If the distance between the plane αx-2y+z=k and the plane containing the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and \(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\) is \(\sqrt{6}\), then|k| is :
- 36
- 12
- 6
- 2\(\sqrt3\)
The Correct Option is C
Solution and Explanation
vector (\(n_1\times n_2\))=|(vector i, vector j, vector k),(2,3,4),(3,4,5)|=-vector i+2vectorj-vector k
\(\bar{n_1}\times \bar{n_2}=\begin{vmatrix} \vec{i} &\vec{j} &\vec{k} \\ 2&3 &4 \\ 3&4 &5 \end{vmatrix}=-\vec{i}+2\vec{j}-\vec{k}\)
Hence the equation of the plane containing the given lines is
-1(x-1)+2(y-2)-1(z-3)=0
\(\Rightarrow\) x-2y+z=0
If a=1, The distance between the plane is \(\frac{|d|}{\sqrt{1^2+2^2+1^2}}=\sqrt6\)
Thus |d|=6
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Concepts Used:
Angle between Two Planes
A plane is demarcated as two-dimensional in nature and the one which has a flat surface that prolongs infinitely far in two dimensions. It is set up by some stack of lines that are kept together.
Angles between two planes refer to the acute angle which is manifest by the standard vectors of the planes. If the standard vectors of two planes are rectangular, then we can say they are perpendicular. This specific portion tells us what a plane is and what is the angle between the two planes, it also shows us how to calculate the angles between the two planes in a Cartesian plane.
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