If L1 is the line of intersection of the planes 2x−2y+3z−2=0, x−y+z+1=0, and L2 is the line of intersection of the planes x+2y−z−3=0, 3x−y+2z−1=0, then the distance of the origin from the plane containing the lines L1 and L2 is:
Solutions · 1
Official solution
L1 is the line of intersection of plane 1 and plane 2.
L2 is the line of intersection of plane 3 and plane 4.
Line L1 and L2 are present on plane 5.
Now we have to find the distance of origin from the plane 5.
The plane 5 is passing through the Line L1,
and L1 is the line of intersection of first two planes.
\[\therefore\] Equation of plane 5,
(2x−2y+3z−2) + \[\lambda \left( {x - y + z + 1} \right)0\]
Now find a point on L2 which is line of intersection of
x + 2y - z - 3 = 0 \[\,\,\,\,\] .....(1)
and 3x - y + 2z - 1 = 0 \[\,\,\,\,\] .....(2)
Put x = 0 at both (1) and (2),
2y - z = 3 \[\,\,\,\,.....(3)
-y + 2z = 1 \[\,\,\,\,....\] (4)
by solving (3) and (4) we get
\[y = {7 \over 3},\,z = {5 \over 3}\]
So, the point on line L2 is \[ = \left( {0,{7 \over 3},{5 \over 3}} \right)\]
This point is also present on the plane 5.
So, putting this point on equation of plane 5,
\[ - {{14} \over 5} + 5 - 2 + \lambda \left( { - {7 \over 3} + {5 \over 3}} \right) = 0\]
\[ \Rightarrow - {5 \over 3} + {\lambda \over 3} = 0\]
\[ \Rightarrow \lambda = 5\]
\[\therefore\] equation of plane 5 is
2x - 2y + 3z - 2 + \[\lambda (x - y + z + 1) = 0
\[ \Rightarrow \] 2x - 2y + 3z - 2 + 5 (x - y + z + 1) = 0
\[ \Rightarrow \] 7x - 7y + 8z + 3 = 0
Distance of this plane 5 from (0, 0, 0)
\[ = {3 \over {\sqrt {49 + 49 + 64} }}\]
\[ = {3 \over {\sqrt {162} }}\]
\[ = {3 \over {9\sqrt 2 }}\]
\[ = {1 \over {3\sqrt 2 }}\]