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The direction cosine of a vector a=3i+4j+5k in the direction of positive x-axis is
- a)3/√50
- b)2/√50
- c)-3/√50
- d)4/√50
Correct answer is option 'A'. Can you explain this answer?
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The direction cosine of a vector a=3i+4j+5k in the direction of positi...
Solution:
We have a vector a=3i + 4j + 5k. To find the direction cosine of the vector a in the direction of positive x-axis, we need to find the angle between the vector a and the positive x-axis.
Let θ be the angle between vector a and the positive x-axis. Then, we have:
cos(θ) = (a . i) / (|a| . |i|)
where a . i is the dot product of vectors a and i, and |a| and |i| are the magnitudes of vectors a and i, respectively.
We have:
a . i = (3i + 4j + 5k) . i = 3
|i| = 1 (since i is a unit vector)
|a| = √(3^2 + 4^2 + 5^2) = √50
Therefore, we have:
cos(θ) = 3 / (√50 * 1) = 3 / √50
Hence, the direction cosine of vector a in the direction of positive x-axis is 3 / √50.
Therefore, the correct answer is option 'A'.
We have a vector a=3i + 4j + 5k. To find the direction cosine of the vector a in the direction of positive x-axis, we need to find the angle between the vector a and the positive x-axis.
Let θ be the angle between vector a and the positive x-axis. Then, we have:
cos(θ) = (a . i) / (|a| . |i|)
where a . i is the dot product of vectors a and i, and |a| and |i| are the magnitudes of vectors a and i, respectively.
We have:
a . i = (3i + 4j + 5k) . i = 3
|i| = 1 (since i is a unit vector)
|a| = √(3^2 + 4^2 + 5^2) = √50
Therefore, we have:
cos(θ) = 3 / (√50 * 1) = 3 / √50
Hence, the direction cosine of vector a in the direction of positive x-axis is 3 / √50.
Therefore, the correct answer is option 'A'.
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The direction cosine of a vector a=3i+4j+5k in the direction of positive x-axis isa)3/√50b)2/√50c)-3/√50d)4/√50Correct answer is option 'A'. Can you explain this answer?
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