Question

If a vector having magnitude of 5 units, makes equal angle with each of the three mutually perpendicular axes, then the sum of the magnitude of the projections on each of the axis is

• A. \( \frac{15}{3} \) unit
• B. \( 5\sqrt{3} \) unit
• C. \( \frac{15\sqrt{3}}{2} \) units
• D. None of these

Hint:

For vectors making equal angles with all axes, use the Pythagorean theorem in 3D space to find the projections along each axis and then sum them up.

Answer 1

The Correct Option is B

Step 1: Let the vector be \( \vec{V} \) with a magnitude of 5 units, and let it make an equal angle \( \theta \) with the \( x \)-axis, \( y \)-axis, and \( z \)-axis. The vector's components along the axes are:
\[ V_x = V \cos(\theta), V_y = V \cos(\theta), V_z = V \cos(\theta) \] where \( V = 5 \) units is the magnitude of the vector. Step 2: The magnitude of the projection of the vector on any axis is \( V \cos(\theta) \). Since the vector makes equal angles with each of the axes, the sum of the magnitudes of the projections on each of the axes is:
\[ \text{Sum of projections} = 3 \times V \cos(\theta) \] Step 3: Since \( V = 5 \), we need to calculate \( \cos(\theta) \). By the Pythagorean theorem in 3D, the sum of the squares of the projections must equal the square of the vector's magnitude: \[ V^2 = (V \cos(\theta))^2 + (V \cos(\theta))^2 + (V \cos(\theta))^2 \] \[ V^2 = 3 \times (V \cos(\theta))^2 \] \[ 25 = 3 \times (5 \cos(\theta))^2 \] \[ 25 = 3 \times 25 \cos^2(\theta) \] \[ 1 = 3 \cos^2(\theta) \] \[ \cos^2(\theta) = \frac{1}{3} \] \[ \cos(\theta) = \frac{1}{\sqrt{3}} \] Step 4: Now substitute \( \cos(\theta) = \frac{1}{\sqrt{3}} \) into the sum of projections formula: \[ \text{Sum of projections} = 3 \times 5 \times \frac{1}{\sqrt{3}} = 5 \sqrt{3} \] Thus, the sum of the magnitudes of the projections is \( 5\sqrt{3} \) units, which corresponds to option (b).