Question

find the equations to the line which intersects the lines 2x+y-4=0=y+2z and x+3z-4=0=2x+5z-8 and parallel to the line x/1=y/2=z/3

Answer 1

Answer

Since the plane passes through the line of intersection of the two given planes, the points lying on both the planes also lie on the new plane.

Let x=0, we get −2y+3z=4 and −y+z=3 for both the given planes.

Solving these two equations simultaneously, we get

−2y+3z−4−2(−y+z−3)=0

⇒z=−2 and y=−5

The point becomes (0,−5,−2)

Now, if we take y to be 0, we get

x+3z=4 and x+z=3

Solving them simultaneously, we get x+3z−4−(x+z−3)=0

⇒2z−1=0  or  z=21​ and thus x=25​

The point becomes (25​,0,21​)

The vector joining two of those gives

2i^+6j^​+2k^ and  21​i^−j^​+21​

Normal to the plane is given by their cross product, which gives

(3+2)i^−(1−1)j^​+(−2−3)k^=5i^−5k^

Thus, the equation of the plane can be written as

5x−5z+d=0

Substituting (2,1,0) we get

10−0+d=0  or  d=−10

The equation of the plane is x−z=2.