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    				| 1. | The shortest distance between the lines `2x+y+z-1=0=3x+y+2z-2` and `x=y=z`, isA. `1/sqrt(2)` unitsB. `sqrt(2)` unitsC. `3/sqrt(2)` unitsD. `sqrt(3)/2` units | 
| Answer» Correct Answer - A 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)`. | |