If $Q$ is the foot of perpendicular from the point $P(4, -5, 3)$ on the line $\frac{x-5}{3} = \frac{y+2}{-4} = \frac{z-6}{5}$, then $[PQ] = $ __________. (where $[.]$ denote greatest integer function)
Step-by-Step Solution
Key Concept: Use the perpendicularity condition $\vec{PQ} \cdot \vec{d} = 0$ where $\vec{d}$ is the direction vector of the line to find parameter $t$ for foot of perpendicular $Q$.
The line is parameterized as $(5+3t, -2-4t, 6+5t)$. Point $Q$ on the line has direction vector $(3, -4, 5)$. Since $PQ$ is perpendicular to the line, $\vec{PQ} \cdot (3, -4, 5) = 0$. We have $\vec{PQ} = (5+3t-4, -2-4t+5, 6+5t-3) = (1+3t, 3-4t, 3+5t)$. Setting $(1+3t, 3-4t, 3+5t) \cdot (3, -4, 5) = 0$: $3(1+3t) - 4(3-4t) + 5(3+5t) = 3 + 9t - 12 + 16t + 15 + 25t = 6 + 50t = 0$, so $t = -\frac{3}{25}$. Thus $Q = (5-\frac{9}{25}, -2+\frac{12}{25}, 6-\frac{15}{25}) = (\frac{116}{25}, -\frac{38}{25}, \frac{135}{25})$. Computing $PQ = \sqrt{(\frac{116}{25}-4)^2 + (-\frac{38}{25}+5)^2 + (\frac{135}{25}-3)^2} = \sqrt{(\frac{16}{25})^2 + (\frac{87}{25})^2 + (\frac{60}{25})^2} = \frac{1}{25}\sqrt{256 + 7569 + 3600} = \frac{\sqrt{11425}}{25} \approx 3.377$, so $[PQ] = 3$.
Correct Answer: 3