Find the equation of the plane which contains two parallel lines and
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