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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the value of \( \lambda \) so that the line $\large \frac{1-x}{3}=\frac{7y-14}{2\lambda}=\frac{5z-10}{11} $$\: and \; \large\frac{7-7x}{3\lambda}=\frac{y-5}{1}=\frac{6-z}{5}$ are perpendicular to each other.

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Toolbox:
  • If two lines are perpendicular then the sum of the product of their direction ratios is 0
  • (i.e) $a_1a_2+b_1b_2+c_1c_2=0$
Step 1:
Let $L_1:\large\frac{1-x}{3}=\frac{7y-14}{2\lambda}=\frac{5z-10}{11}$ and
$L_2:\large\frac{7-7x}{3\lambda}=\frac{y-5}{1}=\frac{6-z}{5}$
It is given that the two lines are perpendicular to each other.
The two lines can be written in the standard form as
$\large\frac{1-x}{-3}=\frac{y-2}{2\lambda/7}=\frac{z-2}{11/5}$ and
$\large\frac{x-1}{-3\lambda/7}=\frac{y-5}{1}=\frac{z-6}{-5}$
Step 2:
If these lines are perpendicular then,
$-3\large\frac{-3\lambda}+\frac{2\lambda}{7}$$\times 1+\large\frac{11}{5}$$\times -5=0$
$\Rightarrow \large\frac{9\lambda}{7}+\frac{2\lambda}{7}$$-11=0$
$\Rightarrow \large\frac{11\lambda}{7}$$=11$
$\Rightarrow \lambda=7$
answered Sep 23, 2013 by sreemathi.v
 

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