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Find $λ$ and $μ$ if $ ( 2\hat i + 6\hat j + 27\hat k)\times (\hat i + λ\hat j + μ\hat k) = \overrightarrow 0$

$\begin{array}{1 1}(A) \lambda=3,\:and\:\mu=\large\frac{27}{2} \\(B) \lambda=\large\frac{27}{2},\:and\: \normalsize \mu=3 \\ (C) \lambda=-3,\:and\:\mu=\large\frac{27}{2} \\(D) \lambda=3,\:and\:\mu=-\large\frac{27}{2} \end{array} $

1 Answer

  • If two vectors are parallel then $\overrightarrow a\times \overrightarrow b$=0.
  • If two vectors are parallel then their coefficients are proportional.
Step 1:
Let $\overrightarrow a=2\hat i+6\hat j+27\hat k$
$\quad\overrightarrow b=\hat i+\lambda\hat j+\mu\hat k$
It is given that $\overrightarrow a\times\overrightarrow b=0$
$\Rightarrow (2\hat i+6\hat j+27\hat k)\times (\hat i+\lambda\hat j+\mu\hat k)=0$
$\overrightarrow a\times\overrightarrow b=0$,it implies $\overrightarrow a$ is parallel to $\overrightarrow b$
Step 2:
Now equating the coefficients of the like terms
(i.e)$\large\frac{2}{1}=\frac{6}{\lambda}$$\Rightarrow \lambda=3.$
Similarly $\large\frac{2}{1}=\frac{27}{\mu}$$\Rightarrow \mu=\large\frac{27}{2}.$
Step 3:
$\lambda=3$ and $\mu=\large\frac{27}{2}$
answered May 22, 2013 by sreemathi.v

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