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Find the angle that the vector (i - j + sqrt(2)*k) makes with the y-axis.

1. 60 deg
2. 120 deg
3. 150 deg
4. arctan(-1/2)

Correct answer: 120 deg

Solution

Vector v = i - j + sqrt(2)*k. The y-component is -1. Magnitude |v| = sqrt(1+1+2) = sqrt(4) = 2. The angle with the y-axis: cos(theta) = (v. j) / |v| = -1/2. Therefore theta = arccos(-1/2) = 120 degrees.

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