Search any question & find its solution
Question:
Answered & Verified by Expert
If the direction ratios of the normal to a plane are $ < 1, \mathrm{~m}, \mathrm{n}>$ and the length of the normal is $\mathrm{p}$, then what is the sum of intercepts cut-off by the plane from the coordinate axes?
Options:
Solution:
2299 Upvotes
Verified Answer
The correct answer is:
$\mathrm{p}\left(\frac{1}{\ell}+\frac{1}{\mathrm{~m}}+\frac{1}{\mathrm{n}}\right)$
Since, the direction ratio's of normal to a plane are $ < l, \mathrm{~m}, \mathrm{n}>$ and the length of normal is $\mathrm{p}$, then intercept on $\mathrm{x}$ -axis is $\mathrm{p} / /$ and that on $\mathrm{y}$ -axis is $\mathrm{p} / \mathrm{m}$ and on $\mathrm{z}$ -axis it is $\mathrm{p} / \mathrm{n}$, hence sum of intercepts cut off by the plane from the coordinate axes
$=p\left(\frac{1}{\ell}+\frac{1}{m}+\frac{1}{n}\right)$
$=p\left(\frac{1}{\ell}+\frac{1}{m}+\frac{1}{n}\right)$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.