Q6:please answer the question that i posted second time because the prevous answ

Question

Q6:please answer the question that i posted second time because the prevous answer incorect

A third plane can be found that passes through the line of intersection of two existing planes

a) Two planes are given by equations 2x-3y+z-6=0 and -3x+5y+4z+6=0 . find the scalar equation of the plane that passes through the line of intersection of those two planes and also passes through the point (1,3.-1)

b) Give the equations of two planes .Create a third plane that passes through the line of intersection of the original two and which is parallel to the z-axis .Explain your resoning and include a LanGraph of your planes

Explanation / Answer

Solution:

a) Each plane through the line of intersection of the given two planes has the following equation:

m(2x - 3y + z - 6) + n(-3x + 5y + 4z + 6) = 0,

here m, n are arbitrary real parameters, or, assuming n 0 and denoting the ratio m/n = k:

k(2x - 3y + z - 6) - 3x + 5y + 4z + 6 = 0, or
(2k - 3)x + (5 - 3k)y + (4 + k)z + 6 - 6k = 0.

The coordinates (1, 3, -1) must satisfy the latter equation in order the given point to lie on the plane, So

2k - 3 + 15 - 9k - 4 - k + 6 - 6k = 0,
or -14k + 14 = 0, k = 1,

so the required plane equation is

1*(2x - 3y + z - 6) - 3x + 5y + 4z + 6 = 0

-x + 2y + 5z = 0, or

x - 2y - 5z = 0.

b) To be parallel to the z-axis is necessary the z-coefficient in the plane equation to be 0, So

4 + k = 0 i.e. k = -4, and the equation becomes

-4*(2x - 3y + z - 6) - 3x + 5y + 4z + 6 = 0

-11x + 17y + 30 = 0, or

11x - 17y - 30 = 0

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