Question

If \( \mathbf{f}, \mathbf{g}, \mathbf{h} \) are mutually orthogonal vectors of equal magnitudes, then find the angle between the vectors \( \mathbf{f} + \mathbf{g} + \mathbf{h} \) and \( \mathbf{h} \).

• A. \( \cos^{-1} \left( \frac{\sqrt{3}}{4} \right) \)
• B. \( \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) \)
• C. \( \pi - \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) \)
• D. \( \pi - \cos^{-1} \left( \frac{\sqrt{3}}{4} \right) \)

Hint:

For mutually perpendicular vectors of equal magnitude, their sum forms an equilateral configuration, leading to angles derived using dot products.

Answer 1

The Correct Option is B

Step 1: Finding the Dot Product
Since \( \mathbf{f}, \mathbf{g}, \mathbf{h} \) are mutually perpendicular and have equal magnitudes, let: \[ |\mathbf{f}| = |\mathbf{g}| = |\mathbf{h}| = r \] \[ \mathbf{a} = \mathbf{f} + \mathbf{g} + \mathbf{h} \] \[ \mathbf{a} \cdot \mathbf{h} = r^2 \] Step 2: Calculating the Angle
Using the dot product formula: \[ \cos \theta = \frac{\mathbf{a} \cdot \mathbf{h}}{|\mathbf{a}||\mathbf{h}|} \] \[ = \frac{r^2}{\sqrt{3r^2} \cdot r} = \frac{1}{\sqrt{3}} \] \[ \theta = \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) \] Thus, the correct answer is \( \cos^{-1} \left( \frac{1}{\sqrt{3}} \right) \).