StudentAidUnit 14: Vectors in Space — Exercise 14 2
11th Class Mathematics · Unit 14: Vectors in Space
13.Find the work done, if the point at which the constant force F = 2i+5j+3k is applied to an object, moves it from P(2,-3,1) to P(7,5,3)
Given
Force
F = 2i + 5j + 3k
Initial point
P1 = (2, -3, 1)
Final point
P2 = (7, 5, 3)
Formula
Displacement
d = P2 - P1
Working
Calculate displacement
d = (7-2)i + (5-(-3))j + (3-1)k = 5i + 8j + 2k
Formula
Work
W = F · d
Working
Substituting
W = (2i + 5j + 3k) · (5i + 8j + 2k)
= 2(5) + 5(8) + 3(2) = 10 + 40 + 6
Result
Work done
W = 56 units
14.A particle, acted by constant forces F₁ = 3i+4j-3k and F₂ = i+4j-k is displaced from A(2,1,3) to B(5,4,4) Find the work done.
Given
Force 1
F1 = 3i + 4j - 3k
Force 2
F2 = i + 4j - k
Initial point
A = (2, 1, 3)
Final point
B = (5, 4, 4)
Formula
Displacement
d = B - A
Working
Calculate displacement
d = (5-2)i + (4-1)j + (4-3)k = 3i + 3j + k
Formula
Net force
F = F1 + F2
Working
Calculate net force
F = (3+1)i + (4+4)j + (-3-1)k = 4i + 8j - 4k
Formula
Work
W = F · d
Working
Substituting
W = (4i + 8j - 4k) · (3i + 3j + k)
= 4(3) + 8(3) + (-4)(1) = 12 + 24 - 4
Result
Work done
W = 32 units
15.A particle is displaced from the point A(5,-5,-7) to the point B(6,2,-2) under the action of constant forces defined by 10i-j+11k, 4i+5j+9k and -2i+j-9k Show that the total work done by the force is 102 units.
Given
Force 1
F1 = 10i - j + 11k
Force 2
F2 = 4i + 5j + 9k
Force 3
F3 = -2i + j - 9k
Initial point
A = (5, -5, -7)
Final point
B = (6, 2, -2)
Formula
Displacement
d = B - A
Working
Calculate displacement
d = (6-5)i + (2-(-5))j + (-2-(-7))k = i + 7j + 5k
Formula
Total force
F = F1 + F2 + F3
Working
Calculate total force
F = (10+4-2)i + (-1+5+1)j + (11+9-9)k = 12i + 5j + 11k
Formula
Work
W = F · d
Working
Substituting
W = (12i + 5j + 11k) · (i + 7j + 5k)
= 12(1) + 5(7) + 11(5) = 12 + 35 + 55
Result
Work done
W = 102 units
16.A force of magnitude 6 acting parallel to 4i+3j-k displaces the point of application from A(2,-1,3) to B(7,3,2) Find the work done.
Given
Force magnitude
left|Fright| = 6
Direction of force
4i + 3j - k
Initial point
A = (2, -1, 3)
Final point
B = (7, 3, 2)
Formula
Magnitude of direction vector
left|4i + 3j - kright| = sqrt{42 + 32 + (-1)2}
= sqrt{16 + 9 + 1} = sqrt{26}
Force vector
F' = 6 · frac{4i + 3j - k}{sqrt{26}
Working
Calculate force
F' = frac{24i + 18j - 6k}{sqrt{26}
Formula
Displacement
d = B - A
Working
Calculate displacement
d = (7-2)i + (3-(-1))j + (2-3)k = 5i + 4j - k
Formula
Work
W = F' · d
Working
Substituting
W = frac{24i + 18j - 6k}{sqrt{26} · (5i + 4j - k)
= frac{24(5) + 18(4) + (-6)(-1)}{sqrt{26} = frac{120 + 72 + 6}{sqrt{26}
Result
Work done
W = frac{198}{sqrt{26} units