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# If the lines $\large\frac{2x-1} {-3} = \frac{y-2}{2k} = \frac{z-3}{2}$ and $\large\frac{x-1}{3k} = \frac{1-2y}{1} = \frac{z-6}{-5}$ are perpendicular, find the value of k.

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• If two lines are $\perp$ then $a_1a_2+b_1b_2+c_1c_2=0$Where $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ are the direction ratios of the two lines.
Step 1:
Let $L_1=\large\frac{2x-1}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}$
$\quad\;\; L_2=\large\frac{x-1}{3k}=\frac{1-2y}{1}=\frac{z-6}{-5}$
Rewriting the equations of $L_1\:and\:L_2$ in standard form we get
Let $L_1=\large\frac{x-1/2}{-3/2}=\frac{y-2}{2k}=\frac{z-3}{2}$
$\quad\;\; L_2=\large\frac{x-1}{3k}=\frac{y-1/2}{-1/2}=\frac{z-6}{-5}$
$\Rightarrow\:$The direction ratios (d.r) of $L_1$ are $(-3/2,2k,2)$ and
The direction ratios of $L_2$ are $(3k,-1/2-5)$
We know that if the lines are $\perp$ then
$a_1a_2+b_1b_2+c_1c_2=0$
Step 2:
Now substituting for $(a_1,b_1,c_1)$ and $(a_2,b_2,c_2)$ we get
$\Rightarrow\:(-3/2)(3k)+(2k)(-1/2)+(2)(-5)=0$
$\Rightarrow\:-9k/2-k-10=0$
$-11k=20$
Therefore $k=-\large\frac{20}{11}$
edited Feb 3, 2014