If a plane has $X -$ intercept $l, Y -$ intercept $m$ and $Z -$ intercept $n,$ and perpendicular distance of plane from origin is $k,$ then

\frac{1}{l^{2}} + \frac{1}{m^{2}} + \frac{1}{n^{2}} = k^{2}

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\frac{1}{l^{2}} + \frac{1}{m^{2}} + \frac{1}{n^{2}} = \frac{1}{k^{2}}

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l^{2} + m^{2} + n^{2} = \frac{1}{k^{2}}

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l^{2} + m^{2} + n^{2} = k^{2}

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Solution

The correct option is B $\frac{1}{l^{2}} + \frac{1}{m^{2}} + \frac{1}{n^{2}} = \frac{1}{k^{2}}$
Equation of plane is $\frac{x}{l} + \frac{y}{m} + \frac{z}{n} = 1$
Perpendicular distance from origin $O (0, 0, 0)$ is $k$
$∴ k = \frac{1}{\sqrt{\frac{1}{l^{2}} + \frac{1}{m^{2}} + \frac{1}{n^{2}}}}$
$\Rightarrow \frac{1}{l^{2}} + \frac{1}{m^{2}} + \frac{1}{n^{2}} = \frac{1}{k^{2}}$

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Similar questions

{lim}_{n \to \infty} ⎡ ⎢ ⎢ ⎣ \frac{1}{n} + \frac{1}{\sqrt{n^{2} - 1^{2}}} + \frac{1}{\sqrt{n^{2} - 2^{2}}} + . . . . + \frac{1}{\sqrt{n^{2} - {(n - 1)}^{2}}} ⎤ ⎥ ⎥ ⎦

lim n \to \infty (\frac{1}{\sqrt{n^{2}}} + \frac{1}{\sqrt{n^{2} - 1^{2}}} + \frac{1}{\sqrt{n^{2} - 2^{2}}} + \dots + \frac{1}{\sqrt{n^{2} - (n - 1)^{2}}})

m = cos A - sin A

n = cos A + sin A

\frac{m^{2} + n^{2}}{m^{2} - n^{2}} = - \frac{1}{2} sec A . cos e c A = - \frac{(cot A + tan A)}{2}

m^{2} + {m^{'}}^{2} + 2 m m^{'} cos θ = 1

n^{2} + {n^{'}}^{2} + 2 n n^{'} cos θ = 1

m n + m^{'} n^{'} + (m n^{'} + m^{'} n) = cos θ = θ

m^{2} + n^{2} = {csc}^{2} θ

lim n \to \infty {\frac{1}{n} + \frac{1}{\sqrt{n^{2} - 1^{2}}} + \frac{1}{\sqrt{n^{2} - 2^{2}}} + . . . + \frac{1}{\sqrt{n^{2} - {(n - 1)}^{2}}}}

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