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A vector of magnitude 20 makes an angle of 30 deg with the x-y plane, and its projection on the x-y plane makes an angle of 60 deg with the x-axis. Find the vector.

1. 15i + 5*sqrt(3)*j + 10k
2. 15i + 10j + 5*sqrt(3)*k
3. 10i + 5*sqrt(3)*j + 15k
4. 5*sqrt(3)*i + 15j + 10k

Correct answer: 5*sqrt(3)*i + 15j + 10k

Solution

Let the angle with x-y plane be alpha = 30 deg and the projection's angle with x-axis be beta = 60 deg. z-component: V_z = 20*sin(30) = 20*(1/2) = 10. Magnitude of projection in x-y plane: V_xy = 20*cos(30) = 20*(sqrt(3)/2) = 10*sqrt(3). x-component: Vₓ = V_xy*cos(60) = 10*sqrt(3)*(1/2) = 5*sqrt(3). y-component: V_y = V_xy*sin(60) = 10*sqrt(3)*(sqrt(3)/2) = 15. Vector = 5*sqrt(3)*i + 15j + 10k. Verification: magnitude = sqrt((5*sqrt(3))² + 15² + 10²) = sqrt(75 + 225 + 100) = sqrt(400) = 20. Correct.

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