Find the equation of the plane which passes through the line of intersection of the planes 2x + 3y - 4z + 5 = 0 and x - 5y + z + 6 = 0 and is parallel to the line x + y - z + 8 = 0 = x - y - z + 2.

Solution:

The equation of a plane which passes through the line of intersection of the planes is

2x + 3y - 4z + 5 = 0

x - 5y + z + 6 = 0

It can be written as

2x + 3y - 4z + 5 + a(x - 5y + z + 6) = 0 where a is a scalar

Using the distributive property

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

d.rs of the plane are (2 + a, 3 - 5a, -4 + a)

Line (1) is parallel to the line represented by the planes x + y - z + 8 = 0 = x - y - z + 2 --- (2)

d.rs of the line represented by (2) is l, m, n

l + m - n = 0

l - m - n = 0

l/0 = m/0 = n/-2 = k where k is a scalar

(l, m, n) = (0, 0, -2k)

As (1) is parallel to (2) then the normal (1) is perpendicular to (2)

0 (2 + a) + 0 (3 - 5a) - 2k (-4 + a) = 0

a = 4

So the plane is 6x - 17y + 0z + 29 = 0.

Therefore, the equation of the plane is 6x - 17y + 0z + 29 = 0.

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Find the equation of the plane which passes through the line of intersection of the planes 2x + 3y - 4z + 5 = 0 and x - 5y + z + 6 = 0 and is parallel to the line x + y - z + 8 = 0 = x - y - z + 2.

Summary:

The equation of the plane which passes through the line of intersection of the planes 2x + 3y - 4z + 5 = 0 and x - 5y + z + 6 = 0 and is parallel to the line x + y - z + 8 = 0 = x - y - z + 2 is 6x - 17y + 0z + 29 = 0.