The shortest distance between the lines {{x - 1} \over 0} = {{y + 1} \over { - 1}} = {z \over 1} and x + y + z + 1 = 0, 2x – y + z + 3 = 0 is :

Question

Accepted Answer

Answer: D. {1 \over {\sqrt 3 }} Plane through line of intersection is x + y + z + 1 + \lambda (2x –y + z + 3) = 0 It should be parallel to given line {{x - 1} \over 0} = {{y + 1} \over { - 1}} = {z \over 1} 	herefore 0(1 + 2\lambda ) - 1(1 - \lambda ) + 1(1 + \lambda ) = 0 \Rightarrow \lambda = 0 	herefore Required Plane : x + y + z + 1 = 0 Shortest distance of (1, –1, 0) from this plane = {{\left| {1 - 1 + 0 + 1} ight|} \over {\sqrt {{1^2} + {1^2} + {1^2}} }} = {1 \over {\sqrt 3 }}

The shortest distance between the lines $${{x - 1} \over 0} = {{y + 1} \over { - 1}} = {z \over 1}$$ and x + y + z + 1 = 0, 2x – y + z + 3 = 0 is :

Answer

Explanation

More JEE Math questions