Question

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 :

• A. 36
• B. 12
• C. 6
• D. 2$\sqrt3$

Answer 1

Answer: (C)

6

Answer 2

The correct answer is option (C) : 6
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