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Home  >>  CBSE XII  >>  Math  >>  Model Papers
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Find the value of \( \lambda \) for which the vectors \( \overrightarrow \alpha = 3\hat i + \hat j - 2\hat k \: and \: \overrightarrow b = \hat i + \lambda\hat j - 3\hat k \) are perpendicular to each other.

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  • If the vectors are $\perp$ ,$\overrightarrow a.\overrightarrow b=0$
  • $\hat i.\hat i=1$
  • $\hat j.\hat j=1$
  • $\hat k.\hat k=1$
$\overrightarrow a.\overrightarrow b=(3\hat i+\hat j-2\hat k).(\hat i+\lambda \hat j-3\hat k)$
$\qquad\;=3+\lambda+6$
Since the vectors are $\perp$,then $\overrightarrow a.\overrightarrow b=0$
$\Rightarrow 3+\lambda+6=0$
$\lambda=-9$
answered Oct 3, 2013 by sreemathi.v
 

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