CBSE 2022 · Region 5 · Set 3 · Q10 · 3 marks
Find the distance of the point ($\displaystyle 2$, $\displaystyle 3$, $\displaystyle 4$) measured along the line $\displaystyle \frac{\mathrm{x}-4}{3}=\frac{\mathrm{y}+5}{6}=\frac{\mathrm{z}+1}{2}$ from the plane $\displaystyle 3 \mathrm{x}+2 \mathrm{y}+2 \mathrm{z}+5=0$. $\displaystyle 2$ $\displaystyle 2$ $\displaystyle 3$ $\displaystyle 3$ $\displaystyle 3$ $\displaystyle 3$ $\displaystyle 3$ $\displaystyle 3$ [
Marking-scheme solution
Equation of line through ($\displaystyle 2$, $\displaystyle 3$, $\displaystyle 4$) and parallel to the given line is
\[\frac{x-2}{3}=\frac{y-3}{6}=\frac{z-4}{2}
\]
Any point on this line is \(\displaystyle (3 \lambda+2,6 \lambda+3,2 \lambda+4)\)
If it lies on the plane , then
\[\begin{gathered}
3(3 \lambda+2)+2(6 \lambda+3)+2(2 \lambda+4)+5=0 \\
\Rightarrow \lambda=-1
\end{gathered}
\]
Hence, the point is (-$\displaystyle 1$, -$\displaystyle 3$, $\displaystyle 2$)
Distance \(\displaystyle =\sqrt{9+36+4}=7\)
Three Dimensional GeometryEquation of a Line in SpaceApplyshort_answerhard
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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.