Question

1) Derive the equation of a plane in a normal form (5 marks)

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Answer 1

☆ ΛПƧЩΣЯ ☆

Let L be the line whose perpendicular distance from origin is P, angle XOA =ѳ

Draw AM perpendicular to x-axis.

Then AM=Psinѳ, OM=Pcosѳ

Point A=(Pcosѳ Psinѳ)

Now slope of OA =tanѳ

OA is perpendicular to Line L.

$slope \: of \: line \: l \: = \frac{ - 1}{ \tan(ѳ) } \\ l = \frac{ - \cos(ѳ) }{ \sin(ѳ) }$

Equation of line is

✈ y-y1=m(x-x1) [Point slope form]

$y - p \sin(ѳ) = \frac{ - \cos(ѳ) }{ \sin(ѳ) } (x -pcosѳ) \\ y \sin(ѳ) - p {sin}^{2} ѳ = - x \cos ѳ + p {cos}^{2} ѳ \\ x \cos(ѳ ) + y \sin(ѳ) = p {cos}^{2} ѳ + p {sin}^{2} ѳ \\ x \cos(ѳ) + y \sin(ѳ) = p( {sin}^{2} ѳ + {cos}^{2}ѳ ) \\ x \cos(ѳ) + y \sin(ѳ) = p(1)$

$x \cos(ѳ) + y \sin(ѳ) = p$

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