1.

The shortest distance between the lines 2x+y+z-1=0=3x+y+2z-2 and x=y=z, is

Answer»

`1/sqrt(2)` UNITS
`sqrt(2)` units
`3/sqrt(2)` units
`sqrt(3)/2` units

Solution :Any plane passing through FIRST line is `2x+y=z-1+lambda(3x+y+2z-2)=0`
Line x=y=z is passing through the point O(0,0,).
Required shortest distance = distance of O from the member plane of above family which is parallel to the line x=y=z
If plane is parallel to the line,
`(2x+3lambda)1+(1+lambda)1+(1+2lambda)1=0`
`RARR lambda=-2/3`
Equation of plane is
`3(2x+y+z-1)-2(3x+y+2z-2)=0`
or `y-z+1=0`
Its distance from (0,0,0) is `1/sqrt(2)`.


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