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Question

Question: The equation of the perpendicular from the point \((\alpha,\beta,\gamma)\) to the plane \(ax + by + ...

The equation of the perpendicular from the point (α,β,γ)(\alpha,\beta,\gamma) to the plane ax+by+cz+d=0ax + by + cz + d = 0is

A

a(xα)+b(yβ)+c(zγ)=0a(x - \alpha) + b(y - \beta) + c(z - \gamma) = 0

B

xαa=yβb=zγc\frac{x - \alpha}{a} = \frac{y - \beta}{b} = \frac{z - \gamma}{c}

C

a(xα)+b(yβ)+c(zγ)=abca(x - \alpha) + b(y - \beta) + c(z - \gamma) = abc

D

None of these

Answer

xαa=yβb=zγc\frac{x - \alpha}{a} = \frac{y - \beta}{b} = \frac{z - \gamma}{c}

Explanation

Solution

It is a line passing through (α,β,γ)( \alpha , \beta , \gamma ) and whose direction cosines are a,b,ca , b , c