VECTORS QUESTIONS
1)
Given the position vectors of the points L, M and N are respectively L = 2i + 3j - 4k, M = 5i - j + 2k and N = 11i + λj + 14k. Find:
2)
The position vector r of a particle of mass 5kg moving in space at any time 't' seconds is given by:
r(t) = (2t² - 7t - 65/8)i + 4j + 3k
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VECTORS QUESTIONS WITH DETAILED SOLUTIONS
1) Given the position vectors:
L = 2i + 3j - 4k, M = 5i - j + 2k, N = 11i + λj + 14k
(a) The unit vector parallel to LM
Solution:
First find vector LM = M - L = (5-2)i + (-1-3)j + (2-(-4))k = 3i - 4j + 6k
Magnitude of LM = √(3² + (-4)² + 6²) = √(9 + 16 + 36) = √61
Unit vector = LM/|LM| = (3/√61)i - (4/√61)j + (6/√61)k
(b) The value of λ if L, M and N are collinear
Solution:
For collinearity, vectors must be scalar multiples: MN = k × LM
MN = N - M = (11-5)i + (λ-(-1))j + (14-2)k = 6i + (λ+1)j + 12k
From LM = 3i - 4j + 6k, we see MN = 2 × LM
Thus: 6 = 2×3, 12 = 2×6, and λ+1 = 2×(-4) ⇒ λ = -9
(c) Position vector of Q where LQ:QM = 2:1
Solution:
Using section formula: Q = (2M + 1L)/(2+1) = (2(5i-j+2k) + 1(2i+3j-4k))/3
= (10i-2j+4k + 2i+3j-4k)/3 = (12i + j)/3 = 4i + (1/3)j
2) Particle motion with r(t) = (2t² - 7t - 65/8)i + 4j + 3k
(a) Verify constant acceleration
Solution:
Velocity v(t) = dr/dt = (4t - 7)i + 0j + 0k
Acceleration a(t) = dv/dt = 4i (constant)
(b) Calculations at rest and at t=5s
(i) Time and distance when at rest
At rest when v(t) = 0 ⇒ 4t - 7 = 0 ⇒ t = 7/4 = 1.75s
Position at t=1.75s: r(1.75) = (2(1.75)² - 7(1.75) - 65/8)i + 4j + 3k
= (6.125 - 12.25 - 8.125)i + 4j + 3k = -14.25i + 4j + 3k
Distance from origin = √((-14.25)² + 4² + 3²) = √(203.0625 + 16 + 9) ≈ 15.09m
(ii) Momentum and force at t=5s
Mass m = 5kg
Velocity at t=5s: v(5) = (4×5 - 7)i = 13i m/s
Momentum p = mv = 5×13i = 65i kg·m/s
Force F = ma = 5×4i = 20i N (constant)
3) Vector operations and proofs
(a) Verify vector formulas
(i) (a - b) × (a + b) = 2(a × b)
Proof:
Expand using distributive property of cross product:
(a - b) × (a + b) = a×a + a×b - b×a - b×b
= 0 + a×b - (-a×b) - 0 = 2(a × b) ✓
(ii) (a - b) · (a + b) = 0
Note: This is only true if |a| = |b| (for orthogonal vectors)
General case: (a - b)·(a + b) = a·a + a·b - b·a - b·b = |a|² - |b|²
Only equals 0 if |a| = |b|
(b) Given a = 2i + j + k, b = i + 2j - 2k, c = 3i + 4j + 2k
(i) Scalar projection of a + c onto b
a + c = (2+3)i + (1+4)j + (1+2)k = 5i + 5j + 3k
Scalar projection = (a + c)·b / |b|
= (5×1 + 5×2 + 3×(-2))/√(1² + 2² + (-2)²) = (5 + 10 - 6)/3 = 9/3 = 3
(ii) Angle between a and b
cosθ = (a·b)/(|a||b|)
a·b = 2×1 + 1×2 + 1×(-2) = 2 + 2 - 2 = 2
|a| = √(2² + 1² + 1²) = √6, |b| = √(1² + 2² + (-2)²) = 3
cosθ = 2/(3√6) ⇒ θ ≈ 74.21°
(c) Particle motion with a = 12cos2t i - 8sin2t j + 16t k
(i) Velocity v(t)
v(t) = ∫a dt = (6sin2t + C₁)i + (4cos2t + C₂)j + (8t² + C₃)k
At t=0, v=0 ⇒ C₁=0, C₂=-4, C₃=0
Thus v(t) = 6sin2t i + (4cos2t - 4)j + 8t² k
(ii) Displacement r(t)
r(t) = ∫v dt = (-3cos2t + D₁)i + (2sin2t - 4t + D₂)j + (8t³/3 + D₃)k
At t=0, r=0 ⇒ D₁=3, D₂=0, D₃=0
Thus r(t) = (3 - 3cos2t)i + (2sin2t - 4t)j + (8t³/3)k
(d) Prove |CB|² = |OB|² + |OC|² - 2|OB||OC|cosθ
Proof: This is the vector form of the cosine rule.
Let CB = OB - OC
|CB|² = (OB - OC)·(OB - OC) = |OB|² + |OC|² - 2(OB·OC)
Since OB·OC = |OB||OC|cosθ, the result follows.
Note:
For more of Topic Test Questiona.
https://mitihanipopote.blogspot.com/search/label/Topic%20Test.
4) Vector Geometry Problems
(a) Points A(4,0,0), B(2,6,0), C(1,4,2)
(i) Cosine of angle between BA and BC
Solution:
BA = A - B = (4-2,0-6,0-0) = 2i - 6j
BC = C - B = (1-2,4-6,2-0) = -i - 2j + 2k
BA·BC = (2)(-1) + (-6)(-2) + (0)(2) = -2 + 12 + 0 = 10
|BA| = √(2² + (-6)²) = √40 = 2√10
|BC| = √((-1)² + (-2)² + 2²) = √9 = 3
cosθ = BA·BC / (|BA||BC|) = 10 / (2√10 × 3) = √10/6 ≈ 0.527
(ii) Area of triangle ABC
Area = ½|BA × BC|
BA × BC = |i j k| |2 -6 0| |-1 -2 2|
= i[(-6)(2)-(0)(-2)] - j[(2)(2)-(0)(-1)] + k[(2)(-2)-(-6)(-1)]
= -12i - 4j - 10k
|BA × BC| = √((-12)² + (-4)² + (-10)²) = √(144+16+100) = √260 = 2√65
Area = ½ × 2√65 = √65 square units
(b) Resultant of two forces (5N and 7N at 90°)
Solution:
Let F₁ = 5i, F₂ = 7j (assuming they act along x and y axes)
Resultant R = F₁ + F₂ = 5i + 7j
Magnitude |R| = √(5² + 7²) = √74 ≈ 8.60N
Direction θ = tan⁻¹(7/5) ≈ 54.46° from x-axis
(c) Vector perpendicular to both a+b and a-b
Given a = 3i + 2j + 2k, b = i + 2j - 2k
a + b = (3+1)i + (2+2)j + (2-2)k = 4i + 4j
a - b = (3-1)i + (2-2)j + (2-(-2))k = 2i + 4k
Perpendicular vector = (a+b) × (a-b) = |i j k| |4 4 0| |2 0 4|
= i(16-0) - j(16-0) + k(0-8) = 16i - 16j - 8k
Unit vector = (16i - 16j - 8k)/√(16² + (-16)² + (-8)²) = (16i - 16j - 8k)/24
= (2/3)i - (2/3)j - (1/3)k
5) Vector Applications
(a) Triangle Area Formula
(i) Derive area formula for vertices (x₁,y₁,z₁), (x₂,y₂,z₂), (x₃,y₃,z₃)
Derivation:
Let vectors AB = (x₂-x₁)i + (y₂-y₁)j + (z₂-z₁)k
AC = (x₃-x₁)i + (y₃-y₁)j + (z₃-z₁)k
Area = ½|AB × AC|
AB × AC = |i j k| |x₂-x₁ y₂-y₁ z₂-z₁| |x₃-x₁ y₃-y₁ z₃-z₁|
Magnitude gives the area of the parallelogram, half gives triangle area
(ii) Apply to vertices (1,1,2), (2,3,5), (1,5,5)
AB = (2-1)i + (3-1)j + (5-2)k = i + 2j + 3k
AC = (1-1)i + (5-1)j + (5-2)k = 4j + 3k
AB × AC = |i j k| |1 2 3| |0 4 3|
= i(6-12) - j(3-0) + k(4-0) = -6i - 3j + 4k
|AB × AC| = √((-6)² + (-3)² + 4²) = √(36+9+16) = √61
Area = ½√61 ≈ 3.905 square units
(b) Work done by resultant force
Given F₁ = 2i + 2j - 3k, F₂ = i + 3k
Resultant F = F₁ + F₂ = 3i + 2j
Displacement: dr/dt = i + 2tj + k ⇒ r(t) = ti + t²j + tk + C
At t=0, r=0 ⇒ C=0 ⇒ r(5) = 5i + 25j + 5k
Work W = F·r(5) = (3)(5) + (2)(25) + (0)(5) = 15 + 50 = 65 Joules
(c) Vector Proofs and Projections
(i) Prove |a × b|² + |a · b|² = |a|²|b|²
Proof:
Let θ be angle between a and b
|a × b|² = (|a||b|sinθ)² = |a|²|b|²sin²Î¸
|a · b|² = (|a||b|cosθ)² = |a|²|b|²cos²Î¸
Adding: |a|²|b|²(sin²Î¸ + cos²Î¸) = |a|²|b|²(1) = |a|²|b|²
(ii) Projection of b = 5i - 3j + k onto Z-axis
Z-axis unit vector = k
Projection = b·k = (5)(0) + (-3)(0) + (1)(1) = 1
6) Vector Relationships
(a) Angle between a = i - j + k and b = i + j - k
a·b = (1)(1) + (-1)(1) + (1)(-1) = 1 - 1 - 1 = -1
|a| = √(1² + (-1)² + 1²) = √3
|b| = √(1² + 1² + (-1)²) = √3
cosθ = a·b/(|a||b|) = -1/3 ⇒ θ ≈ 109.47°
(b) Find t such that A + tB is perpendicular to C
Given A = 2i + 2j + 3k, B = -i + 2j + 3k, C = 3i + j
(A + tB)·C = 0 for perpendicularity
A + tB = (2-t)i + (2+2t)j + (3+3t)k
Dot product with C: (2-t)(3) + (2+2t)(1) + (3+3t)(0) = 0
6-3t + 2+2t = 0 ⇒ 8 - t = 0 ⇒ t = 8
(c) Area of triangle (same as 5(a)(ii))
As calculated previously, area = ½√61 ≈ 3.905 square units
7) Vector Combinations
(a) Find λ for coplanar vectors
Vectors: u = i + j + 2k, v = λi - j + k, w = 3i - 2j - k
For coplanarity, scalar triple product [u v w] = 0
u·(v × w) = |1 1 2| |λ -1 1| |3 -2 -1| = 0
= 1[(-1)(-1)-(1)(-2)] -1[(λ)(-1)-(1)(3)] +2[(λ)(-2)-(-1)(3)]
= 1(1+2) -1(-λ-3) +2(-2λ+3) = 3 + λ + 3 - 4λ + 6 = 12 - 3λ = 0
⇒ λ = 4
(b) Express r = 10i - 3j - k as linear combination of a,b,c
Given a = 2i - j + 3k, b = 3i + 2j - 4k, c = -i + 3j - 2k
Let r = pa + qb + rc
System of equations:
2p + 3q - r = 10 (i component)
-p + 2q + 3r = -3 (j component)
3p - 4q - 2r = -1 (k component)
Solving (using matrix methods or substitution):
p = 2, q = 3, r = -1
Thus r = 2a + 3b - c
Summary
These solutions demonstrate key vector concepts including:
- Vector operations (addition, subtraction, dot and cross products)
- Geometric applications (angles, areas, projections)
- Physical applications (work, motion)
- Linear combinations and coplanarity conditions
- Coordinate geometry in 3D space
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