Question
Distance between the two planes: $2x + 3y + 4z = 4$ and $4x + 6y + 8z = 12$ is
(A) 2 units
(B) 4 units
(C) 8 units
(D) $\cfrac{2}{{\sqrt {29} }}$ units
Three Dimensional Geometry — Class 12 Maths Solution
Step-by-step Solution
Option D is correct
The given planes are $2x + 3y + 4z = 4$ ….(1)
and $4x + 6y + 8z = 12$ …..(2)
Dividing (2) by 2, $2x + 3y + 4z = 6$
Distance between parallel planes $= \left| {\cfrac{{{d_1} - {d_2}}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|$
$= \left| {\cfrac{{4 - 6}}{{\sqrt {4 + 9 + 16} }}} \right| = \cfrac{2}{{\sqrt {29} }}$ units.
NCERT & Exemplar solution for CBSE Class 12 Mathematics, Three Dimensional Geometry. Curated by Sachin Sharma. Free for all students.