CBSE 2022 · Region 4 · Set 1 · Q3 · 2 marks
The foot of a perpendicular drawn from the point (-$\displaystyle 2$, -$\displaystyle 1$, -$\displaystyle 3$) on a plane is ($\displaystyle 1$, - $\displaystyle 3$, $\displaystyle 3$). Find the equation of the plane.
Marking-scheme solution
D.R.'s of normal vector are $\displaystyle \langle 3, -2, 6 \rangle$
Equation of plane is :$\displaystyle 3(x-1) - 2(y+3) + 6(z-3) = 0$$\displaystyle \Rightarrow 3x - 2y + 6z = 27$
Equation of plane is :$\displaystyle 3(x-1) - 2(y+3) + 6(z-3) = 0$$\displaystyle \Rightarrow 3x - 2y + 6z = 27$Three Dimensional GeometryEquation of a Line in SpaceApplyvery_short_answermedium
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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.