Q. 184.9( 7 Votes )

Find the equation

Answer :

We know that equation of plane passing through the line of intersection of planes

(a1x + b1y + c1z + d1) + k(a2x + b2y + c2z + d2) = 0


So equation of plane passing through the line of intersection of planes


x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0 is


x + 2y + 3z – 4 + k(2x + y – z + 5) = 0


x(1 + 2k) + y(2 + k) + z(3 – k) – 4 + 5k = 0



as given that x–intercept is twice of z intercept


so


3 – k = 2(1 + 2k)


3 – k = 2 + 4k


5k = 1



Put this value in equation (1)


x(1 + 2k) + y(2 + k) + z(3 – k) – 4 + 5k = 0


x(1 + ) + y(2 + ) + z(3 – ) – 4 + = 0


x() + y() + z() – 3 = 0


multiply by 5


7x + 11y + 14z = 15 …… (2)


And equation of the plane passing through the point is


a1(x – x1) + b1(y – y1) + c1(z – z1) = 0


and point is (2,3, – 1) so that,


a1(x – 2) + b1(y – 3) + c1(z + 1) = 0


by equation (2) a1 = 7,b1 = 11,c1 = 14


so 7(x – 2) + 11(y – 3) + 14(z + 1) = 0


7x + 11y + 14z – 14 – 33 + 14 = 0


7x + 11y + 14z – 33 = 0


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