Q. 34.4( 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)


Required plane is passing through (0, 7, – 7) so


a(x – 0) + b(y – 7) + c(z + 7) = 0


ax + b(y – 7) + c(z + 7) = 0 …… (2)


plane (2) also contain line so, it passes through point (– 1, 3, – 2),


a(– 1) + b(3 – 7) + c(– 2 + 7) = 0


– a – 4b + 5c = 0 …… (3)


Also plane (2) will be parallel to line so,


a1a2 + b1b2 + c1c2 = 0


(a)(– 3) + (b)(2) + (c)(1) = 0


– 3a + 2b + c = 0 …… (4)


Solution (3) and (4) by cross – multiplication,





a = – 14, b = – 14, c = – 14


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


ax + b(y – 7) + c(z + 7) = 0


(– 14)x + (– 14)(y – 7) + (– 14)(z + 7) = 0


Dividing by (– 14) we get


x + y – 7 + z + 7 = 0


x + y + z = 0


so, equation of plane containing the given point and line is x + y + z = 0


the other line is


so, a1a2 + b1b2 + c1c2 = 0


(1)(1) + (1)(– 3) + (1)(2) = 0


1 – 3 + 2 = 0


0 = 0


LHS = RHS


So, lie on plane x + y + z = 0


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