Question

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

$3 square root of 10$
$10 square root of 3$
$fraction numerator 10 over denominator square root of 3 end fraction$
$20 over 3$

Solution

10 square root of 3

the equation of the line passing through the point (1,5-9 and parallel to x =y=z is

$fraction numerator straight x minus 1 over denominator 1 end fraction space equals space fraction numerator straight y plus 5 over denominator 1 end fraction space equals space fraction numerator straight z minus 9 over denominator 1 end fraction space equals space straight lambda$
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
therefore coordinates of point P are (-9, -15,-1)
Hence, required distance
= $equals square root of left parenthesis 1 plus 9 right parenthesis squared plus left parenthesis negative 5 plus 15 right parenthesis squared plus left parenthesis 9 plus 1 right parenthesis squared end root equals square root of 10 squared plus space 10 squared plus 10 squared end root space equals space 10 square root of 3$

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Integrals

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

$3 square root of 10$
$10 square root of 3$
$fraction numerator 10 over denominator square root of 3 end fraction$
$20 over 3$

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