If θ is the angle between the vectors and a vector and b vector, then
Let us look into some example problems to understand finding angle between two vectors.
Question 1 :
If a vector and b vector are two vectors such that |a| = 10,|b| = 15 and a . b = 75√2, find the angle between a and b.
Solution :
θ = cos-1[75√2/(10)(15)]
θ = cos-1(√2/2)
θ = cos-1(1/√2)
θ = π/4
Question 2 :
Find the angle between the vectors
(i) 2i vector + 3j vector − 6k vector and 6i vector − 3j vector + 2k vector
Solution :
Let a vector = 2i vector + 3j vector − 6k vector and
b vector = 6i vector − 3j vector + 2k vector
Angle between two vector :
θ = cos-1 [a vector . b vector/ |a| |b|]
a vector . b vector = 2(6) + 3(-3) + (-6)(2)
= 12 - 9 - 12
= -9
|a vector| = |2i vector + 3j vector − 6k vector|
r = √(22 + 32 + (-6)2) = √(4 + 9 + 36)
= √49 = 7
|b vector| = |6i vector - 3j vector + 2k vector|
r = √(62 + (-3)2 + 22) = √(36 + 9 + 4)
= √49 = 7
θ = cos-1 [-9/(7)(7)]
θ = cos-1 [-9/49]
(ii) i vector − j vector and j vector − k vector.
Let a vector = i vector − j vector and
b vector = j vector − k vector
Solution :
Angle between two vector :
θ = cos-1 [a vector . b vector/ |a| |b|]
a vector . b vector = 1(0) + (-1)1 + 0(-1)
= 0 - 1 + 0
= -1
|a vector| = |i vector - j vector|
r = √(12 + (-1)2) = √(1 + 1) = √2
|b vector| = |j vector - k vector|
r = √(12 + (-1)2) = √(1+1) = √2
θ = cos-1 [-1/( √2)( √2)]
θ = cos-1 [-1/2]
θ = 2π/3
Question 3 :
If a vector, b vector, c vector are three vectors such that a vector + 2b vector + c vector = 0 vector and |a|= 3, |b| = 4, |c| = 7, find the angle between a vector and b vector.
Solution :
a vector + 2b vector + c vector = 0 vector
|a + 2b| vector = |-c| vector
|a + 2b|2 = |-c|2
|a|2 + |2b|2 + 2|a||2b| = |c|2
|a|2 + 4|b|2 + 4|a vector| |b vector | = |c|2
|a|2 + 4|b|2 + 4(a. b) = |c|2
|a|2 + 4|b|2 + 4 |a| |b| cos θ = |c|2
32 + 4(4)2 + 4 (3)(4) cos θ = 72
9 + 64 + 48 cos θ = 49
73 + 48 cos θ = 49
48 cos θ = 49 - 73
48 cos θ = -24
θ = cos-1 (-24/48)
θ = cos-1 (-1/2)
θ = 2π/3
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