Topic: Engineering Mathematics
Angle made by 2i - 3j + 4k with the z-axis is
- A. cos^-1(4 / $\sqrt{29}$)
- B. 4 / $\sqrt{29}$
- C. sin^-1(4 / $\sqrt{29}$)
- D. tan^-1(4 / $\sqrt{29}$)
Answer: A. cos^-1(4 / $\sqrt{29}$)
Explanation: Let a = 2i - 3j + 4k and the z-axis unit vector b = k. cos(theta) = (a.b) / (|a||b|) = 4 / $\sqrt{2^2 + (-3}$^2 + 4^2) = 4 / $\sqrt{29}$. Therefore, theta = cos^-1(4 / $\sqrt{29}$).