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The angle between two unit magnitude coplanar vectors P(0.866,0.500,0) and Q(0.259,0.966,0) will be
- a)0°
- b)30°
- c)45°
- d)60°
Correct answer is option 'C'. Can you explain this answer?
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The angle between two unit magnitude coplanar vectors P(0.866,0.500,0)...
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The angle between two unit magnitude coplanar vectors P(0.866,0.500,0)...
The angle between two vectors can be calculated using the dot product formula:
cos(theta) = (P · Q) / (|P| |Q|)
Where P · Q is the dot product of vectors P and Q, and |P| and |Q| are the magnitudes of vectors P and Q.
Let's calculate the dot product:
P · Q = (0.866 * 0.259) + (0.500 * 0.966) + (0 * 0) = 0.224694
Now let's calculate the magnitudes:
|P| = √(0.866^2 + 0.500^2 + 0^2) = 1
|Q| = √(0.259^2 + 0.966^2 + 0^2) = 1
Substituting these values into the formula:
cos(theta) = 0.224694 / (1 * 1) = 0.224694
To find the angle, we need to take the inverse cosine (arccos) of this value:
theta = arccos(0.224694) ≈ 77.36 degrees
Therefore, the angle between the two vectors is approximately 77.36 degrees, which is not equal to 0.
cos(theta) = (P · Q) / (|P| |Q|)
Where P · Q is the dot product of vectors P and Q, and |P| and |Q| are the magnitudes of vectors P and Q.
Let's calculate the dot product:
P · Q = (0.866 * 0.259) + (0.500 * 0.966) + (0 * 0) = 0.224694
Now let's calculate the magnitudes:
|P| = √(0.866^2 + 0.500^2 + 0^2) = 1
|Q| = √(0.259^2 + 0.966^2 + 0^2) = 1
Substituting these values into the formula:
cos(theta) = 0.224694 / (1 * 1) = 0.224694
To find the angle, we need to take the inverse cosine (arccos) of this value:
theta = arccos(0.224694) ≈ 77.36 degrees
Therefore, the angle between the two vectors is approximately 77.36 degrees, which is not equal to 0.
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The angle between two unit magnitude coplanar vectors P(0.866,0.500,0) and Q(0.259,0.966,0) will bea)0°b)30°c)45°d)60°Correct answer is option 'C'. Can you explain this answer?
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