Question:
Find the equations of the line that intersects the lines
2x + y - 4 = 0 = y + 2z ; x + 3z = 4, 2x + 5z = 8 and passes through the point (2, -1, 1)
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What I tried so far:
As I understand, here, the set of lines 2x + y - 4 = 0 = y + 2z and x + 3z = 4, 2x + 5z = 8 are considered as the 2 separate set of lines, which we need to show that are being intersected.
So, the equation of a general line passing through 2x + y - 4 = 0 = y + 2z is given by 2x + y - 4 + k(y + 2z) =0 for some real number k
Also, equation of a general line passing through can be represented by, x + 3z - 4 + r(2x + 5z - 8) for some real number r.
I'm naming these equations as follows for ease of use.
2x + y - 4 + k(y + 2z) = 0 ----(1)
x + 3z - 4 + r(2x + 5z - 8) = 0 -------(2)
If these lines intersect at any point(s), for some value of k and r, (1) - (2) = 0 should exist.
That is, 2x + y - 4 + k(y + 2z) - [ x + 3z - 4 + r(2x + 5z - 8) ] = 0
====> (1 - 2r)x + y(1 + k)y + ( 2k - 3 - 5r)z - 8r = 0
If I'm to solve this, I can equate coefficients of x, y, z and constants each to 0.
Which gives, (1 - 2r) = 0 and -8r = 0
So, in these case above two equations cannot exist simultaneously.
How can I proceed further with this?
To find the equations of the line that intersects the given lines and passes through the point (2, -1, 1), we need to solve the system of equations formed by equating the coefficients of x, y, z, and the constants to zero.
Let's proceed further:
From the equation (1 - 2r) = 0, we find that 1 - 2r = 0, which gives r = 1/2.
Now, substituting the value of r in equation (2), we get:
x + 3z - 4 + (1/2)(2x + 5z - 8) = 0
Simplifying the equation:
2x + 6z - 8 + 2x + 5z - 8 = 0
4x + 11z - 7 - 16 = 0
4x + 11z - 23 = 0
Therefore, the equation of the line that intersects the given lines and passes through the point (2, -1, 1) is:
4x + 11z - 23 = 0.