Revision notes, worked problems, a question bank, assertion and reason questions and answers to the NCERT exercises. Tap a group to open it; opening one closes the others.
A scalar has only magnitude (mass, time, length, speed, work). A vector has magnitude and direction (displacement, velocity, force, acceleration).
| Type | Meaning |
|---|---|
| Zero vector 0 | Magnitude 0, no definite direction. Initial and terminal points coincide. |
| Unit vector â | Magnitude 1. â = a/|a| is the unit vector along a. |
| Coinitial | Vectors with the same initial point. |
| Collinear | Parallel to the same line, whatever their magnitudes or directions. |
| Equal | Same magnitude and same direction. |
| Negative of a | −a has the same magnitude as a and opposite direction. |
| Free vector | Only magnitude and direction matter, not the position. |
Two vectors are equal only if both magnitude and direction agree. Collinear vectors may differ in magnitude.
Triangle law: if AB = a and BC = b, then AC = a + b.
Parallelogram law: if a and b are the adjacent sides OA and OB, the diagonal OC is a + b.
For a scalar k, ka has magnitude |k||a| and is parallel to a. It points the same way if k > 0 and the opposite way if k < 0.
î, ĵ, k̂ are the unit vectors along the x, y and z axes. A vector a = a₁î + a₂ĵ + a₃k̂ has components a₁, a₂, a₃.
For P(x₁, y₁, z₁) and Q(x₂, y₂, z₂):
If a vector makes angles α, β, γ with the axes, then l = cos α, m = cos β, n = cos γ are its direction cosines. For a = a₁î + a₂ĵ + a₃k̂:
For A and B with position vectors a and b, the point R dividing AB in the ratio m : n has position vector:
The scalar product of nonzero vectors a and b with angle θ (0° ≤ θ ≤ 180°) is a · b = |a||b| cos θ. The result is a scalar.
The projection of a on b is (a · b)/|b| = |a| cos θ (a scalar). The projection vector is ((a · b)/|b|²) b.
For a and b with angle θ, a × b = |a||b| sin θ n̂, where n̂ is the unit vector perpendicular to both such that a, b, n̂ form a right-handed system. The result is a vector.
| Want | Use |
|---|---|
| Length of a | √(a₁² + a₂² + a₃²) |
| Angle between a and b | cos θ = (a · b)/(|a||b|) |
| Perpendicular? | a · b = 0 |
| Parallel? | a × b = 0 or proportional components |
| Projection of a on b | (a · b)/|b| |
| Area of triangle | ½|AB × AC| |
| Point dividing AB | (m b + n a)/(m + n) internal |
Try each problem first, then open the solution to check your working.
Find the magnitude and the direction cosines of a = 2î + 3ĵ + 6k̂.
|a| = √(4 + 9 + 36) = 7.
Direction cosines: l = 2/7, m = 3/7, n = 6/7. (Check: l² + m² + n² = 1.)
Answer: |a| = 7; direction cosines 2/7, 3/7, 6/7
Find the unit vector in the direction of b = 3î − 4ĵ + 12k̂.
|b| = √(9 + 16 + 144) = 13.
Divide b by |b|.
Answer: (1/13)(3î − 4ĵ + 12k̂)
Find the vector of magnitude 21 in the direction of 2î + 3ĵ + 6k̂.
|2î + 3ĵ + 6k̂| = 7, so the unit vector is (1/7)(2î + 3ĵ + 6k̂).
Multiply by 21: 21/7 = 3, giving 3(2î + 3ĵ + 6k̂).
Answer: 6î + 9ĵ + 18k̂
Find the values of λ for which λ(î + ĵ + k̂) is a unit vector.
|λ(î + ĵ + k̂)| = |λ|√3 = 1, so |λ| = 1/√3.
Answer: λ = ±1/√3
Find the vector PQ and its magnitude for P(2, −1, 3) and Q(5, 3, −1).
PQ = position vector of Q − position vector of P.
PQ = 3î + 4ĵ − 4k̂.
|PQ| = √(9 + 16 + 16) = √41.
Answer: PQ = 3î + 4ĵ − 4k̂, |PQ| = √41
If a = 2î − ĵ + k̂ and b = î + 3ĵ − 2k̂, find 2a − 3b and its magnitude.
2a = 4î − 2ĵ + 2k̂ and 3b = 3î + 9ĵ − 6k̂.
2a − 3b = î − 11ĵ + 8k̂.
|2a − 3b| = √(1 + 121 + 64) = √186.
Answer: 2a − 3b = î − 11ĵ + 8k̂, magnitude √186
Find the position vector of the point dividing the line joining A(1, −2, 3) and B(5, 4, −3) (i) internally in the ratio 1 : 2 and (ii) externally in the ratio 2 : 1.
Internal, m : n = 1 : 2: R = (m·OB + n·OA)/(m + n) = (OB + 2·OA)/3.
R = (7/3)î + k̂.
External, m : n = 2 : 1: R = (m·OB − n·OA)/(m − n) = 2·OB − OA.
R = 9î + 10ĵ − 9k̂.
Answer: (i) (7/3)î + k̂; (ii) 9î + 10ĵ − 9k̂
Show that A(1, 2, 3), B(3, 5, 7) and C(7, 11, 15) are collinear and find the ratio in which B divides AC.
AB = 2î + 3ĵ + 4k̂ and AC = 6î + 9ĵ + 12k̂ = 3(AB).
AC is a scalar multiple of AB, so A, B, C are collinear.
AB = ⅓AC, so AB : BC = 1 : 2.
Answer: collinear; B divides AC internally in the ratio 1 : 2
A(1, 0, 2), B(3, 1, 4) and C(4, 3, 3) are three vertices of a parallelogram ABCD. Find D.
In a parallelogram AC and BD bisect each other, so OA + OC = OB + OD.
OD = OA + OC − OB = 2î + 2ĵ + k̂.
Check: AB = 2î + ĵ + 2k̂ = DC. ✓
Answer: D(2, 2, 1)
Find the angle between a = î + 2ĵ + 3k̂ and b = 3î − ĵ + 2k̂.
a · b = 7, |a| = √14 and |b| = √14.
cos θ = 7/(√14 · √14) = 1/2, so θ = 60°.
Answer: 60° (π/3)
Find the projection of a = 3î − ĵ + 2k̂ on b = î + 2ĵ + 2k̂, and the projection vector.
a · b = 5 and |b| = 3.
Projection = (a · b)/|b| = 5/3.
Projection vector = ((a · b)/|b|²) b = (5/9)(î + 2ĵ + 2k̂) = (5/9)î + (10/9)ĵ + (10/9)k̂.
Answer: 5/3; vector (5/9)î + (10/9)ĵ + (10/9)k̂
Find λ for which a = 2î + λĵ + k̂ and b = î − 2ĵ + 3k̂ are perpendicular.
a · b = 2 − 2λ + 3 = 5 − 2λ = 0.
Answer: λ = 5/2
A force F = 3î − 2ĵ + 5k̂ moves a particle from P(1, 1, 1) to Q(4, 3, −2). Find the work done.
Displacement d = PQ = 3î + 2ĵ − 3k̂.
W = F · d = 9 + -4 + -15 = −10.
Answer: W = −10 units (the force opposes the motion overall)
For a = î + 2ĵ − k̂ and b = 2î + 3k̂, verify that |a + b|² + |a − b|² = 2(|a|² + |b|²).
a + b = 3î + 2ĵ + 2k̂, so |a + b|² = 17. a − b = −î + 2ĵ − 4k̂, so |a − b|² = 21.
Left side = 38. |a|² = 6 and |b|² = 13, so the right side = 2(19) = 38. ✓
Answer: both sides equal 38
Find a × b for a = 2î + ĵ − k̂ and b = î + 3ĵ + 2k̂, and a unit vector perpendicular to both.
a × b = 5î − 5ĵ + 5k̂. Its magnitude is 5√3.
Unit vector: ±(a × b)/|a × b| = (1/1)((√3/3)î − (√3/3)ĵ + (√3/3)k̂).
Answer: a × b = 5î − 5ĵ + 5k̂; unit vector ±(1/√3)(î − ĵ + k̂)
Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).
AB = î + 2ĵ + 3k̂ and AC = 4ĵ + 3k̂.
AB × AC = −6î − 3ĵ + 4k̂, with magnitude √61.
Area = ½|AB × AC| = √61/2.
Answer: √61/2 square units
Find the area of the parallelogram whose adjacent sides are a = î − ĵ + 3k̂ and b = 2î − 7ĵ + k̂.
a × b = 20î + 5ĵ − 5k̂.
|a × b| = √(400 + 25 + 25) = √450 = 15√2.
Answer: 15√2 square units
For a = î + 2ĵ + 3k̂ and b = 3î − ĵ + 2k̂, find a × b and verify that it is perpendicular to both.
a × b = 7î + 7ĵ − 7k̂.
(a × b) · a = 7 + 14 + -21 = 0 and (a × b) · b = 21 + -7 + -14 = 0. ✓
Answer: a × b = 7î + 7ĵ − 7k̂
Find a vector of magnitude 6 perpendicular to both î + ĵ + k̂ and î + 2ĵ + 3k̂.
a × b = î − 2ĵ + k̂, with magnitude √6.
The vector is ±(6/√6)(î − 2ĵ + k̂) = ±√6(î − 2ĵ + k̂).
Answer: ±√6(î − 2ĵ + k̂)
If |a| = 2, |b| = 5 and a · b = 6, find |a × b|.
|a × b|² = (2)²(5)² − 6² = 100 − 36 = 64.
Answer: |a × b| = 8
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Take the origin as the starting point, with north up and east to the right. Choose a scale, for example 1 cm = 10 km.
Draw an arrow 4 cm long, turned 30° from the north line towards the east. Its head marks the end of the displacement.
Answer: The arrow of length 40 km at 30° east of north (shown above).
A vector has both magnitude and direction. A scalar has magnitude only.
(i) 10 kg: scalar. (ii) 2 meters north-west: vector (a direction is given). (iii) 40°: scalar. (iv) 40 watt: scalar. (v) 10⁻¹⁹ coulomb: scalar. (vi) 20 m/s²: acceleration, a vector.
Answer: (ii) and (vi) are vectors; (i), (iii), (iv), (v) are scalars
(i) Time period: scalar. (ii) Distance: scalar (a length with no direction). (iii) Force: vector. (iv) Velocity: vector. (v) Work done: scalar (the dot product of force and displacement).
Answer: Vectors: force, velocity. Scalars: time period, distance, work done
(i) Coinitial vectors start at the same point. a and d both start at the top-left corner.
(ii) Equal vectors have the same magnitude and direction. b and d both point down and have the side length of the square.
(iii) Collinear vectors lie along parallel lines. a and c are parallel, with the same length but opposite directions, so they are collinear but not equal.
Answer: (i) a and d; (ii) b and d; (iii) a and c
(i) True: −a lies along the same line as a.
(ii) False: 2î and 5î are collinear, with magnitudes 2 and 5.
(iii) False: î and ĵ have equal magnitude 1 but are not collinear.
(iv) False: î and −î are collinear with equal magnitudes, but they are not equal (the directions are opposite).
Answer: (i) True; (ii) False; (iii) False; (iv) False
|a| = √(1² + 1² + 1²) = √3.
|b| = √(2² + (−7)² + (−3)²) = √(4 + 49 + 9) = √62.
|c| = √(1/3 + 1/3 + 1/3) = √1 = 1.
Answer: |a| = √3, |b| = √62, |c| = 1
Take a = î + 2ĵ and b = 2î + ĵ. Then |a| = √5 and |b| = √5.
Answer: a = î + 2ĵ and b = 2î + ĵ (any two with equal length work)
Take a = î + ĵ and b = 2î + 2ĵ = 2a. Since b is a positive multiple of a, they have the same direction but different magnitudes.
Answer: a = î + ĵ and b = 2î + 2ĵ
Equal vectors have equal components, so x = 2 and y = 3.
Answer: x = 2, y = 3
The vector is (−5 − 2)î + (7 − 1)ĵ = −7î + 6ĵ.
Answer: Scalar components −7 and 6; vector components −7î and 6ĵ
Add component by component: î: 1 − 2 + 1 = 0; ĵ: −2 + 4 − 6 = −4; k̂: 1 + 5 − 7 = −1.
Answer: −4ĵ − k̂
|a| = √(1 + 1 + 4) = √6.
â = a/|a| = (î + ĵ + 2k̂)/√6.
Answer: (1/√6)(î + ĵ + 2k̂)
PQ = (4 − 1)î + (5 − 2)ĵ + (6 − 3)k̂ = 3î + 3ĵ + 3k̂.
|PQ| = √27 = 3√3, so the unit vector is (3î + 3ĵ + 3k̂)/(3√3) = (î + ĵ + k̂)/√3.
Answer: (1/√3)(î + ĵ + k̂)
a + b = (2 − 1)î + (−1 + 1)ĵ + (2 − 1)k̂ = î + k̂.
|a + b| = √2, so the unit vector is (î + k̂)/√2.
Answer: (1/√2)(î + k̂)
|5î − ĵ + 2k̂| = √(25 + 1 + 4) = √30.
The required vector is 8 × (5î − ĵ + 2k̂)/√30 = (40/√30)î − (8/√30)ĵ + (16/√30)k̂.
Answer: (8/√30)(5î − ĵ + 2k̂)
−4î + 6ĵ − 8k̂ = −2(2î − 3ĵ + 4k̂). One vector is a scalar multiple of the other, so they are collinear.
Answer: Collinear: the second is −2 times the first
|v| = √(1 + 4 + 9) = √14.
Direction cosines are the components divided by the magnitude.
Answer: 1/√14, 2/√14, 3/√14
AB = (−1 − 1)î + (−2 − 2)ĵ + (1 + 3)k̂ = −2î − 4ĵ + 4k̂.
|AB| = √(4 + 16 + 16) = 6.
Answer: −1/3, −2/3, 2/3
The vector is v = î + ĵ + k̂, with |v| = √3. Its direction cosines are 1/√3, 1/√3, 1/√3.
All three are equal, so it makes the same angle cos⁻¹(1/√3) with each axis.
Answer: Shown: each direction cosine is 1/√3
Internal division in ratio m : n gives (m·OQ + n·OP)/(m + n). With m = 2, n = 1: (2(−î + ĵ + k̂) + (î + 2ĵ − k̂))/3 = −(1/3)î + (4/3)ĵ + (1/3)k̂.
External division gives (m·OQ − n·OP)/(m − n) = (2(−î + ĵ + k̂) − (î + 2ĵ − k̂))/1 = −3î + 3k̂.
Answer: (i) −(1/3)î + (4/3)ĵ + (1/3)k̂; (ii) −3î + 3k̂
Mid point = (OP + OQ)/2 = ((2 + 4)î + (3 + 1)ĵ + (4 − 2)k̂)/2 = 3î + 2ĵ + k̂.
Answer: 3î + 2ĵ + k̂
AB = b − a = −î + 3ĵ + 5k̂, BC = c − b = −î − 2ĵ − 6k̂, CA = a − c = 2î − ĵ + k̂.
|AB|² = 35, |BC|² = 41, |CA|² = 6.
Since |AB|² + |CA|² = 35 + 6 = 41 = |BC|², the triangle is right angled at A.
Answer: Right angled at A
By the triangle law AB + BC = AC, so AB + BC + CA = 0 and AB + BC − AC = 0 are true, and AB − CB + CA = AB + BC + CA = 0 is true.
AB + BC − CA = AC − CA = 2AC ≠ 0, so (C) is not true.
Answer: (C) AB + BC − CA = 0
(A) is correct: collinear vectors are scalar multiples of each other.
(B) is incorrect: 2î and 5î are collinear, but a ≠ ±b.
(C) is incorrect: the components of collinear vectors are proportional.
(D) is incorrect as a general statement: collinear vectors may also have opposite directions (b = −3a), and they may have the same magnitude.
Answer: (B), (C) and (D) are incorrect
cos θ = (a · b)/(|a||b|) = √6/(√3 × 2) = √2/2 = 1/√2.
So θ = π/4.
Answer: θ = π/4
a · b = 3 + 4 + 3 = 10. |a| = |b| = √14.
cos θ = 10/14 = 5/7.
Answer: θ = cos⁻¹(5/7)
Projection of a on b = (a · b)/|b| = (1 − 1)/√2 = 0.
Answer: 0
a · b = 7 − 3 + 56 = 60. |b| = √(49 + 1 + 64) = √114.
Projection = 60/√114.
Answer: 60/√114
|(2, 3, 6)| = √(4 + 9 + 36) = 7, |(3, −6, 2)| = √(9 + 36 + 4) = 7 and |(6, 2, −3)| = √(36 + 4 + 9) = 7, so each vector has magnitude 1.
Dot products: 2·3 + 3·(−6) + 6·2 = 6 − 18 + 12 = 0; 3·6 + (−6)·2 + 2·(−3) = 18 − 12 − 6 = 0; 6·2 + 2·3 + (−3)·6 = 12 + 6 − 18 = 0.
All three dot products are zero, so the vectors are mutually perpendicular.
Answer: Each has magnitude 1 and every pair has dot product 0
(a + b) · (a − b) = |a|² − |b|² = 8.
With |a| = 8|b|: 64|b|² − |b|² = 8, so |b|² = 8/63.
|b| = √(8/63) = 2√14/21 and |a| = 8|b| = 16√14/21.
Answer: |b| = 2√14/21, |a| = 16√14/21
Expand: 6 a·a + 21 a·b − 10 b·a − 35 b·b.
Since a·b = b·a, this is 6|a|² + 11 a·b − 35|b|².
Answer: 6|a|² + 11 a·b − 35|b|²
a · b = |a||b| cos 60° = |a|² × 1/2 = 1/2, so |a|² = 1.
Therefore |a| = |b| = 1.
Answer: |a| = |b| = 1
(x − a) · (x + a) = |x|² − |a|² = 12, and |a| = 1.
So |x|² = 13 and |x| = √13.
Answer: |x| = √13
a + λb = (2 − λ)î + (2 + 2λ)ĵ + (3 + λ)k̂.
(a + λb) · c = 3(2 − λ) + (2 + 2λ) = 8 − λ = 0, so λ = 8.
Answer: λ = 8
(|a|b + |b|a) · (|a|b − |b|a) = |a|²(b·b) − |a||b|(b·a) + |b||a|(a·b) − |b|²(a·a).
The middle terms cancel, and |a|²|b|² − |b|²|a|² = 0. The dot product is zero, so the vectors are perpendicular.
Answer: Shown: the dot product is 0
a · a = |a|² = 0 gives |a| = 0, so a is the zero vector.
Then a · b = 0 holds for every b. Nothing can be concluded about b: it can be any vector.
Answer: b can be any vector (a is the zero vector)
0 = |a + b + c|² = |a|² + |b|² + |c|² + 2(a·b + b·c + c·a) = 3 + 2S.
So S = −3/2.
Answer: −3/2
If a = 0 then |a| = 0, so a · b = |a||b| cos θ = 0. The same holds when b = 0.
Converse: take a = î and b = ĵ, both nonzero. a · b = 0 even though neither vector is zero (they are perpendicular).
Answer: Example: a = î, b = ĵ
BA = (2, 2, 3) and BC = (1, 1, 2).
BA · BC = 2 + 2 + 6 = 10, |BA| = √17, |BC| = √6.
cos∠ABC = 10/√102, so ∠ABC = cos⁻¹(10/√102) ≈ 8.05°.
Answer: ∠ABC = cos⁻¹(10/√102)
AB = (1, 4, −4), BC = (1, 4, −4) and AC = (2, 8, −8).
|AB| = |BC| = √33 and |AC| = 2√33 = |AB| + |BC|.
Since AC = AB + BC in length, the points lie on one line.
Answer: Collinear
Take A(2, −1, 1), B(1, −3, −5), C(3, −4, −4). AB = (−1, −2, −6), BC = (2, −1, 1), CA = (−1, 3, 5).
|AB|² = 41, |BC|² = 6, |CA|² = 35, and |BC|² + |CA|² = 41 = |AB|².
Check: BC · CA = −2 − 3 + 5 = 0, so the right angle is at C.
Answer: Right angled at C
|λa| = |λ||a| = |λ|a = 1 gives a = 1/|λ|.
Answer: (D) a = 1/|λ|
a × b = |î ĵ k̂; 1 −7 7; 3 −2 2| = î(−14 + 14) − ĵ(2 − 21) + k̂(−2 + 21) = 19ĵ + 19k̂.
|a × b| = √(19² + 19²) = 19√2.
Answer: 19√2
a + b = 4î + 4ĵ and a − b = 2î + 4k̂.
(a + b) × (a − b) = 16î − 16ĵ − 8k̂, which is perpendicular to both. Its magnitude is √(256 + 256 + 64) = 24.
Unit vector = (16î − 16ĵ − 8k̂)/24 = (2î − 2ĵ − k̂)/3.
Answer: ±(1/3)(2î − 2ĵ − k̂)
The direction cosines satisfy cos²α + cos²β + cos²γ = 1.
cos²(π/3) + cos²(π/4) + cos²θ = 1/4 + 1/2 + cos²θ = 1, so cos²θ = 1/4 and, for an acute θ, cos θ = 1/2: θ = π/3.
Components: cos(π/3) = 1/2, cos(π/4) = 1/√2, cos(π/3) = 1/2.
Answer: θ = π/3; a = (1/2)î + (1/√2)ĵ + (1/2)k̂
(a − b) × (a + b) = a×a + a×b − b×a − b×b.
a×a = b×b = 0 and b×a = −(a×b), so the result is a×b + a×b = 2(a×b).
Answer: Shown
The cross product is zero when the vectors are parallel: (2, 6, 27) = 2(1, 3, 27/2).
Components of the cross product: î(6μ − 27λ) − ĵ(2μ − 27) + k̂(2λ − 6) = 0.
So 2λ = 6 gives λ = 3, and 2μ = 27 gives μ = 27/2. (Check: 6μ − 27λ = 81 − 81 = 0.)
Answer: λ = 3, μ = 27/2
a · b = |a||b| cos θ = 0 and |a × b| = |a||b| sin θ = 0.
If both vectors were nonzero, cos θ = 0 and sin θ = 0 could not both hold.
So at least one of the vectors is the zero vector (a = 0 or b = 0).
Answer: a = 0 or b = 0
Write b + c = (b₁ + c₁)î + (b₂ + c₂)ĵ + (b₃ + c₃)k̂ and expand the determinant of a with b + c.
Each cross-product component is a bilinear expression, for example the î component is a₂(b₃ + c₃) − a₃(b₂ + c₂) = (a₂b₃ − a₃b₂) + (a₂c₃ − a₃c₂).
The same splitting holds for the ĵ and k̂ components, so a × (b + c) = a × b + a × c.
Answer: Shown (distributive law)
If a = 0 or b = 0 then |a × b| = |a||b| sin θ = 0, so a × b = 0.
The converse is false: take a = î and b = 2î. Then a × b = 2(î × î) = 0, although neither vector is zero (they are parallel).
Answer: The converse is not true
AB = (1, 2, 3), AC = (0, 4, 3).
AB × AC = î(6 − 12) − ĵ(3 − 0) + k̂(4 − 0) = −6î − 3ĵ + 4k̂, with magnitude √(36 + 9 + 16) = √61.
Area = (1/2)|AB × AC| = √61/2.
Answer: √61/2 square units
a × b = î(−1 + 21) − ĵ(1 − 6) + k̂(−7 + 2) = 20î + 5ĵ − 5k̂.
|a × b| = √(400 + 25 + 25) = √450 = 15√2.
Answer: 15√2 square units
|a × b| = |a||b| sin θ = 3 × (√2/3) × sin θ = √2 sin θ = 1, so sin θ = 1/√2 and θ = π/4.
Answer: (B) π/4
AB = 2î and AD = −ĵ (from A(−1, 1/2, 4) to D(−1, −1/2, 4)). AB × AD = 2î × (−ĵ) = −2k̂.
Area = |AB × AD| = 2.
Answer: (C) 2
A unit vector in the XY-plane making angle θ with the x-axis is cos θ î + sin θ ĵ.
For θ = 30°: (√3/2)î + (1/2)ĵ.
Answer: (√3/2)î + (1/2)ĵ
PQ = (x₂ − x₁)î + (y₂ − y₁)ĵ + (z₂ − z₁)k̂.
Answer: Scalar components x₂ − x₁, y₂ − y₁, z₂ − z₁; magnitude √((x₂ − x₁)² + (y₂ − y₁)² + (z₂ − z₁)²)
Take east as î and north as ĵ. The first leg is −4î. The second leg is 3(sin 30° î + cos 30° ĵ) = (3/2)î + (3√3/2)ĵ.
Resultant = (−4 + 3/2)î + (3√3/2)ĵ = −(5/2)î + (3√3/2)ĵ.
Magnitude = √(25/4 + 27/4) = √13 km. Direction: tan φ = (3√3/2)/(5/2) = 3√3/5, measured from west towards north.
Answer: √13 km, at tan⁻¹(3√3/5) north of west
Not always. By the triangle inequality, |b + c| ≤ |b| + |c|.
Counter-example: b = î, c = ĵ. Then a = î + ĵ and |a| = √2 ≠ 2 = |b| + |c|.
Equality holds only when b and c have the same direction.
Answer: Not true in general
|x(î + ĵ + k̂)| = |x|√3 = 1, so x = ±1/√3.
Answer: x = ±1/√3
a + b = 3î + ĵ, with magnitude √10.
Required vector = 5(3î + ĵ)/√10 = (3√10/2)î + (√10/2)ĵ.
Answer: (3√10/2)î + (√10/2)ĵ
2a − b + 3c = (2 − 2 + 3)î + (2 + 1 − 6)ĵ + (2 − 3 + 3)k̂ = 3î − 3ĵ + 2k̂.
Its magnitude is √(9 + 9 + 4) = √22.
Answer: (1/√22)(3î − 3ĵ + 2k̂)
AB = (4, 2, 6), BC = (6, 3, 9), AC = (10, 5, 15).
BC = (3/2)AB, so AB and BC are parallel and share the point B: the points are collinear.
|AB| : |BC| = 2 : 3, so B divides AC in the ratio 2 : 3 (internally).
Answer: Collinear; AB : BC = 2 : 3
External division in ratio m : n: R = (m·OQ − n·OP)/(m − n) with m = 1, n = 2.
R = ((a − 3b) − 2(2a + b))/(1 − 2) = (−3a − 5b)/(−1) = 3a + 5b.
Mid point of RQ = (R + Q)/2 = ((3a + 5b) + (a − 3b))/2 = (4a + 2b)/2 = 2a + b = OP, so P is the mid point of RQ.
Answer: OR = 3a + 5b
Diagonal = sum of the adjacent sides = 3î − 6ĵ + 2k̂, with magnitude √(9 + 36 + 4) = 7. The unit vector is (1/7)(3î − 6ĵ + 2k̂).
Area = |a × b|, where a × b = î(12 + 10) − ĵ(−6 − 5) + k̂(−4 + 4) = 22î + 11ĵ. So the area is √(484 + 121) = √605 = 11√5.
Answer: Unit vector (1/7)(3î − 6ĵ + 2k̂); area 11√5 square units
Equal inclination means l = m = n. Since l² + m² + n² = 1, 3l² = 1, so l = ±1/√3.
Answer: Shown
d is parallel to a × b = î(28 + 4) − ĵ(7 − 6) + k̂(−2 − 12) = 32î − ĵ − 14k̂. Put d = λ(32î − ĵ − 14k̂).
c · d = λ(64 + 1 − 56) = 9λ = 15, so λ = 5/3.
d = (160/3)î − (5/3)ĵ − (70/3)k̂.
Answer: (160/3)î − (5/3)ĵ − (70/3)k̂
The sum is (2 + λ)î + 6ĵ − 2k̂, with magnitude √((2 + λ)² + 40).
The scalar product with the unit vector along it is (λ + 6)/√((λ + 2)² + 40) = 1.
Squaring: λ² + 12λ + 36 = λ² + 4λ + 44, so 8λ = 8 and λ = 1.
Answer: λ = 1
Let |a| = |b| = |c| = k. Perpendicular means a·b = b·c = c·a = 0.
(a + b + c) · a = |a|² = k², and similarly (a + b + c) · b = k², (a + b + c) · c = k².
The cosine of the angle with each is k²/(|a + b + c| k), the same for all three, so the angles are equal.
Answer: Shown
(a + b) · (a + b) = |a|² + 2 a·b + |b|².
This equals |a|² + |b|² exactly when 2 a·b = 0, that is a·b = 0. For nonzero vectors this means a ⟂ b.
Answer: Proved
a · b = |a||b| cos θ ≥ 0 needs cos θ ≥ 0, which holds for 0 ≤ θ ≤ π/2.
Answer: (B) 0 ≤ θ ≤ π/2
|a + b|² = 1 + 1 + 2 cos θ = 1 gives cos θ = −1/2, so θ = 2π/3.
Answer: (D) θ = 2π/3
î · (ĵ × k̂) = î · î = 1. ĵ · (î × k̂) = ĵ · (−ĵ) = −1. k̂ · (î × ĵ) = k̂ · k̂ = 1.
Sum = 1 − 1 + 1 = 1.
Answer: (C) 1
|a||b||cos θ| = |a||b| sin θ gives |cos θ| = sin θ, so tan θ = 1 and θ = π/4 (within 0 to π/2).
Answer: (B) π/4