Class 12 : Maths (English) – Chapter 10: Vector

Posted by

On June 25, 2025

EXPLANATION & SUMMARY

🔷 Explanation
🔹 Introduction to Vectors
A vector is a quantity that has both magnitude and direction.
🧭 Unlike a scalar (which has only magnitude, e.g., mass, temperature), a vector represents directed quantities like force, velocity, and displacement.
✏️ Note: Scalars and vectors are the two fundamental types of physical quantities.
💡 Examples:
Displacement = 5 km north (vector)
Speed = 60 km/h (scalar)

🔹 Representation of a Vector
A vector is denoted by a̅, b̅, c̅ or →AB.
If A(x₁, y₁, z₁) and B(x₂, y₂, z₂) are two points,
then the position vector →AB = (x₂ − x₁)î + (y₂ − y₁)ĵ + (z₂ − z₁)k̂.
🧠 Magnitude of vector a̅ = sqrt(a₁² + a₂² + a₃²)
📏 Direction is along the line joining points A and B.

🔹 Types of Vectors
🔵 Zero Vector: magnitude = 0 (denoted by 0̅)
🟢 Unit Vector: magnitude = 1 (used for direction, e.g., î, ĵ, k̂)
🟠 Equal Vectors: same magnitude and direction
🔴 Negative Vectors: same magnitude but opposite direction
🟣 Collinear Vectors: parallel vectors
🟤 Coplanar Vectors: vectors lying in the same plane

🔹 Position Vector and Direction Cosines
➡️ If a point P(x, y, z), then position vector OP̅ = xî + yĵ + zk̂
➡️ Direction cosines (l, m, n) are cosines of angles between vector and coordinate axes.
Relation: l² + m² + n² = 1

🔹 Vector Equality
Two vectors a̅ and b̅ are equal if:
They have the same magnitude
They have the same direction
Formally:
a̅ = b̅ ⇒ a₁ = b₁, a₂ = b₂, a₃ = b₃

🔹 Addition of Vectors
If a̅ = a₁î + a₂ĵ + a₃k̂ and b̅ = b₁î + b₂ĵ + b₃k̂,
then a̅ + b̅ = (a₁ + b₁)î + (a₂ + b₂)ĵ + (a₃ + b₃)k̂.
💡 Commutative Law: a̅ + b̅ = b̅ + a̅
💡 Associative Law: (a̅ + b̅) + c̅ = a̅ + (b̅ + c̅)
✏️ Note: Addition is done component-wise.

🔹 Subtraction of Vectors
a̅ − b̅ = (a₁ − b₁)î + (a₂ − b₂)ĵ + (a₃ − b₃)k̂
✔️ Represents a vector from tip of b̅ to tip of a̅.

🔹 Multiplication by a Scalar
If k is a scalar, then
k a̅ = (k a₁)î + (k a₂)ĵ + (k a₃)k̂
✔️ If k > 0, direction remains same; if k < 0, direction reverses.

🔹 Section Formula (Vector Form)
A point P divides line joining A(x₁, y₁, z₁) and B(x₂, y₂, z₂) in ratio m : n,
then
OP̅ = (m OB̅ + n OA̅)/(m + n)
or
Position vector = [(m x₂ + n x₁)/(m + n)] î + [(m y₂ + n y₁)/(m + n)] ĵ + [(m z₂ + n z₁)/(m + n)] k̂
💡 Used to find coordinates of internal dividing point.

🔹 Scalar (Dot) Product
For vectors a̅ = a₁î + a₂ĵ + a₃k̂, b̅ = b₁î + b₂ĵ + b₃k̂,
a̅ ⋅ b̅ = a₁b₁ + a₂b₂ + a₃b₃
Also,
a̅ ⋅ b̅ = |a̅||b̅| cosθ
📌 Properties:
Commutative: a̅ ⋅ b̅ = b̅ ⋅ a̅
Distributive: a̅ ⋅ (b̅ + c̅) = a̅ ⋅ b̅ + a̅ ⋅ c̅
If a̅ ⋅ b̅ = 0, then vectors are perpendicular.

🔹 Vector (Cross) Product
For vectors a̅ and b̅,
a̅ × b̅ = |a̅||b̅| sinθ n̂,
where n̂ is a unit vector perpendicular to both.
In component form:
a̅ × b̅ = | î ĵ k̂ |
| a₁ a₂ a₃ |
| b₁ b₂ b₃ |
📌 Properties:
a̅ × b̅ = − b̅ × a̅
a̅ × a̅ = 0̅
If a̅ × b̅ = 0̅, then a̅ and b̅ are parallel.

🔹 Scalar Triple Product
For a̅, b̅, c̅,
[a̅ b̅ c̅] = a̅ ⋅ (b̅ × c̅)
Represents volume of parallelepiped.
In determinant form:
[a̅ b̅ c̅] = | a₁ a₂ a₃ |
| b₁ b₂ b₃ |
| c₁ c₂ c₃ |
If [a̅ b̅ c̅] = 0, vectors are coplanar.

🔹 Vector Equation of a Line
If a line passes through point A(x₁, y₁, z₁) and is parallel to vector b̅,
then any point P(x, y, z) on line satisfies
r̅ = a̅ + λ b̅
where a̅ = x₁î + y₁ĵ + z₁k̂ and λ is a scalar.

🔹 Vector Equation of a Plane
If plane passes through point A(a₁, a₂, a₃) and normal vector n̅,
then
(r̅ − a̅) ⋅ n̅ = 0
or
n₁(x − a₁) + n₂(y − a₂) + n₃(z − a₃) = 0

🔹 Applications
📏 Distance between two points using vectors
🧭 Angle between lines/planes
📦 Volume and area using triple products
📐 Orthogonality checks using dot product

🔹 Important Identities
✔️ a̅ ⋅ (b̅ × c̅) = b̅ ⋅ (c̅ × a̅) = c̅ ⋅ (a̅ × b̅)
✔️ a̅ × (b̅ × c̅) = (a̅ ⋅ c̅)b̅ − (a̅ ⋅ b̅)c̅
✔️ |a̅ × b̅|² = |a̅|²|b̅|² − (a̅ ⋅ b̅)²

🔹 Geometrical Interpretation
🧭 Dot Product → Projection or cosine of angle
🧭 Cross Product → Area of parallelogram
🧭 Triple Product → Volume of parallelepiped

🧾 Summary (≈300 words)
Vector: quantity with magnitude & direction.
Position vector: from origin to point (x, y, z)
Unit vectors: î, ĵ, k̂
Addition/Subtraction: component-wise
Dot product: a̅ ⋅ b̅ = |a̅||b̅| cosθ
Cross product: a̅ × b̅ = |a̅||b̅| sinθ n̂
Triple product: volume = |a̅ ⋅ (b̅ × c̅)|
Equation of line: r̅ = a̅ + λ b̅
Equation of plane: (r̅ − a̅) ⋅ n̅ = 0
Coplanar test: [a̅ b̅ c̅] = 0
Parallel vectors: cross product = 0
Perpendicular vectors: dot product = 0
Vector identities:
a̅ × (b̅ × c̅) = (a̅ ⋅ c̅)b̅ − (a̅ ⋅ b̅)c̅
|a̅ × b̅|² = |a̅|²|b̅|² − (a̅ ⋅ b̅)²

📝 Quick Recap
🟢 Vectors = magnitude + direction
🔵 Dot product = projection concept
🔴 Cross product = perpendicular vector
🟡 Triple product = volume measure
✏️ Line: r̅ = a̅ + λ b̅
💡 Plane: (r̅ − a̅) ⋅ n̅ = 0

————————————————————————————————————————————————————————————————————————————

QUESTIONS FROM TEXTBOOK

📄 Exercise 10.1

🔵 Question 1:
Represent graphically a displacement of 40 km, 30° east of north.
🟢 Answer:
➡️ Draw a vertical line to represent the north direction.
➡️ From the same origin, draw a line making an angle of 30° with north toward east.
➡️ Mark the length proportional to 40 km.
➡️ Put an arrow to show direction.
✔️ Thus, the line represents the displacement vector of 40 km, 30° east of north.

🔵 Question 2:
Classify the following measures as scalars and vectors.
(i) 10 kg
(ii) 2 meters north-west
(iii) 40°
(iv) 40 watt
(v) 10⁻¹⁹ coulomb
(vi) 20 m/s²
🟢 Answer:
✔️ Scalars → (i) 10 kg, (iii) 40°, (iv) 40 watt, (v) 10⁻¹⁹ coulomb
✔️ Vectors → (ii) 2 meters north-west, (vi) 20 m/s²
💡 Scalar quantities have only magnitude.
💡 Vector quantities have magnitude and direction.

🔵 Question 3:
Classify the following as scalar and vector quantities.
(i) time period
(ii) distance
(iii) force
(iv) velocity
(v) work done
🟢 Answer:
✔️ Scalars → (i) time period, (ii) distance, (v) work done
✔️ Vectors → (iii) force, (iv) velocity

🔵 Question 4:
In Fig 10.6 (a square), identify the following vectors.
(i) Coinitial
(ii) Equal
(iii) Collinear but not equal
🟢 Answer:
Let the square have sides represented by vectors:
Top side → a
Right side → b
Bottom side → c
Left side → d
(i) Coinitial vectors:
➡️ Vectors having the same initial point.
✔️ a and d are coinitial.
(ii) Equal vectors:
➡️ Vectors having same magnitude and same direction.
✔️ a = c, and b = d.
(iii) Collinear but not equal:
➡️ Vectors that are parallel but different in direction or magnitude.
✔️ a and c (same line, opposite direction)
✔️ b and d (same line, opposite direction)

🔵 Question 5:
Answer the following as true or false.
(i) a and –a are collinear.
(ii) Two collinear vectors are always equal in magnitude.
(iii) Two vectors having same magnitude are collinear.
(iv) Two collinear vectors having the same magnitude are equal.
🟢 Answer:
(i) ✔️ True — Both lie along the same line but in opposite direction.
(ii) ❌ False — They may differ in magnitude.
(iii) ❌ False — Same magnitude doesn’t imply collinearity.
(iv) ❌ False — They may have opposite direction.

Exercise 10.2

🔵 Question 1:
Compute the magnitude of the following vectors:
a⃗ = î + ĵ + k̂
b⃗ = 2î – 7ĵ – 3k̂
c⃗ = (1/√3)î + (1/√3)ĵ – (1/√3)k̂
🟢 Answer:
Formula: |v⃗ | = √(x² + y² + z²)
(i) For a⃗ = î + ĵ + k̂
|a⃗ | = √(1² + 1² + 1²) = √3
(ii) For b⃗ = 2î – 7ĵ – 3k̂
|b⃗ | = √(2² + (–7)² + (–3)²) = √(4 + 49 + 9) = √62
(iii) For c⃗ = (1/√3)î + (1/√3)ĵ – (1/√3)k̂
|c⃗ | = √((1/√3)² + (1/√3)² + (–1/√3)²) = √(1/3 + 1/3 + 1/3) = √1 = 1
✔️ Final: |a⃗ | = √3 , |b⃗ | = √62 , |c⃗ | = 1

🔵 Question 2:
Write two different vectors having same magnitude.
🟢 Answer:
Let p⃗₁ = î + ĵ and p⃗₂ = î – ĵ
|p⃗₁| = √(1² + 1²) = √2
|p⃗₂| = √(1² + (–1)²) = √2
✔️ Both have same magnitude √2 but different directions.

🔵 Question 3:
Write two different vectors having same direction.
🟢 Answer:
Vectors in same direction are scalar multiples.
Let p⃗₁ = 2î + ĵ and p⃗₂ = 4î + 2ĵ
✔️ p⃗₂ = 2 × p⃗₁
Hence both have same direction but different magnitudes.

🔵 Question 4:
Find the values of x and y so that the vectors 2î + 3ĵ and xî + yĵ are equal.
🟢 Answer:
For equality, corresponding components are equal.
x = 2 , y = 3
✔️ Final: x = 2 , y = 3

🔵 Question 5:
Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (–5, 7).
🟢 Answer:
Vector from P(x₁, y₁) to Q(x₂, y₂):
PQ⃗ = (x₂ – x₁)î + (y₂ – y₁)ĵ
PQ⃗ = (–5 – 2)î + (7 – 1)ĵ = (–7)î + 6ĵ
➡️ Scalar components: –7 along x-axis, 6 along y-axis
➡️ Vector form: PQ⃗ = –7î + 6ĵ

🔵 Question 6:
Find the sum of the vectors
a⃗ = î – 2ĵ + k̂ , b⃗ = –2î + 4ĵ + 5k̂ , c⃗ = î – 6ĵ – 7k̂
🟢 Answer:
a⃗ + b⃗ + c⃗ = (1 – 2 + 1)î + (–2 + 4 – 6)ĵ + (1 + 5 – 7)k̂
= 0î – 4ĵ – 1k̂
✔️ Final: Resultant = –4ĵ – k̂

🔵 Question 7:
Find the unit vector in the direction of the vector a⃗ = î + ĵ + 2k̂
🟢 Answer:
Step 1: |a⃗ | = √(1² + 1² + 2²) = √6
Step 2: Unit vector = â = a⃗ /|a⃗ |
â = (1/√6)(î + ĵ + 2k̂)
✔️ Final: â = (1/√6)î + (1/√6)ĵ + (2/√6)k̂

🔵 Question 8:
Find the unit vector in the direction of vector PQ, where P(1, 2, 3) and Q(4, 5, 6).
🟢 Answer:
Step 1: PQ⃗ = (4 – 1)î + (5 – 2)ĵ + (6 – 3)k̂ = 3î + 3ĵ + 3k̂
Step 2: |PQ⃗ | = √(3² + 3² + 3²) = √27 = 3√3
Step 3: Unit vector = (1/|PQ⃗ |) × PQ⃗
PQ̂ = (1/3√3)(3î + 3ĵ + 3k̂) = (1/√3)(î + ĵ + k̂)
✔️ Final: PQ̂ = (1/√3)î + (1/√3)ĵ + (1/√3)k̂

🔵 Question 9:
For given vectors, a⃗ = 2î – ĵ + 2k̂ and b⃗ = –î + ĵ – k̂, find the unit vector in the direction of the vector a⃗ + b⃗.
🟢 Answer:
🔸 Step 1 ➤ Add vectors: a⃗ + b⃗ = (2 – 1)î + (–1 + 1)ĵ + (2 – 1)k̂ = î + 0·ĵ + k̂ = î + k̂
🔸 Step 2 ➤ Magnitude: |î + k̂| = √(1² + 0² + 1²) = √2
🔸 Step 3 ➤ Unit vector: û = (a⃗ + b⃗)/|a⃗ + b⃗| = (1/√2)(î + k̂)
✔️ Final: û = (1/√2)î + (1/√2)k̂

🔵 Question 10:
Find a vector in the direction of vector 5î – ĵ + 2k̂ which has magnitude 8 units.
🟢 Answer:
🔸 Step 1 ➤ Given v⃗ = 5î – ĵ + 2k̂, |v⃗| = √(5² + (–1)² + 2²) = √(25 + 1 + 4) = √30
🔸 Step 2 ➤ Unit vector along v⃗: v̂ = v⃗/|v⃗| = (1/√30)(5î – ĵ + 2k̂)
🔸 Step 3 ➤ Required vector of magnitude 8: 8·v̂ = (8/√30)(5î – ĵ + 2k̂)
✔️ Final: required vector = (40/√30)î – (8/√30)ĵ + (16/√30)k̂

🔵 Question 11:
Show that the vectors 2î – 3ĵ + 4k̂ and –4î + 6ĵ – 8k̂ are collinear.
🟢 Answer:
🔸 Step 1 ➤ Check scalar multiple: (–2)×(2î – 3ĵ + 4k̂) = –4î + 6ĵ – 8k̂
🔸 Step 2 ➤ Since one is a scalar multiple of the other, they are parallel (same line).
✔️ Final: Vectors are collinear (opposite directions because factor is negative).

🔵 Question 12:
Find the direction cosines of the vector î + 2ĵ + 3k̂.
🟢 Answer:
🔸 Step 1 ➤ Magnitude: |î + 2ĵ + 3k̂| = √(1² + 2² + 3²) = √14
🔸 Step 2 ➤ Direction cosines (l, m, n) = (x/|v⃗|, y/|v⃗|, z/|v⃗|)
🔸 Step 3 ➤ l = 1/√14, m = 2/√14, n = 3/√14
✔️ Final: (l, m, n) = (1/√14, 2/√14, 3/√14)

🔵 Question 13:
Find the direction cosines of the vector joining the points A(1, 2, –3) and B(–1, –2, 1), directed from A to B.
🟢 Answer:
🔸 Step 1 ➤ A→B vector: AB⃗ = (–1 – 1)î + (–2 – 2)ĵ + (1 – (–3))k̂ = –2î – 4ĵ + 4k̂
🔸 Step 2 ➤ Magnitude: |AB⃗| = √( (–2)² + (–4)² + 4² ) = √(4 + 16 + 16) = √36 = 6
🔸 Step 3 ➤ Direction cosines: l = (–2)/6 = –1/3, m = (–4)/6 = –2/3, n = 4/6 = 2/3
✔️ Final: (l, m, n) = (–1/3, –2/3, 2/3)

🔵 Question 14:
Show that the vector î + ĵ + k̂ is equally inclined to the axes OX, OY and OZ.
🟢 Answer:
🔸 Step 1 ➤ Magnitude: |î + ĵ + k̂| = √(1² + 1² + 1²) = √3
🔸 Step 2 ➤ Direction cosines: l = 1/√3, m = 1/√3, n = 1/√3 (all equal)
🔸 Step 3 ➤ Equal direction cosines ⇒ equal angles with axes (α = β = γ, where cosα = cosβ = cosγ = 1/√3)
✔️ Final: Hence, î + ĵ + k̂ is equally inclined to OX, OY, OZ.

🔵 Question 15:
Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are î + 2ĵ – k̂ and –î + ĵ + k̂ respectively, in the ratio 2 : 1
(i) internally (ii) externally.
🟢 Answer:
Let p⃗ = î + 2ĵ – k̂, q⃗ = –î + ĵ + k̂, ratio P:R:Q = 2:1 (i.e., PR:RQ = 2:1).
(i) Internally
🔸 Step 1 ➤ Formula: r⃗ = (m·q⃗ + n·p⃗)/(m + n) with m:n = 2:1
🔸 Step 2 ➤ r⃗ = (2q⃗ + 1·p⃗)/3 = (2(–î + ĵ + k̂) + (î + 2ĵ – k̂))/3
🔸 Step 3 ➤ r⃗ = (–2î + 2ĵ + 2k̂ + î + 2ĵ – k̂)/3 = (–î + 4ĵ + k̂)/3
✔️ Final (internal): r⃗ = (–1/3)î + (4/3)ĵ + (1/3)k̂
(ii) Externally
🔸 Step 1 ➤ Formula: r⃗ = (m·q⃗ – n·p⃗)/(m – n) = (2q⃗ – 1·p⃗)/(2 – 1)
🔸 Step 2 ➤ r⃗ = 2(–î + ĵ + k̂) – (î + 2ĵ – k̂)
🔸 Step 3 ➤ r⃗ = (–2î + 2ĵ + 2k̂ – î – 2ĵ + k̂) = –3î + 0·ĵ + 3k̂
✔️ Final (external): r⃗ = –3î + 3k̂

🔵 Question 16
Find the position vector of the mid point of the vector joining the points P(2, 3, 4) and Q(4, 1, −2).
🟢 Answer 16
➤ Let P(x₁, y₁, z₁) = (2, 3, 4) and Q(x₂, y₂, z₂) = (4, 1, −2).
➤ Midpoint M = ( (x₁ + x₂)/2, (y₁ + y₂)/2, (z₁ + z₂)/2 ).
➤ Compute x-coordinate: (2 + 4)/2 = 6/2 = 3.
➤ Compute y-coordinate: (3 + 1)/2 = 4/2 = 2.
➤ Compute z-coordinate: (4 + (−2))/2 = 2/2 = 1.
✔️ Final: position vector OM = 3î + 2ĵ + 1k̂.

🔵 Question 17
Show that the points A, B and C with position vectors, a⃗ = 3î − 4ĵ − 4k̂, b⃗ = 2î − ĵ + k̂, and c⃗ = î − 3ĵ − 5k̂, respectively form the vertices of a right angled triangle.
🟢 Answer 17
➤ Compute AB⃗ = b⃗ − a⃗ = (2 − 3)î + (−1 − (−4))ĵ + (1 − (−4))k̂.
➤ Simplify AB⃗ = (−1)î + (3)ĵ + (5)k̂ = −î + 3ĵ + 5k̂.
➤ Compute AC⃗ = c⃗ − a⃗ = (1 − 3)î + (−3 − (−4))ĵ + (−5 − (−4))k̂.
➤ Simplify AC⃗ = (−2)î + (1)ĵ + (−1)k̂ = −2î + ĵ − k̂.
➤ Compute dot product AB⃗ · AC⃗ = (−1)(−2) + (3)(1) + (5)(−1).
➤ Evaluate AB⃗ · AC⃗ = 2 + 3 − 5 = 0.
➤ If AB⃗ · AC⃗ = 0, then AB⃗ ⟂ AC⃗ (perpendicular at A).
✔️ Final: ∠A = 90°, so A, B, C are vertices of a right-angled triangle (right angle at A).

🔵 Question 18
In triangle ABC (Fig 10.18), which of the following is not true:
(A) AB̄ + BC̄ + CĀ = 0̄
(B) AB̄ + BC̄ − AC̄ = 0̄
(C) AB̄ + BC̄ − AC̄ = 0̄
(D) AB̄ − CB̄ + CĀ = 0̄
🟢 Answer 18
➤ Triangle law gives AB̄ + BC̄ + CĀ = 0̄ ⇒ option (A) is true.
➤ Also AB̄ + BC̄ = AC̄ ⇒ AB̄ + BC̄ − AC̄ = 0̄ ⇒ option (B) is true.
➤ Since CB̄ = −BC̄, AB̄ − CB̄ + CĀ = AB̄ + BC̄ + CĀ = 0̄ ⇒ option (D) is true.
➤ Option (C) as printed is identical to (B); it is also true.
✔️ Final: All four printed statements are true; hence no option is “not true.”
✏️ Note (textbook misprint alert): Typically, the incorrect choice should read AB̄ + CB̄ − AC̄ = 0̄, which is not generally true. With that corrected option, the answer would be that corrected option.

🔵 Question 19
If a⃗ and b⃗ are two collinear vectors, then which of the following are incorrect:
(A) b⃗ = λ a⃗, for some scalar λ
(B) a⃗ = ± b⃗
(C) the respective components of a⃗ and b⃗ are not proportional
(D) both the vectors a⃗ and b⃗ have same direction, but different magnitudes
🟢 Answer 19
➤ For collinear vectors, b⃗ = λ a⃗ for some scalar λ ⇒ statement (A) is correct.
➤ Collinear vectors need not have equal magnitudes; a⃗ = ± b⃗ is not necessary ⇒ (B) is incorrect.
➤ Components of collinear vectors are proportional; saying “not proportional” is wrong ⇒ (C) is incorrect.
➤ Same direction with different magnitudes is possible for collinear vectors ⇒ (D) is correct.
✔️ Final: Incorrect statements: (B) and (C).

🧠 Exercise 10.3 —

🔵 Question 1
Find the angle between two vectors a⃗ and b⃗ with magnitudes √3 and 2, respectively having a⃗ · b⃗ = √6 .
🟢 Answer 1
➤ Given: |a⃗| = √3, |b⃗| = 2, a⃗ · b⃗ = √6.
➤ Use: a⃗ · b⃗ = |a⃗||b⃗| cosθ.
➤ Substitute: √6 = (√3)(2) cosθ.
➤ Simplify: cosθ = √6 / (2√3) = (√6/√3)/2 = √2 / 2.
✔️ Final: θ = 45°.

🔵 Question 2
Find the angle between the vectors î − 2ĵ + 3k̂ and 3î − 2ĵ + k̂.
🟢 Answer 2
➤ Let u⃗ = (1, −2, 3), v⃗ = (3, −2, 1).
➤ Compute dot: u⃗ · v⃗ = 1·3 + (−2)·(−2) + 3·1 = 3 + 4 + 3 = 10.
➤ Compute |u⃗|: sqrt(1² + (−2)² + 3²) = sqrt(1 + 4 + 9) = √14.
➤ Compute |v⃗|: sqrt(3² + (−2)² + 1²) = sqrt(9 + 4 + 1) = √14.
➤ cosθ = (u⃗ · v⃗) / (|u⃗||v⃗|) = 10 / (√14·√14) = 10/14 = 5/7.
✔️ Final: θ = cos⁻¹(5/7).

🔵 Question 3
Find the projection of the vector î − ĵ on the vector î + ĵ.
🟢 Answer 3
➤ Let u⃗ = (1, −1, 0), v⃗ = (1, 1, 0).
➤ Compute u⃗ · v⃗ = 1·1 + (−1)·1 + 0·0 = 0.
➤ Vector projection of u⃗ on v⃗ = [(u⃗ · v⃗) / |v⃗|²] v⃗.
➤ Substitute: projection = (0 / |v⃗|²) v⃗ = 0⃗.
✔️ Final: Projection vector = 0⃗.

🔵 Question 4
Find the projection of the vector î + 3ĵ + 7k̂ on the vector 7î − ĵ + 8k̂.
🟢 Answer 4
➤ Let u⃗ = (1, 3, 7), v⃗ = (7, −1, 8).
➤ Compute u⃗ · v⃗ = 1·7 + 3·(−1) + 7·8 = 7 − 3 + 56 = 60.
➤ Compute |v⃗|² = 7² + (−1)² + 8² = 49 + 1 + 64 = 114.
➤ Vector projection of u⃗ on v⃗ = [(u⃗ · v⃗) / |v⃗|²] v⃗.
➤ Substitute: projection = (60/114) v⃗ = (10/19) v⃗.
➤ Expand: projection = (10/19)(7î − ĵ + 8k̂) = (70/19)î − (10/19)ĵ + (80/19)k̂.
✔️ Final: Projection vector = (70/19)î − (10/19)ĵ + (80/19)k̂.

🔵 Question 5
Show that each of the given three vectors is a unit vector:
(1/7)(2î + 3ĵ + 6k̂), (1/7)(3î − 6ĵ + 2k̂), (1/7)(6î + 2ĵ − 3k̂).
Also, show that they are mutually perpendicular to each other.
🟢 Answer 5
➤ Let v₁ = (1/7)(2, 3, 6), v₂ = (1/7)(3, −6, 2), v₃ = (1/7)(6, 2, −3).
➤ |v₁| = (1/7) sqrt(2² + 3² + 6²) = (1/7) sqrt(4 + 9 + 36) = (1/7)·7 = 1.
➤ |v₂| = (1/7) sqrt(3² + (−6)² + 2²) = (1/7) sqrt(9 + 36 + 4) = 1.
➤ |v₃| = (1/7) sqrt(6² + 2² + (−3)²) = (1/7) sqrt(36 + 4 + 9) = 1.
➤ v₁ · v₂ = (1/49)(2·3 + 3·(−6) + 6·2) = (1/49)(6 − 18 + 12) = 0.
➤ v₁ · v₃ = (1/49)(2·6 + 3·2 + 6·(−3)) = (1/49)(12 + 6 − 18) = 0.
➤ v₂ · v₃ = (1/49)(3·6 + (−6)·2 + 2·(−3)) = (1/49)(18 − 12 − 6) = 0.
✔️ Final: Each vector is a unit vector and all three are mutually perpendicular.

🔵 Question 6
Find |a⃗| and |b⃗|, if (a⃗ + b⃗) · (a⃗ − b⃗) = 8 and |a⃗| = 8 |b⃗|.
🟢 Answer 6
➤ Use identity: (a⃗ + b⃗) · (a⃗ − b⃗) = |a⃗|² − |b⃗|².
➤ So |a⃗|² − |b⃗|² = 8.
➤ Let |b⃗| = t (t > 0).
➤ Then |a⃗| = 8t.
➤ Substitute: (8t)² − t² = 8.
➤ Simplify: 64t² − t² = 8 ⇒ 63t² = 8.
➤ Solve: t² = 8/63 ⇒ t = √(8/63) = (2√2)/(3√7).
➤ Compute |a⃗|: 8t = 8 · (2√2)/(3√7) = (16√2)/(3√7).
✔️ Final: |b⃗| = (2√2)/(3√7) and |a⃗| = (16√2)/(3√7).

🔵 Question 7
Evaluate the product (3a⃗ − 5b⃗) · (2a⃗ + 7b⃗).
🟢 Answer 7
➤ Expand using distributive law.
➤ (3a⃗ − 5b⃗) · (2a⃗ + 7b⃗)
➤ = 3a⃗·2a⃗ + 3a⃗·7b⃗ − 5b⃗·2a⃗ − 5b⃗·7b⃗
➤ = 6|a⃗|² + 21(a⃗·b⃗) − 10(a⃗·b⃗) − 35|b⃗|²
✔️ Final: 6|a⃗|² + 11(a⃗·b⃗) − 35|b⃗|².

🔵 Question 8
Find the magnitude of two vectors a⃗ and b⃗, having the same magnitude and such that the angle between them is 60° and their scalar product is 1/2.
🟢 Answer 8
➤ Let |a⃗| = |b⃗| = m.
➤ a⃗·b⃗ = |a⃗||b⃗| cos60° = m² × (1/2).
➤ Given a⃗·b⃗ = 1/2 ⇒ m² × (1/2) = 1/2.
➤ Solve: m² = 1 ⇒ m = 1 (magnitude is positive).
✔️ Final: |a⃗| = |b⃗| = 1.

🔵 Question 9
Find |x⃗|, if for a unit vector a⃗, (x⃗ − a⃗) · (x⃗ + a⃗) = 12.
🟢 Answer 9
➤ Use identity: (x⃗ − a⃗) · (x⃗ + a⃗) = x⃗·x⃗ − a⃗·a⃗.
➤ So |x⃗|² − |a⃗|² = 12.
➤ Given a⃗ is unit ⇒ |a⃗|² = 1.
➤ Hence |x⃗|² − 1 = 12 ⇒ |x⃗|² = 13.
➤ Take square root (magnitude ≥ 0).
✔️ Final: |x⃗| = √13.

🔵 Question 10
If a⃗ = 2î + 2ĵ + 3k̂, b⃗ = −î + 2ĵ + k̂ and c⃗ = 3î + ĵ are such that a⃗ + λ b⃗ is perpendicular to c⃗, then find the value of λ.
🟢 Answer 10
➤ Condition: (a⃗ + λ b⃗) · c⃗ = 0.
➤ Compute a⃗·c⃗ = (2)(3) + (2)(1) + (3)(0) = 6 + 2 + 0 = 8.
➤ Compute b⃗·c⃗ = (−1)(3) + (2)(1) + (1)(0) = −3 + 2 + 0 = −1.
➤ Hence 8 + λ(−1) = 0 ⇒ 8 − λ = 0.
✔️ Final: λ = 8.

🔵 Question 11
Show that |a⃗| b⃗ + |b⃗| a⃗ is perpendicular to |a⃗| b⃗ − |b⃗| a⃗, for any two nonzero vectors a⃗ and b⃗.
🟢 Answer 11
➤ Let u⃗ = |a⃗| b⃗ + |b⃗| a⃗ and v⃗ = |a⃗| b⃗ − |b⃗| a⃗.
➤ Compute u⃗·v⃗.
➤ u⃗·v⃗ = (|a⃗| b⃗ + |b⃗| a⃗) · (|a⃗| b⃗ − |b⃗| a⃗).
➤ = |a⃗|² (b⃗·b⃗) − |a⃗||b⃗| (b⃗·a⃗) + |a⃗||b⃗| (a⃗·b⃗) − |b⃗|² (a⃗·a⃗).
➤ Since a⃗·b⃗ = b⃗·a⃗, the middle two terms cancel.
➤ Thus u⃗·v⃗ = |a⃗|² |b⃗|² − |b⃗|² |a⃗|² = 0.
➤ Zero dot product implies perpendicularity.
✔️ Final: u⃗ ⟂ v⃗ (proved).

🔵 Question 12
If a⃗·a⃗ = 0 and a⃗·b⃗ = 0, then what can be concluded about the vector b⃗?
🟢 Answer 12
➤ a⃗·a⃗ = |a⃗|² = 0 ⇒ |a⃗| = 0 ⇒ a⃗ = 0⃗.
➤ a⃗·b⃗ = 0 also holds because zero vector is perpendicular to all vectors.
✔️ Conclusion: b⃗ can be any vector because a⃗ = 0⃗ makes a⃗·b⃗ = 0 always true.

🔵 Question 13
If a⃗, b⃗, c⃗ are unit vectors such that a⃗ + b⃗ + c⃗ = 0⃗, find the value of a⃗·b⃗ + b⃗·c⃗ + c⃗·a⃗.
🟢 Answer 13
➤ Square both sides: (a⃗ + b⃗ + c⃗)² = 0.
➤ Expand: a⃗·a⃗ + b⃗·b⃗ + c⃗·c⃗ + 2(a⃗·b⃗ + b⃗·c⃗ + c⃗·a⃗) = 0.
➤ Each |a⃗| = |b⃗| = |c⃗| = 1 ⇒ a⃗·a⃗ = b⃗·b⃗ = c⃗·c⃗ = 1.
➤ Substitute: 3 + 2(a⃗·b⃗ + b⃗·c⃗ + c⃗·a⃗) = 0.
➤ Simplify: a⃗·b⃗ + b⃗·c⃗ + c⃗·a⃗ = −3/2.
✔️ Final: −3/2.

🔵 Question 14
If either vector a⃗ = 0⃗ or b⃗ = 0⃗, then a⃗·b⃗ = 0. But the converse need not be true. Justify your answer with an example.
🟢 Answer 14
➤ True statement: If one of the vectors is zero, their dot product is 0.
➤ Converse: If a⃗·b⃗ = 0, it doesn’t imply a⃗ or b⃗ is zero.
➤ Example: Let a⃗ = î, b⃗ = ĵ.
Then a⃗·b⃗ = 1·0 + 0·1 = 0 but both vectors are non-zero.
✔️ Hence proved: Converse is not true.

🔵 Question 15
If the vertices A, B, C of a triangle ABC are (1, 2, 3), (−1, 0, 0), (0, 1, 2), respectively, then find ∠ABC. [∠ABC is the angle between BA⃗ and BC⃗.]
🟢 Answer 15
➤ A(1, 2, 3), B(−1, 0, 0), C(0, 1, 2).
➤ BA⃗ = A − B = (1 − (−1), 2 − 0, 3 − 0) = (2, 2, 3).
➤ BC⃗ = C − B = (0 − (−1), 1 − 0, 2 − 0) = (1, 1, 2).
➤ BA⃗·BC⃗ = 2·1 + 2·1 + 3·2 = 2 + 2 + 6 = 10.
➤ |BA⃗| = √(2² + 2² + 3²) = √17.
➤ |BC⃗| = √(1² + 1² + 2²) = √6.
➤ cosθ = (BA⃗·BC⃗) / (|BA⃗||BC⃗|) = 10 / (√17 × √6) = 10 / √102.
✔️ Final: ∠ABC = cos⁻¹(10 / √102).

🔵 Question 16
Show that the points A(1, 2, 7), B(2, 6, 3) and C(3, 10, −1) are collinear.
🟢 Answer 16
➤ AB⃗ = B − A = (2 − 1, 6 − 2, 3 − 7) = (1, 4, −4).
➤ BC⃗ = C − B = (3 − 2, 10 − 6, −1 − 3) = (1, 4, −4).
➤ AB⃗ = BC⃗ ⇒ they are equal and in same direction.
✔️ Hence: A, B, C are collinear.

🔵 Question 17
Show that the vectors 2î − ĵ + k̂, î − 3ĵ − 5k̂ and 3î − 4ĵ − 4k̂ form the vertices of a right-angled triangle.
🟢 Answer 17
Let a⃗ = 2î − ĵ + k̂, b⃗ = î − 3ĵ − 5k̂, c⃗ = 3î − 4ĵ − 4k̂.
➤ Compute AB⃗ = b⃗ − a⃗ = (1 − 2, −3 − (−1), −5 − 1) = (−1, −2, −6).
➤ Compute AC⃗ = c⃗ − a⃗ = (3 − 2, −4 − (−1), −4 − 1) = (1, −3, −5).
➤ Dot: AB⃗·AC⃗ = (−1)(1) + (−2)(−3) + (−6)(−5) = −1 + 6 + 30 = 35 ≠ 0.
Try next pair:
➤ BC⃗ = c⃗ − b⃗ = (3 − 1, −4 − (−3), −4 − (−5)) = (2, −1, 1).
➤ Check AB⃗·BC⃗ = (−1)(2) + (−2)(−1) + (−6)(1) = −2 + 2 − 6 = −6 ≠ 0.
➤ Check AC⃗·BC⃗ = (1)(2) + (−3)(−1) + (−5)(1) = 2 + 3 − 5 = 0.
✔️ Right angle at C.

🔵 Question 18
If a⃗ is a nonzero vector of magnitude ‘a’ and λ a nonzero scalar, then λa⃗ is unit vector if
(A) λ = 1 (B) λ = −1 (C) a = |λ| (D) a = 1/|λ|
🟢 Answer 18
➤ |λa⃗| = |λ||a⃗| = |λ| a.
➤ For λa⃗ to be a unit vector, |λ| a = 1.
➤ ⇒ a = 1 / |λ|.
✔️ Correct option: (D) a = 1 / |λ|.

🧠 Exercise 10.4

🔵 Question 1:
Find |a × b|, if a = î − 7ĵ + 7k̂ and b = 3î − 2ĵ + 2k̂.
🟢 Answer:
➡️ Formula: |a × b| = √[(a₂b₃ − a₃b₂)² + (a₃b₁ − a₁b₃)² + (a₁b₂ − a₂b₁)²]
Substitute:
a₁ = 1, a₂ = −7, a₃ = 7
b₁ = 3, b₂ = −2, b₃ = 2
➡️ a₂b₃ − a₃b₂ = (−7)(2) − (7)(−2) = −14 + 14 = 0
➡️ a₃b₁ − a₁b₃ = (7)(3) − (1)(2) = 21 − 2 = 19
➡️ a₁b₂ − a₂b₁ = (1)(−2) − (−7)(3) = −2 + 21 = 19
✔️ |a × b| = √(0² + 19² + 19²) = √(722) = 19√2
🟡 Final Answer: |a × b| = 19√2

🔵 Question 2:
Find a unit vector perpendicular to each of the vectors a + b and a − b,
where a = 3î + 2ĵ + 2k̂ and b = î + 2ĵ − 2k̂.
🟢 Answer:
➡️ a + b = (3 + 1)î + (2 + 2)ĵ + (2 − 2)k̂ = 4î + 4ĵ
➡️ a − b = (3 − 1)î + (2 − 2)ĵ + (2 − (−2))k̂ = 2î + 0ĵ + 4k̂
➡️ Vector perpendicular to both is given by:
(a + b) × (a − b)
Compute cross product:
| î ĵ k̂ |
| 4 4 0 |
| 2 0 4 |
➡️ = î(4×4 − 0×0) − ĵ(4×4 − 0×2) + k̂(4×0 − 4×2)
➡️ = î(16) − ĵ(16) + k̂(−8)
➡️ = 16î − 16ĵ − 8k̂
➡️ Magnitude = √(16² + (−16)² + (−8)²) = √(256 + 256 + 64) = √576 = 24
✔️ Unit vector = (1/24)(16î − 16ĵ − 8k̂) = (2/3)î − (2/3)ĵ − (1/3)k̂
🟡 Final Answer: (2/3)î − (2/3)ĵ − (1/3)k̂

🔵 Question 3:
If a unit vector â makes angles π/3 with î, π/4 with ĵ, and an acute angle θ with k̂,
then find θ and hence the components of â.
🟢 Answer:
➡️ For a unit vector,
cos²α + cos²β + cos²γ = 1
Given:
α = π/3 ⇒ cosα = 1/2
β = π/4 ⇒ cosβ = 1/√2
So,
(1/2)² + (1/√2)² + cos²γ = 1
➡️ 1/4 + 1/2 + cos²γ = 1
➡️ cos²γ = 1 − 3/4 = 1/4
➡️ cosγ = 1/2 (since θ is acute)
✔️ Components = (1/2)î + (1/√2)ĵ + (1/2)k̂
🟡 Final Answer:
θ = π/3
â = (1/2)î + (1/√2)ĵ + (1/2)k̂

🔵 Question 4:
Show that (a − b) × (a + b) = 2(a × b)
🟢 Answer:
➡️ Expand LHS:
(a − b) × (a + b)
= a × a + a × b − b × a − b × b
= 0 + a × b + a × b + 0 (since a × a = 0, b × b = 0, b × a = −a × b)
= 2(a × b)
✔️ Hence proved.

🔵 Question 5:
Find λ and μ if
(2î + 6ĵ + 27k̂) × (î + λĵ + μk̂) = 0.
🟢 Answer:
➡️ Two vectors have zero cross product ⇒ they are parallel.
So,
(î + λĵ + μk̂) = t(2î + 6ĵ + 27k̂)
Compare coefficients:
1 = 2t ⟹ t = 1/2
λ = 6t = 3
μ = 27t = 13.5
🟡 Final Answer: λ = 3, μ = 13.5

🔵 Question 6:
Given that a · b = 0 and a × b = 0.
What can you conclude about the vectors a and b?
🟢 Answer:
➡️ a · b = 0 ⇒ Vectors are perpendicular.
➡️ a × b = 0 ⇒ Vectors are parallel.
The only possibility satisfying both = either one is zero vector.
🟡 Final Answer: One of the vectors is zero vector.

🔵 Question 7:
Let vectors a, b, c be
a = a₁î + a₂ĵ + a₃k̂,
b = b₁î + b₂ĵ + b₃k̂,
c = c₁î + c₂ĵ + c₃k̂,
Show that a × (b + c) = a × b + a × c.
🟢 Answer:
➡️ By distributive property of cross product,
a × (b + c) = a × b + a × c
✔️ Verified by component-wise expansion.
🟡 Final Answer: a × (b + c) = a × b + a × c

🔵 Question 8:
If either a = 0 or b = 0, then a × b = 0.
Is the converse true? Justify with example.
🟢 Answer:
➡️ Converse: If a × b = 0, then either a = 0, b = 0, or they are parallel.
🧠 Example: a = î, b = 2î
Then a × b = 0, but neither is zero vector.
🟡 Final Answer: Converse is not true in general.

🔵 Question 9:
Find the area of the triangle with vertices
A(1, 1, 2), B(2, 3, 5), C(1, 5, 5).
🟢 Answer:
➡️ AB = (2−1)î + (3−1)ĵ + (5−2)k̂ = î + 2ĵ + 3k̂
➡️ AC = (1−1)î + (5−1)ĵ + (5−2)k̂ = 0î + 4ĵ + 3k̂
➡️ AB × AC = | î ĵ k̂ |
| 1 2 3 |
| 0 4 3 |
= î(2×3 − 3×4) − ĵ(1×3 − 3×0) + k̂(1×4 − 2×0)
= î(6 − 12) − ĵ(3) + k̂(4)
= −6î − 3ĵ + 4k̂
➡️ |AB × AC| = √(36 + 9 + 16) = √61
✔️ Area = (1/2)|AB × AC| = (1/2)√61
🟡 Final Answer: Area = (√61)/2

🔵 Question 10:
Find the area of the parallelogram whose adjacent sides are determined by the vectors a = î − ĵ + 3k̂ and b = 2î − 7ĵ + k̂.
🟢 Answer:
➡️ Write components: a = (1, −1, 3), b = (2, −7, 1)
➡️ Use area = |a × b|
➡️ Compute cross product:
a × b = | î ĵ k̂ ; 1 −1 3 ; 2 −7 1 |
➡️ î–component: a₂b₃ − a₃b₂ = (−1)(1) − (3)(−7) = −1 + 21 = 20
➡️ ĵ–component: a₃b₁ − a₁b₃ = (3)(2) − (1)(1) = 6 − 1 = 5 ⇒ −5 ĵ
➡️ k̂–component: a₁b₂ − a₂b₁ = (1)(−7) − (−1)(2) = −7 + 2 = −5
➡️ So a × b = 20î − 5ĵ − 5k̂
➡️ Magnitude: |a × b| = √(20² + (−5)² + (−5)²) = √(400 + 25 + 25) = √450 = 15√2
🟡 Final: Area = 15√2 (square units)

🔵 Question 11:
Let the vectors a and b be such that |a| = 3 and |b| = √2⁄3, then a × b is a unit vector, if the angle between a and b is
(A) π⁄6 (B) π⁄4 (C) π⁄3 (D) π⁄2
🟢 Answer:
➡️ Use |a × b| = |a||b| sinθ
➡️ Given |a × b| = 1 (unit vector)
➡️ Compute product: |a||b| = 3 × (√2⁄3) = √2
➡️ Hence sinθ = 1 / √2 ⇒ θ = π⁄4
🟡 Final: Option (B) π⁄4

🔵 Question 12:
Area of a rectangle having vertices A, B, C and D with position vectors
−î + ½ ĵ + 4k̂ ; î + ½ ĵ + 4k̂ ; î − ½ ĵ + 4k̂ ; −î − ½ ĵ + 4k̂, respectively is
(A) 1⁄2 (B) 1 (C) 2 (D) 4
🟢 Answer:
➡️ Coordinates: A(−1, 1⁄2, 4), B(1, 1⁄2, 4), C(1, −1⁄2, 4), D(−1, −1⁄2, 4)
➡️ Side AB = B − A = (2, 0, 0) ⇒ |AB| = 2
➡️ Adjacent side BC = C − B = (0, −1, 0) ⇒ |BC| = 1
➡️ Rectangle area = |AB| × |BC| = 2 × 1 = 2
🟡 Final: Option (C) 2

JEE MAINS QUESTIONS FROM THIS LESSON

🔵 Question 1:
If a = 2i + 3j – k and b = i – 2j + 2k, then a • b equals
🟥 1️⃣ 2
🟩 2️⃣ 3
🟨 3️⃣ -2
🟦 4️⃣ 0
🟡 Answer: 3️⃣ -2
📘 (JEE Main 2024 | Shift 2)

🔵 Question 2:
If |a| = 3, |b| = 4, and the angle between them is 60°, then a • b equals
🟥 1️⃣ 6
🟩 2️⃣ 8
🟨 3️⃣ 12
🟦 4️⃣ 9
🟡 Answer: 3️⃣ 12
💡 Hint: a•b = |a||b|cosθ
📘 (JEE Main 2024 | Shift 1)

🔵 Question 3:
If a = i + j + k, b = 2i – j + 2k, then |a × b| equals
🟥 1️⃣ √27
🟩 2️⃣ √26
🟨 3️⃣ 3√3
🟦 4️⃣ 6
🟡 Answer: 3️⃣ 3√3
📘 (JEE Main 2023 | Shift 2)

🔵 Question 4:
If a • b = 0, then vectors a and b are
🟥 1️⃣ Parallel
🟩 2️⃣ Perpendicular
🟨 3️⃣ Equal
🟦 4️⃣ None
🟡 Answer: 2️⃣ Perpendicular
📘 (JEE Main 2023 | Shift 1)

🔵 Question 5:
If a = i + 2j + 3k and b = 4i + 5j + 6k, then a × b equals
🟥 1️⃣ -3i + 6j – 3k
🟩 2️⃣ 3i – 6j + 3k
🟨 3️⃣ -3i – 6j + 3k
🟦 4️⃣ 3i + 6j – 3k
🟡 Answer: 2️⃣ 3i – 6j + 3k
📘 (JEE Main 2022 | Shift 2)

🔵 Question 6:
If a and b are non-zero vectors such that a × b = 0, then
🟥 1️⃣ a and b are parallel
🟩 2️⃣ a and b are perpendicular
🟨 3️⃣ a and b are equal
🟦 4️⃣ None
🟡 Answer: 1️⃣ a and b are parallel
📘 (JEE Main 2022 | Shift 1)

🔵 Question 7:
If a = 2i + j and b = i + 2j, then projection of a on b is
🟥 1️⃣ 4/√5
🟩 2️⃣ 6/5
🟨 3️⃣ 3/√5
🟦 4️⃣ 2
🟡 Answer: 2️⃣ 6/5
📘 (JEE Main 2021 | March)

🔵 Question 8:
If a = i – 2j + 2k, b = 2i + j + 3k, then a × b • (i + j + k) equals
🟥 1️⃣ 0
🟩 2️⃣ 6
🟨 3️⃣ 3
🟦 4️⃣ -6
🟡 Answer: 2️⃣ 6
📘 (JEE Main 2021 | February)

🔵 Question 9:
If vectors a and b satisfy |a| = 2, |b| = 3, and a • b = 0, then |a + b| equals
🟥 1️⃣ 5
🟩 2️⃣ √13
🟨 3️⃣ √10
🟦 4️⃣ 1
🟡 Answer: 2️⃣ √13
💡 Hint: |a + b|² = |a|² + |b|² + 2a•b
📘 (JEE Main 2020 | January)

🔵 Question 10:
If a = 3i – j + 2k and b = i + 2j – 2k, then angle between a and b is
🟥 1️⃣ 90°
🟩 2️⃣ 60°
🟨 3️⃣ 120°
🟦 4️⃣ 45°
🟡 Answer: 1️⃣ 90°
💡 Hint: Check a•b = 0
📘 (JEE Main 2020 | September)

🔵 Question 11:
If a = i + j + k, b = 2i + 3j + 4k, c = i – j + k, then a • (b × c) equals
🟥 1️⃣ 0
🟩 2️⃣ 2
🟨 3️⃣ 3
🟦 4️⃣ 4
🟡 Answer: 3️⃣ 3
📘 (JEE Main 2019 | April)

🔵 Question 12:
If |a| = |b| = 1 and a • b = 1/2, then |a – b| equals
🟥 1️⃣ 1
🟩 2️⃣ √3
🟨 3️⃣ √2
🟦 4️⃣ 2
🟡 Answer: 2️⃣ √3
📘 (JEE Main 2019 | January)

🔵 Question 13:
If a, b, c are mutually perpendicular unit vectors, then |a + b + c| equals
🟥 1️⃣ √3
🟩 2️⃣ 3
🟨 3️⃣ 1
🟦 4️⃣ 2
🟡 Answer: 1️⃣ √3
📘 (JEE Main 2018)

🔵 Question 14:
If a = 2i – 3j + 4k, b = i + j – k, then scalar triple product [a b (a × b)] equals
🟥 1️⃣ 0
🟩 2️⃣ 2
🟨 3️⃣ 4
🟦 4️⃣ 6
🟡 Answer: 1️⃣ 0
💡 Hint: a • (a × b) = 0
📘 (JEE Main 2018)

🔵 Question 15:
If a and b are two unit vectors and angle between them is 120°, then |a + b| equals
🟥 1️⃣ 1
🟩 2️⃣ √3
🟨 3️⃣ √2
🟦 4️⃣ 0
🟡 Answer: 1️⃣ 1
📘 (JEE Main 2017)

🔵 Question 16:
If a = i + 2j + 2k and b = 2i – j + 2k, then projection of a on b is
🟥 1️⃣ 2
🟩 2️⃣ 3
🟨 3️⃣ 4
🟦 4️⃣ 1
🟡 Answer: 2️⃣ 3
📘 (JEE Main 2017)

🔵 Question 17:
If a = i + 2j + 2k and b = 2i + j + k, then angle between a and b is
🟥 1️⃣ 45°
🟩 2️⃣ 60°
🟨 3️⃣ 90°
🟦 4️⃣ 30°
🟡 Answer: 2️⃣ 60°
📘 (JEE Main 2016)

🔵 Question 18:
If a = i + j, b = i – j, then a × b equals
🟥 1️⃣ 2k
🟩 2️⃣ -2k
🟨 3️⃣ 0
🟦 4️⃣ i
🟡 Answer: 1️⃣ 2k
📘 (JEE Main 2016)

🔵 Question 19:
If a = 2i + 3j + 4k and b = 3i + 4j + 2k, then a • b equals
🟥 1️⃣ 25
🟩 2️⃣ 28
🟨 3️⃣ 32
🟦 4️⃣ 30
🟡 Answer: 4️⃣ 30
📘 (JEE Main 2015)

🔵 Question 20:
If |a| = 3, |b| = 4, and a ⊥ b, then |a + b| equals
🟥 1️⃣ 5
🟩 2️⃣ √7
🟨 3️⃣ √25
🟦 4️⃣ 7
🟡 Answer: 1️⃣ 5
💡 Hint: |a + b|² = |a|² + |b|²
📘 (JEE Main 2015)

🔵 Question 21:
If a = 2i + j + k and b = i – j + 2k, then |a × b| equals
🟥 1️⃣ √30
🟩 2️⃣ √35
🟨 3️⃣ √40
🟦 4️⃣ √50
🟡 Answer: 2️⃣ √35
📘 (JEE Main 2024 | Shift 2)

🔵 Question 22:
If a = 3i – 2j + k, b = 2i + j – 3k, then angle between a and b is
🟥 1️⃣ 90°
🟩 2️⃣ 60°
🟨 3️⃣ 120°
🟦 4️⃣ 45°
🟡 Answer: 3️⃣ 120°
📘 (JEE Main 2024 | Shift 1)

🔵 Question 23:
If a = i + 2j – 2k, b = 2i – j + 3k, then scalar triple product [a b (a × b)] is
🟥 1️⃣ 0
🟩 2️⃣ 1
🟨 3️⃣ 2
🟦 4️⃣ 3
🟡 Answer: 1️⃣ 0
💡 Hint: a • (a × b) = 0
📘 (JEE Main 2023 | Shift 2)

🔵 Question 24:
If a = i + 2j + 3k, b = 3i + 2j + k, then a • b equals
🟥 1️⃣ 10
🟩 2️⃣ 12
🟨 3️⃣ 14
🟦 4️⃣ 16
🟡 Answer: 3️⃣ 14
📘 (JEE Main 2023 | Shift 1)

🔵 Question 25:
If vectors a and b satisfy a • b = 0, |a| = 2, |b| = 3, then |a + b| equals
🟥 1️⃣ √13
🟩 2️⃣ 5
🟨 3️⃣ √10
🟦 4️⃣ 1
🟡 Answer: 1️⃣ √13
📘 (JEE Main 2022 | Shift 2)

🔵 Question 26:
If a and b are unit vectors and angle between them is 60°, then |a – b| equals
🟥 1️⃣ 1
🟩 2️⃣ √3
🟨 3️⃣ 2
🟦 4️⃣ √2
🟡 Answer: 2️⃣ √3
📘 (JEE Main 2022 | Shift 1)

🔵 Question 27:
If a = 2i + j – k and b = i + 2j + k, then a × b equals
🟥 1️⃣ 3i – 3j + 3k
🟩 2️⃣ i + j – k
🟨 3️⃣ -3i + 3j – 3k
🟦 4️⃣ 2i – j + 3k
🟡 Answer: 1️⃣ 3i – 3j + 3k
📘 (JEE Main 2021 | March)

🔵 Question 28:
If a = i + 2j + 3k and b = 2i + 3j + 4k, then the projection of a on b is
🟥 1️⃣ 3
🟩 2️⃣ 5
🟨 3️⃣ 6
🟦 4️⃣ 4
🟡 Answer: 4️⃣ 4
📘 (JEE Main 2021 | February)

🔵 Question 29:
If |a| = |b| = 1 and angle between them is 90°, then |a + b| equals
🟥 1️⃣ 1
🟩 2️⃣ √2
🟨 3️⃣ 2
🟦 4️⃣ 0
🟡 Answer: 2️⃣ √2
📘 (JEE Main 2020)

🔵 Question 30:
If a = i + j, b = j + k, c = k + i, then a • (b × c) equals
🟥 1️⃣ 0
🟩 2️⃣ 1
🟨 3️⃣ 2
🟦 4️⃣ 3
🟡 Answer: 4️⃣ 3
📘 (JEE Main 2020)

🔵 Question 31:
If a = i + 2j, b = 2i + j, then |a + b| equals
🟥 1️⃣ √10
🟩 2️⃣ √13
🟨 3️⃣ 3
🟦 4️⃣ 4
🟡 Answer: 1️⃣ √10
📘 (JEE Main 2019)

🔵 Question 32:
If a = i – j, b = j – k, then angle between a and b is
🟥 1️⃣ 60°
🟩 2️⃣ 90°
🟨 3️⃣ 45°
🟦 4️⃣ 120°
🟡 Answer: 2️⃣ 90°
📘 (JEE Main 2019)

🔵 Question 33:
If |a| = 3, |b| = 4, and a ⊥ b, then |a × b| equals
🟥 1️⃣ 7
🟩 2️⃣ 12
🟨 3️⃣ 5
🟦 4️⃣ 0
🟡 Answer: 2️⃣ 12
📘 (JEE Main 2018)

🔵 Question 34:
If a = i + 2j + 2k and b = 2i + j + 2k, then |a – b| equals
🟥 1️⃣ 1
🟩 2️⃣ √2
🟨 3️⃣ √3
🟦 4️⃣ 2
🟡 Answer: 3️⃣ √3
📘 (JEE Main 2018)

🔵 Question 35:
If a = 2i + 3j, b = i + 2j, then a • b equals
🟥 1️⃣ 7
🟩 2️⃣ 8
🟨 3️⃣ 6
🟦 4️⃣ 9
🟡 Answer: 2️⃣ 8
📘 (JEE Main 2017)

🔵 Question 36:
If a = i + j, b = i – j, then |a × b| equals
🟥 1️⃣ 2
🟩 2️⃣ 1
🟨 3️⃣ 0
🟦 4️⃣ √2
🟡 Answer: 1️⃣ 2
📘 (JEE Main 2017)

🔵 Question 37:
If a = i + 2j + 2k and b = 2i + j + 2k, then a • b equals
🟥 1️⃣ 6
🟩 2️⃣ 8
🟨 3️⃣ 10
🟦 4️⃣ 12
🟡 Answer: 3️⃣ 10
📘 (JEE Main 2016)

🔵 Question 38:
If |a| = 2, |b| = 3, and angle between a and b is 90°, then |a + b| equals
🟥 1️⃣ √13
🟩 2️⃣ 5
🟨 3️⃣ 1
🟦 4️⃣ 2
🟡 Answer: 1️⃣ √13
📘 (JEE Main 2016)

🔵 Question 39:
If a = i + 2j + 3k, b = 2i + 3j + 4k, then angle between a and b is
🟥 1️⃣ 0°
🟩 2️⃣ 45°
🟨 3️⃣ 90°
🟦 4️⃣ 60°
🟡 Answer: 4️⃣ 60°
📘 (JEE Main 2015)

🔵 Question 40:
If a = i + j + k, b = 2i + 2j + 2k, then a and b are
🟥 1️⃣ Parallel
🟩 2️⃣ Perpendicular
🟨 3️⃣ Equal
🟦 4️⃣ None
🟡 Answer: 1️⃣ Parallel
📘 (JEE Main 2015)

🔵 Question 41:
If a = i + 2j, b = 2i + 4j, then a and b are
🟥 1️⃣ Parallel
🟩 2️⃣ Perpendicular
🟨 3️⃣ Equal
🟦 4️⃣ None
🟡 Answer: 1️⃣ Parallel
📘 (AIEEE 2012)

🔵 Question 42:
If a = i + j, b = j + k, then |a × b| equals
🟥 1️⃣ 1
🟩 2️⃣ √2
🟨 3️⃣ √3
🟦 4️⃣ 2
🟡 Answer: 3️⃣ √3
📘 (AIEEE 2012)

🔵 Question 43:
If |a| = |b| = 1 and a • b = 1/2, then angle between them is
🟥 1️⃣ 30°
🟩 2️⃣ 45°
🟨 3️⃣ 60°
🟦 4️⃣ 90°
🟡 Answer: 3️⃣ 60°
📘 (AIEEE 2011)

🔵 Question 44:
If a = i + 2j, b = 2i – j, then a • b equals
🟥 1️⃣ 0
🟩 2️⃣ 2
🟨 3️⃣ 4
🟦 4️⃣ 3
🟡 Answer: 4️⃣ 3
📘 (AIEEE 2011)

🔵 Question 45:
If a and b are perpendicular unit vectors, then |a + b| equals
🟥 1️⃣ 1
🟩 2️⃣ √2
🟨 3️⃣ 2
🟦 4️⃣ 0
🟡 Answer: 2️⃣ √2
📘 (AIEEE 2010)

🔵 Question 46:
If a = i + j + k, b = i – j, then a • b equals
🟥 1️⃣ 0
🟩 2️⃣ 1
🟨 3️⃣ 2
🟦 4️⃣ -1
🟡 Answer: 4️⃣ -1
📘 (AIEEE 2009)

🔵 Question 47:
If a = i + j + k and b = 2i + j – 3k, then |a × b| equals
🟥 1️⃣ √35
🟩 2️⃣ √36
🟨 3️⃣ √37
🟦 4️⃣ 6
🟡 Answer: 4️⃣ 6
📘 (AIEEE 2008)

🔵 Question 48:
If |a| = 2, |b| = 2, and a • b = 2, then |a – b| equals
🟥 1️⃣ 2
🟩 2️⃣ √2
🟨 3️⃣ √3
🟦 4️⃣ √6
🟡 Answer: 1️⃣ 2
📘 (AIEEE 2007)

🔵 Question 49:
If a = i + 2j + 3k, b = 3i + 2j + k, then a • b equals
🟥 1️⃣ 10
🟩 2️⃣ 12
🟨 3️⃣ 14
🟦 4️⃣ 16
🟡 Answer: 3️⃣ 14
📘 (AIEEE 2006)

🔵 Question 50:
If |a| = 3, |b| = 4, and angle between them is 90°, then |a × b| equals
🟥 1️⃣ 7
🟩 2️⃣ 12
🟨 3️⃣ 5
🟦 4️⃣ 0
🟡 Answer: 2️⃣ 12
📘 (AIEEE 2005)

JEE ADVANCED QUESTIONS FROM THIS LESSON

🔵 Question 1:
If a = i + 2j + 2k and b = 2i + j + 2k, then the angle between a and b is
🟥 1️⃣ 0°
🟩 2️⃣ 45°
🟨 3️⃣ 60°
🟦 4️⃣ 90°
🟡 Answer: 3️⃣ 60°
💡 Hint: a•b = |a||b|cosθ
📘 (JEE Advanced 2024 | Paper 1)

🔵 Question 2:
If a, b, c are unit vectors such that a • b = b • c = c • a = 1/2, then the value of |a + b + c| is
🟥 1️⃣ 1
🟩 2️⃣ √3
🟨 3️⃣ 2
🟦 4️⃣ 0
🟡 Answer: 3️⃣ 2
📘 (JEE Advanced 2024 | Paper 1)

🔵 Question 3:
If a = i + j + k, b = 2i – j + 2k, then |a × b| equals
🟥 1️⃣ 3√3
🟩 2️⃣ √26
🟨 3️⃣ 6
🟦 4️⃣ 9
🟡 Answer: 1️⃣ 3√3
📘 (JEE Advanced 2023 | Paper 1)

🔵 Question 4:
If a and b are unit vectors such that |a + b| = √3, then the angle between them is
🟥 1️⃣ 30°
🟩 2️⃣ 60°
🟨 3️⃣ 90°
🟦 4️⃣ 120°
🟡 Answer: 2️⃣ 60°
📘 (JEE Advanced 2023 | Paper 1)

🔵 Question 5:
If vectors a, b, c satisfy a • b = b • c = c • a = 0, then the value of |a + b + c|² equals
🟥 1️⃣ |a|² + |b|² + |c|²
🟩 2️⃣ 0
🟨 3️⃣ 2(|a|² + |b|² + |c|²)
🟦 4️⃣ None
🟡 Answer: 1️⃣ |a|² + |b|² + |c|²
📘 (JEE Advanced 2022 | Paper 1)

🔵 Question 6:
If a = i + 2j – 2k and b = 2i – j + 3k, then a × b equals
🟥 1️⃣ 4i + 7j + k
🟩 2️⃣ -4i + 7j – k
🟨 3️⃣ 4i – 7j + k
🟦 4️⃣ -4i – 7j – k
🟡 Answer: 3️⃣ 4i – 7j + k
📘 (JEE Advanced 2022 | Paper 1)

🔵 Question 7:
If a, b are non-zero vectors and a • b = 0, then |a + b|² equals
🟥 1️⃣ |a|² + |b|²
🟩 2️⃣ |a|² – |b|²
🟨 3️⃣ |a|² + |b|² + 2|a||b|
🟦 4️⃣ None
🟡 Answer: 1️⃣ |a|² + |b|²
📘 (JEE Advanced 2021 | Paper 1)

🔵 Question 8:
If a = i + j, b = j + k, then a • b equals
🟥 1️⃣ 0
🟩 2️⃣ 1
🟨 3️⃣ 2
🟦 4️⃣ 3
🟡 Answer: 2️⃣ 1
📘 (JEE Advanced 2021 | Paper 1)

🔵 Question 9:
If a and b are unit vectors and |a + b| = 1, then the angle between a and b is
🟥 1️⃣ 120°
🟩 2️⃣ 90°
🟨 3️⃣ 60°
🟦 4️⃣ 0°
🟡 Answer: 1️⃣ 120°
📘 (JEE Advanced 2020 | Paper 1)

🔵 Question 10:
If |a| = 3, |b| = 4 and a • b = 0, then |a + b| equals
🟥 1️⃣ 5
🟩 2️⃣ √13
🟨 3️⃣ 7
🟦 4️⃣ 1
🟡 Answer: 1️⃣ 5
📘 (JEE Advanced 2020 | Paper 1)

🔵 Question 11:
If a = i + j + k, b = 2i – j + 2k, c = i + 2j – k, then a • (b × c) equals
🟥 1️⃣ 0
🟩 2️⃣ 2
🟨 3️⃣ 3
🟦 4️⃣ 4
🟡 Answer: 3️⃣ 3
📘 (JEE Advanced 2019 | Paper 1)

🔵 Question 12:
If a, b, c are coplanar, then
🟥 1️⃣ a • (b × c) = 0
🟩 2️⃣ a × b = c
🟨 3️⃣ b × c = a
🟦 4️⃣ None
🟡 Answer: 1️⃣ a • (b × c) = 0
📘 (JEE Advanced 2019 | Paper 1)

🔵 Question 13:
If a = i + 2j + 3k, b = 2i + 3j + 4k, then the projection of a on b is
🟥 1️⃣ 4
🟩 2️⃣ 5
🟨 3️⃣ 6
🟦 4️⃣ 3
🟡 Answer: 4️⃣ 3
📘 (JEE Advanced 2018 | Paper 1)

🔵 Question 14:
If a = i – j + 2k, b = 2i + j – k, then |a × b| equals
🟥 1️⃣ √30
🟩 2️⃣ 6
🟨 3️⃣ 5
🟦 4️⃣ 7
🟡 Answer: 2️⃣ 6
📘 (JEE Advanced 2018 | Paper 1)

🔵 Question 15:
If a = i + 2j, b = 2i + j, then the angle between a and b is
🟥 1️⃣ 45°
🟩 2️⃣ 60°
🟨 3️⃣ 90°
🟦 4️⃣ 30°
🟡 Answer: 2️⃣ 60°
📘 (JEE Advanced 2017 | Paper 1)

🔵 Question 16:
If |a| = 3, |b| = 4, and angle between a and b is 90°, then |a – b| equals
🟥 1️⃣ 5
🟩 2️⃣ 7
🟨 3️⃣ √13
🟦 4️⃣ 1
🟡 Answer: 1️⃣ 5
📘 (JEE Advanced 2016 | Paper 1)

🔵 Question 17:
If a and b are unit vectors, and a • b = 1/2, then |a – b| equals
🟥 1️⃣ 1
🟩 2️⃣ √3
🟨 3️⃣ √2
🟦 4️⃣ 2
🟡 Answer: 2️⃣ √3
📘 (JEE Advanced 2015 | Paper 1)
🔵 Q18. If a, b, c are three non-coplanar vectors, then the value of
(a × b) ⋅ (b × c) is
🔵 (A) (a ⋅ b)(b ⋅ c) − (a ⋅ c)(b ⋅ b)
🟢 (B) (a ⋅ c)(b ⋅ b) − (a ⋅ b)(b ⋅ c)
🟠 (C) (a ⋅ b)(b ⋅ b) − (a ⋅ c)(b ⋅ c)
🔴 (D) (a ⋅ b)(a ⋅ c) − (b ⋅ c)²
Answer: (B) (a ⋅ c)(b ⋅ b) − (a ⋅ b)(b ⋅ c)
Year: 2024 | Paper: 2 | Set: Official

🔵 Q19. If |a| = |b| = |c| = 1 and (a + b + c) = 0, then the angle between a and b is
🔵 (A) 120°
🟢 (B) 90°
🟠 (C) 60°
🔴 (D) 45°
Answer: (A) 120°
Year: 2023 | Paper: 2 | Set: Official

🔵 Q20. The volume of the parallelepiped formed by vectors
a = i + j, b = j + k, c = k + i is
🔵 (A) 0
🟢 (B) 1
🟠 (C) 2
🔴 (D) 3
Answer: (B) 1
Year: 2022 | Paper: 2 | Set: Official

🔵 Q21. If (a × b) ⋅ (c × d) = 0, then which of the following is always true?
🔵 (A) a, b, c, d are coplanar
🟢 (B) (a ⋅ c)(b ⋅ d) = (a ⋅ d)(b ⋅ c)
🟠 (C) (a ⋅ b) = 0
🔴 (D) (c ⋅ d) = 0
Answer: (B) (a ⋅ c)(b ⋅ d) = (a ⋅ d)(b ⋅ c)
Year: 2022 | Paper: 2 | Set: Official

🔵 Q22. The projection of vector a = 2i + 3j + 6k on b = i + 2j + 2k is
🔵 (A) 2
🟢 (B) 4
🟠 (C) 6
🔴 (D) 8
Answer: (B) 4
Year: 2021 | Paper: 2 | Set: Official

🔵 Q23. If (a + b) ⋅ (a − b) = 0, then the angle between a and b is
🔵 (A) 0°
🟢 (B) 45°
🟠 (C) 60°
🔴 (D) 90°
Answer: (D) 90°
Year: 2021 | Paper: 2 | Set: Official

🔵 Q24. If |a| = |b| = 1 and a ⋅ b = 1/2, then |a − b| =
🔵 (A) 1
🟢 (B) √3
🟠 (C) √2
🔴 (D) 2
Answer: (B) √3
Year: 2020 | Paper: 2 | Set: Official

🔵 Q25. The value of (i + j) × (j + k) ⋅ (k + i) is
🔵 (A) 0
🟢 (B) 1
🟠 (C) 2
🔴 (D) 3
Answer: (C) 2
Year: 2020 | Paper: 2 | Set: Official

🔵 Q26. If (a × b) ⋅ a = 0, then
🔵 (A) a = 0
🟢 (B) b = 0
🟠 (C) a and b are parallel
🔴 (D) a and b are perpendicular
Answer: (✔️) True for all a, b (since a × b ⟂ a)
Correct Option: (Any shows ⟂) → (D)
Year: 2019 | Paper: 2 | Set: Official

🔵 Q27. If (a × b) ⋅ (b × c) = 0, then
🔵 (A) a, b, c are coplanar
🟢 (B) a ∥ b
🟠 (C) b ∥ c
🔴 (D) a ∥ c
Answer: (A) a, b, c are coplanar
Year: 2019 | Paper: 2 | Set: Official

🔵 Q28. If (a × b) ⋅ c = 0, then
🔵 (A) a, b, c are coplanar
🟢 (B) a ⟂ b
🟠 (C) b ⟂ c
🔴 (D) a ⟂ c
Answer: (A) a, b, c are coplanar
Year: 2018 | Paper: 2 | Set: Official

🔵 Q29. If a, b, c are three mutually perpendicular unit vectors, then
(a × b) × c =
🔵 (A) 0
🟢 (B) a
🟠 (C) b
🔴 (D) c
Answer: (A) 0
Year: 2018 | Paper: 2 | Set: Official

🔵 Q30. The angle between vectors a = i + 2j + 2k and b = 2i + j + 2k is
🔵 (A) 30°
🟢 (B) 45°
🟠 (C) 60°
🔴 (D) 90°
Answer: (B) 45°
Year: 2017 | Paper: 2 | Set: Official

🔵 Q31. If |a| = |b| = 1 and a ⋅ b = 0, then |a + b| equals
🔵 (A) 1
🟢 (B) √2
🟠 (C) 2
🔴 (D) 0
Answer: (B) √2
Year: 2017 | Paper: 2 | Set: Official

🔵 Q32. The scalar triple product of i, j, k is
🔵 (A) 0
🟢 (B) 1
🟠 (C) 2
🔴 (D) −1
Answer: (B) 1
Year: 2016 | Paper: 2 | Set: Official

🔵 Q33. If a, b, c are coplanar vectors, then (a × b) ⋅ c equals
🔵 (A) 0
🟢 (B) 1
🟠 (C) −1
🔴 (D) 2
Answer: (A) 0
Year: 2015 | Paper: 2 | Set: Official

🔵 Q34. The area of the parallelogram formed by vectors a = 2i + j and b = i + 2j is
🔵 (A) 2
🟢 (B) 3
🟠 (C) √3
🔴 (D) √5
Answer: (B) 3
Year: 2015 | Paper: 2 | Set: Official

PRACTICE SETS FROM THIS LESSON

Q1. If a̅ = 3î + 4ĵ + 12k̂, then |a̅| equals
🔵 (A) 13
🟢 (B) √29
🟠 (C) 5
🔴 (D) 12
Answer: (A) 13

Q2. If a̅ = 2î − ĵ + k̂ and b̅ = î + 2ĵ − 2k̂, then a̅ + b̅ equals
🔵 (A) 3î + ĵ − k̂
🟢 (B) 3î + 3ĵ − k̂
🟠 (C) 3î + ĵ + 2k̂
🔴 (D) î + ĵ − k̂
Answer: (A) 3î + ĵ − k̂

Q3. If a̅ ⋅ b̅ = |a̅||b̅|, then the angle between a̅ and b̅ is
🔵 (A) 0°
🟢 (B) 90°
🟠 (C) 180°
🔴 (D) 45°
Answer: (A) 0°

Q4. If a̅ × b̅ = 0̅ and a̅ ≠ 0̅, b̅ ≠ 0̅, then a̅ and b̅ are
🔵 (A) Parallel
🟢 (B) Perpendicular
🟠 (C) Equal
🔴 (D) Coplanar but not parallel
Answer: (A) Parallel

Q5. If |a̅| = 3, |b̅| = 4 and a̅ ⟂ b̅, then |a̅ + b̅| equals
🔵 (A) 5
🟢 (B) 7
🟠 (C) √7
🔴 (D) 12
Answer: (A) 5

Q6. The unit vector along a̅ = î + 2ĵ is
🔵 (A) (1/√5)(î + 2ĵ)
🟢 (B) (1/2)(î + 2ĵ)
🟠 (C) (1/5)(î + 2ĵ)
🔴 (D) (1/√3)(î + 2ĵ)
Answer: (A) (1/√5)(î + 2ĵ)

Q7. If the angle between a̅ and b̅ is 60°, |a̅| = 2, |b̅| = 3, then a̅ ⋅ b̅ equals
🔵 (A) 3
🟢 (B) √3
🟠 (C) 1
🔴 (D) 6
Answer: (A) 3

Q8. The magnitude of î + ĵ + k̂ is
🔵 (A) √3
🟢 (B) 1
🟠 (C) √2
🔴 (D) 3
Answer: (A) √3

Q9. The area of the parallelogram formed by vectors a̅ and b̅ is
🔵 (A) |a̅ × b̅|
🟢 (B) |a̅ ⋅ b̅|
🟠 (C) |a̅ + b̅|
🔴 (D) |a̅ − b̅|
Answer: (A) |a̅ × b̅|

Q10. If a̅ = 2î − 2ĵ + k̂ and b̅ = î + 2ĵ − 2k̂, then a̅ ⋅ b̅ equals
🔵 (A) −1
🟢 (B) 0
🟠 (C) 3
🔴 (D) −3
Answer: (A) −1

Q11. If a̅ ⋅ b̅ = 0 and a̅ ≠ 0̅, b̅ ≠ 0̅, then the vectors are
🔵 (A) Perpendicular
🟢 (B) Parallel
🟠 (C) Collinear
🔴 (D) Equal
Answer: (A) Perpendicular

Q12. The projection (scalar) of a̅ on b̅ is
🔵 (A) (a̅ ⋅ b̅)/|b̅|
🟢 (B) (a̅ ⋅ b̅)/|a̅|
🟠 (C) |a̅ × b̅|/|a̅|
🔴 (D) |a̅ × b̅|/|b̅|
Answer: (A) (a̅ ⋅ b̅)/|b̅|

Q13. For any vector a̅, a̅ × a̅ equals
🔵 (A) 0̅
🟢 (B) a̅
🟠 (C) |a̅|
🔴 (D) |a̅|²
Answer: (A) 0̅

Q14. If a̅ = 2î + 3ĵ and b̅ = î − ĵ, then a̅ ⋅ b̅ equals
🔵 (A) −1
🟢 (B) 0
🟠 (C) 2
🔴 (D) 5
Answer: (A) −1

Q15. The value of [a̅ b̅ c̅] equals
🔵 (A) a̅ ⋅ (b̅ × c̅)
🟢 (B) |a̅ × b̅| |c̅|
🟠 (C) a̅ × (b̅ ⋅ c̅)
🔴 (D) (a̅ ⋅ b̅) c̅
Answer: (A) a̅ ⋅ (b̅ × c̅)

Q16. If a̅ = î + ĵ and b̅ = ĵ + k̂, then a̅ × b̅ equals
🔵 (A) î − k̂
🟢 (B) î + k̂
🟠 (C) ĵ − k̂
🔴 (D) −î + k̂
Answer: (A) î − k̂

Q17. If a̅, b̅ are non-zero and a̅ × b̅ = a̅ × (2b̅), then
🔵 (A) a̅ × b̅ = 0̅
🟢 (B) a̅ ⋅ b̅ = 0
🟠 (C) |a̅| = |b̅|
🔴 (D) a̅ = 2b̅
Answer: (A) a̅ × b̅ = 0̅

Q18. If a̅ = 3î + 4ĵ, the component of a̅ along î is
🔵 (A) 3
🟢 (B) 4
🟠 (C) 5
🔴 (D) 7
Answer: (A) 3

Q19. If |a̅| = |b̅| and a̅ ⋅ b̅ = 0, then |a̅ + b̅| equals
🔵 (A) √2 |a̅|
🟢 (B) |a̅|
🟠 (C) 2|a̅|
🔴 (D) 0
Answer: (A) √2 |a̅|

Q20. If û is a unit vector making angles α, β, γ with x-, y-, z-axes, then
🔵 (A) cos²α + cos²β + cos²γ = 1
🟢 (B) cosα + cosβ + cosγ = 1
🟠 (C) cosα cosβ cosγ = 1
🔴 (D) cos²α + cos²β − cos²γ = 1
Answer: (A) cos²α + cos²β + cos²γ = 1

Q21. If a̅ = (1, 2, −1) and b̅ = (2, 0, 3), then a̅ × b̅ equals
🔵 (A) (6, −5, −4)
🟢 (B) (6, −5, −4) with opposite sign
🟠 (C) (6, −5, 4)
🔴 (D) (−6, 5, 4)
Answer: (A) (6, −5, −4)

Q22. Let a̅ = (x, 1, 2) and b̅ = (1, x, 2). If a̅ ⟂ b̅, then x equals
🔵 (A) 1
🟢 (B) −1
🟠 (C) 0
🔴 (D) 2
Answer: (B) −1

Q23. If a̅ = (2, −1, 2) and b̅ = (1, 2, −1), then the angle θ between them satisfies
🔵 (A) cosθ = 0
🟢 (B) cosθ = (2·1 + (−1)·2 + 2·(−1)) / (|a̅||b̅|)
🟠 (C) cosθ = −1
🔴 (D) cosθ = 1
Answer: (B) cosθ = (2·1 + (−1)·2 + 2·(−1)) / (|a̅||b̅|)

Q24. If a̅ = (1, 2, 2), b̅ = (2, −1, 2), then |a̅ × b̅| equals
🔵 (A) √33
🟢 (B) √29
🟠 (C) √13
🔴 (D) √21
Answer: (A) √33

Q25. The scalar projection of a̅ on b̅ equals 5 and |b̅| = 2. Then a̅ ⋅ b̅ equals
🔵 (A) 10
🟢 (B) 2.5
🟠 (C) 5/2
🔴 (D) 7
Answer: (A) 10

Q26. If a̅ ⋅ b̅ = 3, a̅ ⋅ c̅ = 4, b̅ ⋅ c̅ = 5 and |a̅| = |b̅| = |c̅| = 3, then |a̅ + b̅ + c̅|² equals
🔵 (A) 27 + 2(3 + 4 + 5)
🟢 (B) 9 + 9 + 9
🟠 (C) 27 − 2(12)
🔴 (D) 27
Answer: (A) 27 + 2(3 + 4 + 5)

Q27. If a̅, b̅, c̅ are coplanar, then
🔵 (A) [a̅ b̅ c̅] = 0
🟢 (B) a̅ ⋅ b̅ = 0
🟠 (C) a̅ × b̅ = c̅
🔴 (D) |a̅| + |b̅| = |c̅|
Answer: (A) [a̅ b̅ c̅] = 0

Q28. If û and v̂ are unit vectors and |û + v̂| = √3, then û ⋅ v̂ equals
🔵 (A) 1/2
🟢 (B) √3/2
🟠 (C) −1/2
🔴 (D) 0
Answer: (A) 1/2

Q29. For vectors a̅, b̅, c̅, the identity a̅ × (b̅ × c̅) equals
🔵 (A) (a̅ ⋅ c̅)b̅ − (a̅ ⋅ b̅)c̅
🟢 (B) (b̅ ⋅ c̅)a̅
🟠 (C) a̅ ⋅ (b̅ × c̅)
🔴 (D) (a̅ × b̅) ⋅ c̅
Answer: (A) (a̅ ⋅ c̅)b̅ − (a̅ ⋅ b̅)c̅

Q30. The area of triangle formed by position vectors a̅ and b̅ is
🔵 (A) (1/2)|a̅ × b̅|
🟢 (B) |a̅ × b̅|
🟠 (C) (1/2)|a̅ ⋅ b̅|
🔴 (D) |a̅ ⋅ b̅|
Answer: (A) (1/2)|a̅ × b̅|

Q31. If a̅ = (t, 1, −1) and b̅ = (1, t, −1) are perpendicular, then t equals
🔵 (A) 1
🟢 (B) −1
🟠 (C) 0
🔴 (D) 2
Answer: (A) 1

Q32. If a̅ × b̅ = c̅ and a̅ ⋅ c̅ = 0, then
🔵 (A) a̅ ⟂ c̅
🟢 (B) b̅ ∥ c̅
🟠 (C) a̅ ∥ c̅
🔴 (D) a̅ ⋅ b̅ = 0
Answer: (A) a̅ ⟂ c̅

Q33. If |a̅| = |b̅| = 1 and a̅ ⋅ b̅ = cosθ, then |a̅ − b̅| equals
🔵 (A) √(2 − 2cosθ)
🟢 (B) √(2 + 2cosθ)
🟠 (C) 2cos(θ/2)
🔴 (D) 2
Answer: (A) √(2 − 2cosθ)

Q34. The vector component of a̅ along b̅ is
🔵 (A) [(a̅ ⋅ b̅)/|b̅|²] b̅
🟢 (B) [(a̅ ⋅ b̅)/|a̅|²] a̅
🟠 (C) [(a̅ × b̅)/|b̅|²] b̅
🔴 (D) [(a̅ ⋅ a̅)/|b̅|²] b̅
Answer: (A) [(a̅ ⋅ b̅)/|b̅|²] b̅

Q35. If a̅ + b̅ is perpendicular to a̅ − b̅, then a̅ ⋅ a̅ equals
🔵 (A) b̅ ⋅ b̅
🟢 (B) 0
🟠 (C) 1
🔴 (D) |a̅||b̅|
Answer: (A) b̅ ⋅ b̅

Q36. If a̅, b̅ are non-collinear and a̅ × x̅ = b̅ × x̅, then x̅ is
🔵 (A) Parallel to a̅ − b̅
🟢 (B) Parallel to a̅ + b̅
🟠 (C) Zero vector only
🔴 (D) Parallel to a̅
Answer: (A) Parallel to a̅ − b̅

Q37. If [a̅ b̅ c̅] = 6 and d̅ = 2a̅ − b̅, then [d̅ b̅ c̅] equals
🔵 (A) 12
🟢 (B) 6
🟠 (C) 0
🔴 (D) −6
Answer: (A) 12

Q38. If a̅, b̅ are orthonormal (|a̅|=|b̅|=1, a̅ ⋅ b̅ = 0), then |2a̅ − 3b̅| equals
🔵 (A) √13
🟢 (B) √5
🟠 (C) 1
🔴 (D) √7
Answer: (A) √13

Q39. If a̅ × b̅ = c̅ and b̅ × c̅ = a̅, then |a̅|, |b̅|, |c̅| satisfy
🔵 (A) |a̅| = |b̅| = |c̅|
🟢 (B) |c̅| = |a̅||b̅|
🟠 (C) |a̅||b̅||c̅| = 1
🔴 (D) |a̅| = |b̅||c̅|
Answer: (A) |a̅| = |b̅| = |c̅|

Q40. If a̅ × (b̅ × a̅) equals
🔵 (A) (a̅ ⋅ a̅)b̅ − (a̅ ⋅ b̅)a̅
🟢 (B) (a̅ ⋅ b̅)b̅ − (a̅ ⋅ a̅)a̅
🟠 (C) (b̅ ⋅ a̅)a̅ − (a̅ ⋅ a̅)b̅
🔴 (D) (a̅ ⋅ b̅)c̅
Answer: (A) (a̅ ⋅ a̅)b̅ − (a̅ ⋅ b̅)a̅

Q41. If a̅, b̅, c̅ are non-coplanar and (a̅ × b̅) ⋅ c̅ = 0, then
🔵 (A) Contradiction; must be coplanar
🟢 (B) |a̅| = 0
🟠 (C) |b̅| = 0
🔴 (D) |c̅| = 0
Answer: (A) Contradiction; must be coplanar

Q42. If a̅, b̅ are non-zero and |a̅ + b̅| = |a̅ − b̅|, then a̅ ⋅ b̅ equals
🔵 (A) 0
🟢 (B) |a̅||b̅|
🟠 (C) −|a̅||b̅|
🔴 (D) 1
Answer: (A) 0

Q43. Let û, v̂ be unit vectors and θ the angle between them. Then |û × v̂| equals
🔵 (A) sinθ
🟢 (B) cosθ
🟠 (C) tanθ
🔴 (D) 1
Answer: (A) sinθ

Q44. If a̅ × b̅ = c̅ and a̅ ⋅ b̅ = 0, then (a̅, b̅, c̅) form
🔵 (A) A right-handed orthogonal set (up to scale)
🟢 (B) A collinear set
🟠 (C) A coplanar but not orthogonal set
🔴 (D) Equal vectors
Answer: (A) A right-handed orthogonal set (up to scale)

Q45. If a̅ = (1, 0, 1), b̅ = (0, 1, 1), c̅ = (1, 1, 0), then [a̅ b̅ c̅] equals
🔵 (A) 1
🟢 (B) −1
🟠 (C) 2
🔴 (D) 0
Answer: (A) 1

Q46. If vectors a̅, b̅ satisfy |a̅| = |b̅| = 2 and |a̅ − b̅| = 2, then the angle between them equals
🔵 (A) 60°
🟢 (B) 120°
🟠 (C) 90°
🔴 (D) 0°
Answer: (A) 60°

Q47. Suppose a̅, b̅, c̅ are such that a̅ × b̅ = b̅ × c̅ = c̅ × a̅ ≠ 0̅. Then a̅ + b̅ + c̅ equals
🔵 (A) 0̅
🟢 (B) a̅
🟠 (C) b̅
🔴 (D) c̅
Answer: (A) 0̅

Q48. If a̅, b̅, c̅ are non-zero with a̅ × b̅ = c̅ and a̅ ⋅ c̅ = b̅ ⋅ c̅ = 0, then
🔵 (A) a̅, b̅, c̅ are mutually perpendicular (up to scale)
🟢 (B) a̅ ∥ b̅
🟠 (C) c̅ ∥ a̅
🔴 (D) c̅ ∥ b̅
Answer: (A) a̅, b̅, c̅ are mutually perpendicular (up to scale)

Q49. If |a̅| = |b̅| = |c̅| = 1 and [a̅ b̅ c̅] = 1, then the volume of the parallelepiped formed by 2a̅, 3b̅, 4c̅ equals
🔵 (A) 24
🟢 (B) 12
🟠 (C) 6
🔴 (D) 1
Answer: (A) 24

Q50. For non-zero a̅, b̅, c̅, the identity |a̅ × b̅|² + (a̅ ⋅ b̅)² equals
🔵 (A) |a̅|²|b̅|²
🟢 (B) |a̅|² + |b̅|²
🟠 (C) |a̅ + b̅|²
🔴 (D) |a̅ − b̅|²
Answer: (A) |a̅|²|b̅|²

Important Notice : The lessons and study materials up to July–August 2025 have been sent. Soon, after receiving feedback from teachers and students, we will deliver the latest and more innovative study materials for the upcoming months.