Question

The two planes x-2y+4z = 10 and 18 x + 17y + kz = 50 are perpendicular ,if k is equal to

Answer 1

$\text{The normal vector to the plane x-2y+4z-10=0 is}\;\overrightarrow{n_1}=\overrightarrow{i}-2\overrightarrow{j}+4\overrightarrow{k}$

$\text{The normal vector to the plane 18x+17y+kz-50=0 is}\;\overrightarrow{n_2}=18\overrightarrow{i}+17\overrightarrow{j}+k\overrightarrow{k}$

$\text{Since the planes are perpendicular, we have}$

$\overrightarrow{n_1}.\overrightarrow{n_2}=0$

$(\overrightarrow{i}-2\overrightarrow{j}+4\overrightarrow{k}).(18\overrightarrow{i}+17\overrightarrow{j}+k\overrightarrow{k})=0$

$\implies\;18-34+4k=0$

$\implies\;4k=16$

$\implies\boxed{\bd\;k=4}$

$\therefore\textbf{The value of k is 4}$

Answer 2

Given,

Equations of planes are  x-2y+4z = 10 and 18 x + 17y + kz = 50

and , they are perpendicular.

We have to find the value of k such that the both planes are perpendicular .

Solution :

The normal vector of x-2y+4z = 10 is a⟨1,−2,4⟩

and , the normal vector of 18 x + 17y + kz = 50 is b⟨18,17,k⟩

Planes are perpendicular .  Then ,

a.b = 0

=> ( 1 * 18 ) + ( -2 * 17 ) + ( 4 * k ) = 0

=> 18 - 34 + 4k = 0

=> 4k = 16

=> k = 4

The value of k = 4