Problem#1
Vector A has a magnitude of 5.00 units, and B has a magnitude of 9.00 units. The two vectors make an angle of 50.0° with each other. Find A.B.Answer:
A.B = AB cosθ
A.B = (5.00)(9.00) cos50.00 = 28.9
Problem#2
For any two vectors A and B, show that A.B = AxBx + AyBy + AzBz. (Suggestion: Write A and B in unit vector form and use Equations 7.4 and 7.5.)
Answer:
If, A = Axi + Ayj + Azk and B = Bxi + Byj + Bzk, then
A.B = (Axi + Ayj + Azk).(Bxi + Byj + Bzk)
= AxBx(i.i) + AxBy(i.j) + AxBz(i.k) + AyBx(j.i) + AyBy(j.j) + AyBz(j.k) + AzBx(k.i) + AzBy(k.j) + AzBz(k.k)
A.B = AxBx + AyBy + AzBz
Problem#3
A force F = (6i – 2j) N acts on a particle that undergoes a displacement Δr = (3i + j) m. Find (a) the work done by the force on the particle and (b) the angle between F and Δr.
Answer:
(a) the work done by the force on the particle, given by
W = F.Δr
W = (6i – 2j).(3i + j) = 18 – 2 = 16 J
(b) the angle between F and Δr is
W = F.Δr cos θ
With
F = [(6 N)2 + (–2 N)2]1/2 = 2√10 N, and
Δr = [(3 m)2 + (–1 m)2]1/2 = √10 m
Then
16 J = (2√10 N)(√10 m) cosθ
cosθ = 16/20 = 0.8
θ = cos-1(0.8) = 36.90
Problem#4
Find the scalar product of the vectors in Figure 1.
Answer:
F.v = Fv cosθ
We must first find the angle between the two vectors. It is:
θ = 3600 – 1180 – 1320 – 90.00 = 20.00
then
F.v = (32.8 N)(0.173 m/s) cos20.00 = 5.33 J/s
Problem#5
Using the definition of the scalar product, find the angles between (a) A = 3i – 2j and B = 4i – 4j; (b) A = –2i + 4j and B = 3i – 4j + 2k; (c) A = i – 2j + 2k and B = 3j + 4k.
Answer:
(a) For A = 3i – 2j and B = 4i – 4j;
A = [(3)2 + (–2)2]1/2 = √13
B = [(4)2 + (–4)2]1/2 = 4√2, and
A.B = (3).(4) + (–2)(–4) = 20
then
θ = cos-1(A.B/AB)
θ = cos-1[20/(√13)(4√2)] = 11.30
(b) For A = –2i + 4j and B = 3i – 4j + 2k;
A = [(–2)2 + (4)2]1/2 = √20
B = [(3)2 + (–4)2 + (2)2]1/2 = √29, and
A.B = (–2).(3) + (4)(–4) + 0 = –22
then
θ = cos-1(A.B/AB)
θ = cos-1[22/(√20)(√29)] = 1560
(c) For A = i – 2j + 2k and B = 3j + 4k.
A = [(1)2 + (–2)2 + (2)2]1/2 = 3
B = [(3)2 + (4)2]1/2 = 5, and
A.B = 0 + (–2).(3) + (2)(4) = 2.00
then
θ = cos-1(A.B/AB)
θ = cos-1[2/(3)(5)] = 82.30
Problem#6
For A = 3i + j – k, B = –i + 2j + 5k, and C = 2j – 3k, find C.(A x B).
Answer:
A – B = (3i + j – k) – (–i + 2j + 5k) = 4i – j – 6k
then
C.(A – B) = (2j – 3k).(4i – j – 6k)
C.(A – B) = 0 + (–2) + 18 = 16.0
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