Exams Joint Entrance Examination Mathematics 3D Geometry Question 461 Question 461 of 521 Medium MCQ
If the line, \[{{x - 3} \over 1} = {{y + 2} \over { - 1}} = {{z + \lambda } \over { - 2}}\] lies in the plane, 2x-4y+3z=2, then the shortest distance between this line and the line, \[{{x - 1} \over {12}} = {y \over 9} = {z \over 4}\] is :
Solutions · 1 Post your solution Official solution
Point (3, - 2, - \[\lambda \]) on p line 2x - 4y + 3z - 2 = = = 0
= = = 6 + 8 - 3\[\lambda \] - 2 = 0
= = = 3\[\lambda \] = = = 12
\[\lambda \] = = = 4
Now,
\[{{x - 3} \over 1} = {{y + 2} \over { - 1}} = {{z + 4} \over { - 2}} = {k_1}\] . . .(i)
\[{{x - 1} \over {12}} = {y \over 9} = {z \over 4} = {k_2}\] . . .(ii)
Point on equation (i) P (k1 + 3, - k1 - 2, - 2k1 - 4)
Point on equation (ii) Q(12k2 + 1, 9k2, 4k2)
k1 + 3 = = = 12k2 + 1 \[\left| { - {k_1} - 2 = 9{k_2}} \right|\] - 2k1 - 4 = = = 4k2
k2 = = = 0
k1 = = = - 2
p (1, 0, 0) lie on equation of a line 1
gives shortest distance = = = 0 Answer the question to unlock