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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...
Direction Cosine of a Vector in the Direction of Positive x-axis:
To find the direction cosine of a vector in the direction of the positive x-axis, we need to determine the angle between the vector and the positive x-axis. The direction cosine of a vector in a specific direction is the cosine of the angle between the vector and that direction.
Given Vector:
a = 3i + 4j + 5k
Calculating the Direction Cosine:
The direction cosine of a vector with components (a, b, c) is given by:
l = a/√(a^2 + b^2 + c^2)
m = b/√(a^2 + b^2 + c^2)
n = c/√(a^2 + b^2 + c^2)
In this case, since we are interested in the direction of the positive x-axis, we need to find the direction cosine with respect to the x-axis:
l = 3/√(3^2 + 4^2 + 5^2) = 3/√50 = 3/√50
Final Answer:
Therefore, the direction cosine of the vector a = 3i + 4j + 5k in the direction of the positive x-axis is 3/√50, which corresponds to 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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