aur signs use karte hain.
Scalar triple product ka use kahan hota hai?
Scalar triple product [a b c] = a·(b×c) batata hai ki teen vectors coplanar hain ya nahi (agar zero ho toh coplanar) aur uski magnitude parallelepiped ka volume deta hai jiske edges wo teen vectors hon.
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