Q. 44.0( 5 Votes )

Find the eq

Answer :

we know that equation of plane passing through (x1, y1, z1) is given by

a(x – x1) + b(y – y1) + c(z – z1) = 0 …… (1)


since, required plane contain lines


and


So, required plane passes through (4, 3, 2) and (3, – 2, 0) so equation of required plane is


a(x – 4) + b(y – 3) + c(z – 2) = 0 …… (2)


plane (2) also passes through (3, – 2, 0), so


a(3 – 4) + b(– 2 – 3) + c(0 – 2)


– a – 5b – 2c = 0


a + 5b + 2c = 0 …… (3)


now plane (2) is also parallel to line with direction ratios 1, – 4, 5 so,


a1a2 + b1b2 + c1c2 = 0


(a)(1) + (b)(– 4) + (c)(5) = 0


a – 4b + 5c = 0 …… (4)


solving equation (3) and (4) by cross – multiplication,





Multiplying by 3,



a = 11 λ, b = – λ, c = – 3 λ


put a, b, c in equation (2),


a(x – 4) + b(y – 3) + c(z – 2) = 0


(11 λ)(x – 4) + (– λ)(y – 3) + (– 3 λ)(z – 2) = 0


11 λx – 44 λ – λy + 3 λ – 3 λz + 6 λ = 0


11 λx – λy – 3 λz – 35 λ = 0


Dividing by λ,


11x – y – 3z – 35 = 0


So, equation of required plane is 11x – y – 3z – 35 = 0


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