Q. 134.3( 6 Votes )

# Find the distance

Answer :

Given line, Let, x = 3k+2 , y = 4k-1 ,z = 12k+2

Coordinates of any point on the line are (3k+2 , 4k-1 , 12k+2).

The point of intersection of the line and the plane x – y + z = 5 will be in the form (3k+2 , 4k-1 , 12k+2).

we have, plane x – y + z = 5

3k+2 – (4k-1)+ 12k+2 =5

3k+2 – 4k+1+ 12k+2-5 =0

11k = 0

k = 0

Hence, the point of intersection of the line and the plane x – y + z = 5 is (2 , -1 , 2).

The required distance between the (-1, -5, -10) and (2 , -1 , 2)

= = =13 units

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