Q. 94.5( 2 Votes )

Mark the co

Answer :

Equation of line passing through the line x + y + z + 3=0 and 2x–y + 3z + 1=0 is given by,


(x + y + z + 3) + k(2x–y + 3z + 1)=0 …………………….(1)


x(1 + 2k) + y(1 - k) + z(1 + 3k) + 3 + k=0 [k is a constant]


Again, the required plane is parallel to the line



So, we should have,


[1×(1 + 2k)] + [2×(1 - k)] + [3×(1 + 3k)]=0


1 + 2k + 2 - 2k + 3 + 9k=0


9k= - 6




Putting in equation (1) we get,



3(x + y + z + 3) - 2(2x–y + 3z + 1)=0


3x + 3y + 3z + 9 - 4x + 2y - 6z - 2=0


- x + 5y - 3z + 7=0


x - 5y + 3z - 7=0


x - 5y + 3z=7


The equation of the plane through the line x + y + z + 3 = 0 = 2x – y + 3z + 1 and parallel to the line


is x - 5y + 3z=7.

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