Q92 / 160 · AP EAMCET 2015 · Mathematics
The length of perpendicular from the origin to the plane which makes intercepts $\frac{1}{3}, \frac{1}{4}$ and $\frac{1}{5}$ respectively on the coordinate axes is
- Correct answerA$\frac{1}{5 \sqrt{2}}$
- B$\frac{1}{10}$
- C${5\sqrt{2}}$
- D$5$
Solution
Here, intercepts made with the coordinate axes $\mathrm{X}, \mathrm{Y}$ and $\mathrm{Z}$ are $\frac{1}{3}, \frac{1}{4}$ and $\frac{1}{5}$ respectively. Intercept form of a plane is
$\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$
We have, $a=\frac{1}{3}, b=\frac{1}{4}, c=\frac{1}{5}$
$\therefore \quad$ Equation of plane is
$\begin{aligned}
& 3 x+4 y+5 z=1 \\
& \Rightarrow \quad 3 x+4 y+5 z-1=0
\end{aligned}$
Distance from origin
$=\left|\frac{3 \times 0+4 \times 0+5 \times 0-1}{\sqrt{3^2+4^2+5^2}}\right|=\frac{1}{5 \sqrt{2}}$
Hence, the correct answer is option: 'A'.