MHT CET · Maths · Three Dimensional Geometry
The perpendicular distance of the origin from the plane \(x-3 y+4 z-6=0\) is
- A \(6\)
- B \(\frac{6}{\sqrt{26}}\)
- C \(\frac{1}{\sqrt{26}}\)
- D \(\frac{3}{\sqrt{26}}\)
Answer & Solution
Correct Answer
(B) \(\frac{6}{\sqrt{26}}\)
Step-by-step Solution
Detailed explanation
Length of perpendicular from point \(\mathrm{O}(0,0,0)\) to plane \(x-3 y+4 z-6=0\) is given by
\(\begin{aligned}
d & =\left|\frac{0(1)+0(-3)+0(4)-6}{\sqrt{(1)^2+(-3)^2+(4)^2}}\right| \\
& =\left|\frac{-6}{\sqrt{1+9+16}}\right|=\frac{6}{\sqrt{26}}
\end{aligned}\)
\(\begin{aligned}
d & =\left|\frac{0(1)+0(-3)+0(4)-6}{\sqrt{(1)^2+(-3)^2+(4)^2}}\right| \\
& =\left|\frac{-6}{\sqrt{1+9+16}}\right|=\frac{6}{\sqrt{26}}
\end{aligned}\)
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