The equation of the planes parallel to the plane x $-$ 2y + 2z $-$ 3 = 0 which are at unit distance from the point (1, 2, 3) is ax + by + cz + d = 0. If (b $-$ d) = k(c $-$ a), then the positive value of k is :
Answer (integer)
4
Solution
The equation of the planes parallel to the plane x $-$ 2y + 2z $-$ 3 = 0
<br><br>$x - 2y + 2z + \lambda = 0$<br><br>Now given<br><br>$d = {{\left| {1 - 4 + 6 + \lambda } \right|} \over {\sqrt 9 }} = 1$<br><br>$\left| {\lambda + 3} \right| = 3$<br><br>$\lambda + 3 = \pm 3 \Rightarrow \lambda = 0, - 6$<br><br>So planes are : $x - 2y + 2z - 6 = 0$<br><br>and $x - 2y + 2z = 0$<br><br>$b - d = - 2 + 6 = 4$<br><br>$c - a = 2 - 1 = 1$<br><br>$\therefore$ ${{b - d} \over {c - a}} = k$<br><br>$\Rightarrow k = 4$
About this question
Subject: Mathematics · Chapter: Three Dimensional Geometry · Topic: Direction Cosines and Ratios
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