The equation of the plane containing the straight line \[{x \over 2} = {y \over 3} = {z \over 4}\] and perpendicular to the plane containing the straight lines \[{x \over 3} = {y \over 4} = {z \over 2}\] and \[{x \over 4} = {y \over 2} = {z \over 3}\] is :
Solutions · 1
Official solution
Vector \[ \bot \] to given plane = \[\left| {i34j42k23 } \right|\]
= \[\widehat i\left( {12 - 4} \right) - \widehat j\left( {9 - 8} \right) + \widehat k\left( {6 - 16} \right)\]
= \[8\widehat i - \widehat j - 10\widehat k\,\] . . . . (1)
Vector parallel to given line
= \[2\widehat i + 3\widehat j + 4\widehat k\,\,\,\,\] . . . (2)
Vector \[ \bot \,\,\,\] to both (1) \& (2) vector
= \[\left| {i82j−13k−104 } \right|\]
= \[\widehat i\left( { - 4 + 30} \right) - \widehat j\left( {32 + 20} \right) + \widehat k\left( {24 + 2} \right)\]
= \[26\widehat i - 52\widehat j + 26\widehat k\]
Dr's of normal of required plane is
(26, -52, 26) \[ \Rightarrow \] (1, -2, 1)
Equation of plane whose Dr's of Normal is (1, -2, 1) and passes through origin
1.(x - 0) - 2(y - 0) + 1.(z - 0) = 0
x - 2y + z = 0