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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