    Question

# The distance of the point (1, -5, 9) from the plane x - y + z = 5 measured along the line x = y = z is

A

310

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B

103

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C

103

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D

203

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Solution

## The correct option is B 10√3 Equation of line passing through the point (1, -5, 9) and parallel to x = y = z is x−11=y+51=z−91=λ Thus, any point on this line is of the form λ+1,λ−5,λ+9. Now, if P(λ+1,λ−5,λ+9) is the point of intersection of line and plane, then (λ+1)−(λ−5)+λ+9=5 ⇒λ+15=5⇒λ=−10 ∴ Coordintes of point P are (-9, -15, -1). Hence, required distance =√(1+9)2+(−5+15)2+(9+1)2 =√102+102+102=10√3  Suggest Corrections  0      Similar questions
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