Q. 265.0( 1 Vote )

# Find the image of the point (2, -1, 5) in the line

Given: Point (2, -1, 5)

Equation of line

The equation of line can be re-arranged as

The general point on this line is

(10r + 11, -4r – 2, -11r – 8)

Let N be the foot of the perpendicular drawn from the point P(2, 1, -5) on the given line.

Then, this point is N(10r + 11, -4r – 2, -11r – 8) for some fixed value of r.

D.r.’s of PN are (10r + 9, -4r - 3, -11r - 3)

D.r.’s of the given line is 10, -4, -11.

Since, PN is perpendicular to the given line, we have,

10(10r + 9) – 4(-4r – 3) – 11(-11r – 3) = 0

100r + 90 + 16r + 12 + 121r + 33 = 0

237r = 135

r

Then, the image of the point is

Therefore, the image is (0, 5, 1).

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