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Question

# A perpendicular is drawn from the point P(2,4,−1) to the line x+51=y+31=z−6−9. The equation of the perpendicular from P to the given line is

A
x29=y49=z+12
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B
x+26=y43=z+12
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C
x+26=y43=z+12
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D
x+26=y+43=z+12
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Solution

## The correct option is A x−29=y−49=z+1−2Any general point on the line is given by x=λ−5y=λ−3z=−9λ+6Direction ratios of the line joining this general point with the given point P(2,4,−1) are given by ((λ−5)−2,(λ−3)−4,(−9λ+6)+1)We need to find out that value of λ for which this line is perpendicular to the given line.Condition for two lines to be perpendicular is that the dot product of their Direction ratios is zero.D.Rs of the given line are (1,1,−9)Taking it's dot product with the required line, we get=> (λ−7+λ−7−9(−9λ+7)=0=> λ=7783Therefore, D.R of line are (50483,50483,−11283)The equation of line passing through (2,4,−1) and having direction ratios (50483,50483,−11283) is given by:x−2504=y−4504=z+1−112On further simplifying we get option A.

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