Question

If the sum of the distances of the point (3, 4, 0), $\alpha \in \mathbb{R}$ from X-axis, Y-axis and Z-axis is minimum, then $\sec \alpha$ =
Identify the correct option from the following:

• A. 2
• B. 1
• C. 0
• D. -1

Hint:

To minimize a sum of distances, set the derivative of the expression to zero and solve for the variable, then verify the result with the options.

Answer 1

The Correct Option is B

Step 1: Interpret the distances from the axes
For a point (3, 4, $\alpha$), the distance from the X-axis is the distance to the line $y=0, z=0$, which is $\sqrt{4^2 + \alpha^2}$. The distance from the Y-axis ($x=0, z=0$) is $\sqrt{3^2 + \alpha^2}$, and from the Z-axis ($x=0, y=0$) is $\sqrt{3^2 + 4^2} = 5$. The sum of distances is $S = \sqrt{16 + \alpha^2} + \sqrt{9 + \alpha^2} + 5$. Step 2: Minimize the sum of distances
To minimize $S$, focus on the variable part: $f(\alpha) = \sqrt{16 + \alpha^2} + \sqrt{9 + \alpha^2}$. Take the derivative with respect to $\alpha$: $f'(\alpha) = \frac{\alpha}{\sqrt{16 + \alpha^2}} + \frac{\alpha}{\sqrt{9 + \alpha^2}}$. Set $f'(\alpha) = 0$: $\alpha \left( \frac{1}{\sqrt{16 + \alpha^2}} + \frac{1}{\sqrt{9 + \alpha^2}} \right) = 0$. Since the denominator terms are positive, $\alpha = 0$ is the solution. At $\alpha = 0$, $S = \sqrt{16} + \sqrt{9} + 5 = 4 + 3 + 5 = 12$, which is the minimum. Step 3: Compute $\sec \alpha$
If $\alpha = 0$, then $\sec \alpha = \sec 0 = 1$, which matches option (2).