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+4 votes
156k views
in Mathematics by (80.9k points)
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Find the distance of the point (1,-2,3) from the plane x -y + z = 5 measured parallel to the line

x-1/2 = y-3/3 = z+2/-6

2 Answers

+2 votes
by (17.0k points)
selected by
 
Best answer

Let P(1, - 2, 3).

The d,r.'s of \(\frac{x - 1}2 = \frac{y - 3}3 = \frac {z + 2}{-6}\) are 2, 3, -6.

Since parallel lines have proportionate d.r.'s so, equation of the line through P(1, -2, 3) and parallel to the given line is:

\(\frac{x - 1}2 = \frac{y +2}3 = \frac {z -3}{-6} = \lambda\)    .....(i)

Co-ordinates of any random point on line (i) is Q(2λ + 1, 3λ - 2, 3 - 6λ).

If Q lies on the given equation of plane x - y + z = 5 then, we have

(2λ + 1) - (3λ - 2) + (3 - 6λ) = 5

or λ = \(\frac 17\)

So, co-ordinates of the point Q are Q\(\left(\frac 97, - \frac{11}7, \frac {15}7\right)\).

\(\therefore\) Required distance

PQ = \(\sqrt{\left(\frac 97 - 1\right)^2+\left(-\frac{11}7 + 2\right)^2 + \left(\frac{15}7 - 3\right)^2}\) 

= 1 unit

+6 votes
by (79.8k points)

Let P(1, -2, 3).

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