CBSE 2022 · Region 3 · Set 2 · Q5 · 2 marks
If the distance of the point ($\displaystyle 1,1,1$) from the plane $\displaystyle x-\mathrm{y}+\mathrm{z}+\lambda=0$ is $\displaystyle \frac{5}{\sqrt{3}}$, find the value (s) of $\displaystyle \lambda$.
Marking-scheme solution
Distance of point ($\displaystyle 1$, $\displaystyle 1$, $\displaystyle 1$) from \(\displaystyle x-y+z+\lambda=0\) is
\[\frac{|1+\lambda|}{\sqrt{3}}
\]
\[\frac{|1+\lambda|}{\sqrt{3}}=\frac{5}{\sqrt{3}} \Rightarrow 1+\lambda= \pm 5
\]
\[\lambda=-6 \text { or }+4
\]
Three Dimensional GeometryShortest Distance between Two LinesApplyvery_short_answereasy
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CBSE Class 12 Mathematics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.