Q. 295.0( 1 Vote )

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 + y + z = 1 and 2x + 3y + 4z = 5 is


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


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


so,
as given that twice of its y intercept is equals to the three times its z intercept


so


2(1 + 4k) = 3(1 + 3k)


2 + 8k = 3 + 9k


k = – 1


put this in equation (1)


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


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


– x – 2y – 3z + 4 = 0


x + 2y + 3z = 4

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