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Home  >>  CBSE XII  >>  Math  >>  Three Dimensional Geometry
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If the lines $\large\frac{x-1} {-3} = \frac{y-2}{2k} = \frac{z-3}{2}$ and $\large\frac{x-1}{3k} = \frac{y-1}{1} = \frac{z-6}{-5}$ are perpendicular, find the value of $k$.

$\begin{array}{1 1} k = \large\frac{-10}{7} \\ k = \large\frac{10}{7} \\ k = \large\frac{-7}{10} \\ k = \large\frac{7}{10}\end{array} $

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Toolbox:
  • 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 cosines of the two lines.
Step 1:
Let $L_1=\large\frac{x-1}{-3}=\frac{y-2}{2x}=\frac{z-3}{2}$
$\quad\;\; L_2=\large\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$
The direction ratios of $L_1$ are $-3,2k$ and $2$
The direction ratios of $L_2$ are $3k,1$ and $-5$
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
$(-3)(3k)+(2k)(1)+(2)(-5)=0$
$-9k+2k-10=0$
$-7k=10$
Therefore $k=\large\frac{-10}{7}$
answered Jun 3, 2013 by sreemathi.v
 

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