A vector is given by R=2i+1j+3K. Find the angles between R and the X, Y and Z axis?

Answer 1

R°i = (R)(1) cos a (with the x axe)
R°i = sqrt (2^2 + 1^2 + 3^2) cos a = sqrt 14 cos a
R°i = (2i + 1j + 3k) ° i = 2i ° i + 1j ° i + 3k ° i = 2
Then sqrt 14 cos a = 2
cos a = 2 / sqrt 14
a = 58° aprox.

Similarly,

cos b = 1 / sqrt 14 (y axe)

cos c = 3 / sqrt 14 (z axe)

Answer 2

A vector is given by R = 2i + j + 3k.
Find the angles between R and the x, y and z axes?
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First calculate the magnitude of the vector R. You are going to need it.

|| R || = √(2² + 1² + 3²) = √(4 + 1 + 9) = √14

Take the dot product.
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For the x-axis.

R • i = <2, 1, 3> • <1, 0, 0> = 2 + 0 + 0 = 2

The dot product can also be calculated as:

R • i = || R || || i || cosα
where α is the angle between R and the x-axis

cosα = (R • i) / (|| R || || i ||) = 2 / (√14 * 1) = 2/√14

α = arccos(2/√14) ≈ 57.7°
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For the y-axis.

R • j = <2, 1, 3> • <0, 1, 0> = 0 + 1 + 0 = 1

The dot product can also be calculated as:

R • j = || R || || j || cosβ
where β is the angle between R and the y-axis

cosβ = (R • j) / (|| R || || j ||) = 1 / (√14 * 1) = 1/√14

β = arccos(1/√14) ≈ 74.5°
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For the z-axis.

R • k = <2, 1, 3> • <0, 0, 1> = 0 + 0 + 3 = 3

The dot product can also be calculated as:

R • k = || R || || k || cosγ
where γ is the angle between R and the z-axis

cosγ = (R • k) / (|| R || || k ||) = 3 / (√14 * 1) = 3/√14

γ = arccos(3/√14) ≈ 36.7°
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