Question

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

Answer 1

Answer:

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Answer 2

The distance of the point (1,-5,9) from the plane x-y+z=5 measured along the line x=y=z is 10√3.

• Equation of the line through the point (1,-5,9) along the line x=y=z is

$\frac{x-1}{1} + \frac{y+5}{1} + \frac{z-9}{1} = \lambda$    (Equation 1)

• Let this line intersect the plane x-y+z = 5 at point X.

X is given by X(λ+1,λ-5,λ+9).

X lies on the plane x-y+z = 5.

• Therefore , it will satisfy the equation of plane

λ+1 + λ-5 + λ+9 = 5

λ = -10

Therefore, X = (-9,-15,-1)

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

Dist = $\sqrt{(10^2 + 10^2 + 10^2)}$ = 10√3