Topic: Engineering Mathematics
Angle made between vector c = 2i - 3j + 4k and the Z-axis
- A. sin^-1(4/$\sqrt{29}$)
- B. 4/$\sqrt{29}$
- C. cos^-1(4/$\sqrt{29}$)
- D. tan^-1(4/$\sqrt{29}$)
Answer: C. cos^-1(4/$\sqrt{29}$)
Explanation: The Z-axis is represented by the unit vector k. The dot product of c and k is c.k = 4. The magnitude of c is |c| = $\sqrt{2^2 + (-3}$^2 + 4^2) = $\sqrt{29}$. Using the dot product formula, cos(theta) = (c.k) / (|c||k|) = 4 / $\sqrt{29}$. Therefore, theta = cos^-1(4 / $\sqrt{29}$).