Part A – Multiple Choice Questions
15 Qs1
The magnitude of the position vector of the point P(3, −4, 5) is:
✅ Solution
Answer: (C) 5√2
|OP| = √(x² + y² + z²) = √(9 + 16 + 25) = √50 = 5√2
2
The unit vector in the direction of a⃗ = 2î + 3ĵ + k̂ is:
✅ Solution
Answer: (A) (2î + 3ĵ + k̂)/√14
|a⃗| = √(4 + 9 + 1) = √14
â = a⃗/|a⃗| = (2î + 3ĵ + k̂)/√14
â = a⃗/|a⃗| = (2î + 3ĵ + k̂)/√14
3
If a⃗ = î + 2ĵ − k̂ and b⃗ = 2î + ĵ + k̂, then a⃗ · b⃗ equals:
✅ Solution
Answer: (B) 3
a⃗ · b⃗ = (1)(2) + (2)(1) + (−1)(1) = 2 + 2 − 1 = 3
4
The angle between vectors a⃗ and b⃗ where |a⃗| = 1, |b⃗| = 2 and a⃗ · b⃗ = 1 is:
✅ Solution
Answer: (C) π/3
cos θ = a⃗·b⃗/(|a⃗||b⃗|) = 1/(1×2) = 1/2
∴ θ = cos⁻¹(1/2) = π/3
∴ θ = cos⁻¹(1/2) = π/3
5
The direction cosines of the vector a⃗ = î + ĵ − 2k̂ are:
✅ Solution
Answer: (A) 1/√6, 1/√6, −2/√6
|a⃗| = √(1 + 1 + 4) = √6
Direction cosines: l = 1/√6, m = 1/√6, n = −2/√6
Note: l² + m² + n² = 1/6 + 1/6 + 4/6 = 1 ✓
Direction cosines: l = 1/√6, m = 1/√6, n = −2/√6
Note: l² + m² + n² = 1/6 + 1/6 + 4/6 = 1 ✓
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6
The value of î·(ĵ × k̂) + ĵ·(î × k̂) + k̂·(î × ĵ) is:
✅ Solution
Answer: (C) 1
î·(ĵ × k̂) = î·î = 1
ĵ·(î × k̂) = ĵ·(−ĵ) = −1
k̂·(î × ĵ) = k̂·k̂ = 1
Total = 1 − 1 + 1 = 1
ĵ·(î × k̂) = ĵ·(−ĵ) = −1
k̂·(î × ĵ) = k̂·k̂ = 1
Total = 1 − 1 + 1 = 1
7
If a⃗ × b⃗ = 0⃗ and a⃗ · b⃗ = 0, then which of the following is true?
✅ Solution
Answer: (C) At least one of a⃗, b⃗ is zero vector
a⃗ × b⃗ = 0⃗ implies a⃗ ∥ b⃗ (or one is zero).
a⃗ · b⃗ = 0 implies a⃗ ⊥ b⃗ (or one is zero).
A non-zero vector cannot simultaneously be both parallel and perpendicular to another non-zero vector.
∴ At least one of a⃗, b⃗ must be the zero vector.
a⃗ · b⃗ = 0 implies a⃗ ⊥ b⃗ (or one is zero).
A non-zero vector cannot simultaneously be both parallel and perpendicular to another non-zero vector.
∴ At least one of a⃗, b⃗ must be the zero vector.
8
The area of the triangle with adjacent sides a⃗ = î − ĵ + k̂ and b⃗ = 2î + ĵ − k̂ is:
✅ Solution
Answer: (D) √6/2 sq. units
a⃗ × b⃗ = |î ĵ k̂; 1 −1 1; 2 1 −1|
= î(1−1) − ĵ(−1−2) + k̂(1+2) = 0î + 3ĵ + 3k̂
|a⃗ × b⃗| = √(0 + 9 + 9) = √18 = 3√2
Area of triangle = (1/2)|a⃗ × b⃗| = 3√2/2 = √6/2... wait, 3√2/2 = 3√2/2. Let me recalculate: √(0+9+9) = √18 = 3√2. Area = 3√2/2.
= î(1−1) − ĵ(−1−2) + k̂(1+2) = 0î + 3ĵ + 3k̂
|a⃗ × b⃗| = √(0 + 9 + 9) = √18 = 3√2
Area of triangle = (1/2)|a⃗ × b⃗| = 3√2/2 = √6/2... wait, 3√2/2 = 3√2/2. Let me recalculate: √(0+9+9) = √18 = 3√2. Area = 3√2/2.
9
The projection of vector a⃗ = 2î + 3ĵ + 2k̂ on the vector b⃗ = î + 2ĵ + k̂ is:
✅ Solution
Answer: (D) 5√6/3
Projection = (a⃗·b⃗)/|b⃗|
a⃗·b⃗ = (2)(1) + (3)(2) + (2)(1) = 2 + 6 + 2 = 10
|b⃗| = √(1 + 4 + 1) = √6
Projection = 10/√6 = 10√6/6 = 5√6/3
a⃗·b⃗ = (2)(1) + (3)(2) + (2)(1) = 2 + 6 + 2 = 10
|b⃗| = √(1 + 4 + 1) = √6
Projection = 10/√6 = 10√6/6 = 5√6/3
10
Find |a⃗ − b⃗|, if |a⃗| = 2, |b⃗| = 3 and a⃗·b⃗ = 4:
✅ Solution
Answer: (A) √5
|a⃗ − b⃗|² = |a⃗|² − 2(a⃗·b⃗) + |b⃗|²
= 4 − 2(4) + 9 = 4 − 8 + 9 = 5
∴ |a⃗ − b⃗| = √5
= 4 − 2(4) + 9 = 4 − 8 + 9 = 5
∴ |a⃗ − b⃗| = √5
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11
The position vector of the midpoint of the segment joining P(2, 3, 4) and Q(4, 1, −2) is:
✅ Solution
Answer: (A) 3î + 2ĵ + k̂
Midpoint R = (a⃗ + b⃗)/2
= [(2+4)/2 î + (3+1)/2 ĵ + (4−2)/2 k̂]
= 3î + 2ĵ + k̂
= [(2+4)/2 î + (3+1)/2 ĵ + (4−2)/2 k̂]
= 3î + 2ĵ + k̂
12
If a⃗ is a nonzero vector of magnitude a and λ a nonzero scalar, then λa⃗ is a unit vector if:
✅ Solution
Answer: (D) a = 1/|λ|
|λa⃗| = |λ| · |a⃗| = |λ| · a = 1 (for unit vector)
∴ a = 1/|λ|
∴ a = 1/|λ|
13
If |a⃗ + b⃗| = |a⃗ − b⃗|, then the angle between a⃗ and b⃗ is:
✅ Solution
Answer: (C) π/2
|a⃗+b⃗|² = |a⃗|² + 2a⃗·b⃗ + |b⃗|²
|a⃗−b⃗|² = |a⃗|² − 2a⃗·b⃗ + |b⃗|²
Setting equal: 4a⃗·b⃗ = 0 → a⃗·b⃗ = 0
∴ θ = π/2
|a⃗−b⃗|² = |a⃗|² − 2a⃗·b⃗ + |b⃗|²
Setting equal: 4a⃗·b⃗ = 0 → a⃗·b⃗ = 0
∴ θ = π/2
14
The area of the parallelogram whose adjacent sides are a⃗ = 3î + ĵ + 4k̂ and b⃗ = î − ĵ + k̂ is:
✅ Solution
Answer: (A) √42 sq. units
a⃗ × b⃗ = |î ĵ k̂; 3 1 4; 1 −1 1|
= î(1+4) − ĵ(3−4) + k̂(−3−1)
= 5î + ĵ − 4k̂
|a⃗ × b⃗| = √(25 + 1 + 16) = √42 sq. units
= î(1+4) − ĵ(3−4) + k̂(−3−1)
= 5î + ĵ − 4k̂
|a⃗ × b⃗| = √(25 + 1 + 16) = √42 sq. units
15
The point R divides segment PQ internally in the ratio 2:1 where OP⃗ = 3a⃗ − 2b⃗ and OQ⃗ = a⃗ + b⃗. Then OR⃗ is:
✅ Solution
Answer: (A) 5a⃗/3
OR⃗ = (m·OQ⃗ + n·OP⃗)/(m+n) = (2(a⃗+b⃗) + 1(3a⃗−2b⃗))/(2+1)
= (2a⃗ + 2b⃗ + 3a⃗ − 2b⃗)/3 = 5a⃗/3
∴ OR⃗ = 5a⃗/3
= (2a⃗ + 2b⃗ + 3a⃗ − 2b⃗)/3 = 5a⃗/3
∴ OR⃗ = 5a⃗/3
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Part B – One-Word / One-Value Answer
10 Qs16
Find the magnitude of a⃗ = 2î − 7ĵ − 3k̂.
✅ Solution
Answer: √62
|a⃗| = √(4 + 49 + 9) = √62
17
Find the scalar product a⃗·b⃗ if a⃗ = î + ĵ − k̂ and b⃗ = î − ĵ + k̂.
✅ Solution
Answer: −1
a⃗·b⃗ = (1)(1) + (1)(−1) + (−1)(1) = 1 − 1 − 1 = −1
18
Find the vector joining P(2, 3, 0) to Q(−1, −2, −4). What is the magnitude of PQ⃗?
✅ Solution
Answer: 5√2
PQ⃗ = (−1−2)î + (−2−3)ĵ + (−4−0)k̂ = −3î − 5ĵ − 4k̂
|PQ⃗| = √(9 + 25 + 16) = √50 = 5√2
|PQ⃗| = √(9 + 25 + 16) = √50 = 5√2
19
If a⃗ is a unit vector and (x⃗ − a⃗)·(x⃗ + a⃗) = 8, find |x⃗|.
✅ Solution
Answer: 3
(x⃗−a⃗)·(x⃗+a⃗) = x⃗·x⃗ − a⃗·a⃗ = |x⃗|² − 1 = 8
|x⃗|² = 9 → |x⃗| = 3
|x⃗|² = 9 → |x⃗| = 3
20
Find |a⃗ × b⃗| if a⃗ = 2î + ĵ + 3k̂ and b⃗ = 3î + 5ĵ − 2k̂.
✅ Solution
Answer: √507
a⃗ × b⃗ = |î ĵ k̂; 2 1 3; 3 5 −2|
= î(−2−15) − ĵ(−4−9) + k̂(10−3)
= −17î + 13ĵ + 7k̂
|a⃗ × b⃗| = √(289 + 169 + 49) = √507
= î(−2−15) − ĵ(−4−9) + k̂(10−3)
= −17î + 13ĵ + 7k̂
|a⃗ × b⃗| = √(289 + 169 + 49) = √507
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21
If a⃗, b⃗, c⃗ are unit vectors such that a⃗ + b⃗ + c⃗ = 0⃗, find the value of a⃗·b⃗ + b⃗·c⃗ + c⃗·a⃗.
✅ Solution
Answer: −3/2
|a⃗ + b⃗ + c⃗|² = 0
|a⃗|² + |b⃗|² + |c⃗|² + 2(a⃗·b⃗ + b⃗·c⃗ + c⃗·a⃗) = 0
1 + 1 + 1 + 2µ = 0
µ = −3/2
|a⃗|² + |b⃗|² + |c⃗|² + 2(a⃗·b⃗ + b⃗·c⃗ + c⃗·a⃗) = 0
1 + 1 + 1 + 2µ = 0
µ = −3/2
22
Find the angle in degrees that the vector a⃗ = î + ĵ + k̂ makes with the x-axis.
✅ Solution
Answer: cos⁻¹(1/√3)
|a⃗| = √3. Direction cosine with x-axis: l = 1/√3
α = cos⁻¹(1/√3) ≈ 54.74°
(The vector is equally inclined to all three axes since l = m = n = 1/√3)
α = cos⁻¹(1/√3) ≈ 54.74°
(The vector is equally inclined to all three axes since l = m = n = 1/√3)
23
Find the value of λ if (2î + 6ĵ + 27k̂) × (î + λĵ + µk̂) = 0⃗. (Give the value of λ)
✅ Solution
Answer: λ = 3
For cross product to be zero, vectors must be parallel → b⃗ = k·a⃗
b₁/a₁ = b₂/a₂ = b₃/a₃
1/2 = λ/6 = µ/27
λ = 6/2 = 3, µ = 27/2
b₁/a₁ = b₂/a₂ = b₃/a₃
1/2 = λ/6 = µ/27
λ = 6/2 = 3, µ = 27/2
24
Three vectors a⃗, b⃗, c⃗ satisfy a⃗ + b⃗ + c⃗ = 0⃗ with |a⃗| = 3, |b⃗| = 4, |c⃗| = 2. Find |a⃗ + b⃗ + c⃗| (given µ = a⃗·b⃗ + b⃗·c⃗ + c⃗·a⃗). Find the value of µ.
✅ Solution
Answer: µ = −29/2
a⃗ + b⃗ + c⃗ = 0⃗ → |a⃗ + b⃗ + c⃗|² = 0
|a⃗|² + |b⃗|² + |c⃗|² + 2µ = 0
9 + 16 + 4 + 2µ = 0
2µ = −29 → µ = −29/2
|a⃗|² + |b⃗|² + |c⃗|² + 2µ = 0
9 + 16 + 4 + 2µ = 0
2µ = −29 → µ = −29/2
25
Find the area of the triangle having vertices A(1,1,1), B(1,2,3) and C(2,3,1). Give the answer in simplest radical form.
✅ Solution
Answer: √21/2 sq. units
AB⃗ = (0, 1, 2), AC⃗ = (1, 2, 0)
AB⃗ × AC⃗ = |î ĵ k̂; 0 1 2; 1 2 0|
= î(0−4) − ĵ(0−2) + k̂(0−1)
= −4î + 2ĵ − k̂
|AB⃗ × AC⃗| = √(16 + 4 + 1) = √21
Area = (1/2)√21 = √21/2 sq. units
AB⃗ × AC⃗ = |î ĵ k̂; 0 1 2; 1 2 0|
= î(0−4) − ĵ(0−2) + k̂(0−1)
= −4î + 2ĵ − k̂
|AB⃗ × AC⃗| = √(16 + 4 + 1) = √21
Area = (1/2)√21 = √21/2 sq. units
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