Q. 254.3( 3 Votes )

Find the coordina

Answer :

The equation of the straight line passing through (3, –4, –5) and (2, –3, 1) is




The coordinates of any point on above line is


(-α+3, α-4, 6α-5).


Equation of the plane passing through (2, 2, 1), (3, 0, 1) and (4, –1, 0)



(x-2)(2-0)-(y-2)(-1-0)+(z-1)(-3+4)=0


2x+y+z-7=0


It is given that (-α+3, α-4, 6α-5) point lies on above plane so,


2(-α+3) +α-4+6α-5-7=0


5α-10=0


α=2


Putting α = 2 in (-α+3, α-4, 6α-5), we get


(-2+3, 2-4, 6× 2-5) = (1, -2, 7)


Hence, (1, -2, 7) is the required coordinate that where the line through (3, –4, –5) and (2, –3, 1) crosses the plane, passing through the points (2, 2, 1), (3, 0, 1) and (4, –1, 0).


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