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If vectros $ \overrightarrow a = 2\hat i + 2\hat j +3\hat k , \overrightarrow b = -\hat i + 2\hat j +\hat k \: and \: \overrightarrow c = 3\hat i + \hat j $ are such that \( \overrightarrow a +\lambda \overrightarrow b\) is perpendicular to \( \overrightarrow c \), then find the value of \( \lambda \)

$\begin{array}{1 1}\text{a. 8} \\ \text{b. 6} \\ \text{c. 9} \\ \text{d. 5} \end{array} $

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1 Answer

Let $ \overrightarrow a = 2\hat i + 2\hat j +3\hat k , \overrightarrow b = -\hat i + 2\hat j +\hat k \: and \: \overrightarrow c = 3\hat i + \hat j $
$ \lambda \overrightarrow b = - \lambda \hat i + 2 \lambda \hat j + \lambda \hat k$
$ \overrightarrow a + \lambda \overrightarrow b = (2 - \lambda) \hat i + (2 + 2 \lambda) \hat j + (3 + \lambda) \hat k$
$ (\overrightarrow a + \lambda \overrightarrow b) . \overrightarrow c = 0$
$ (2 - \lambda) 3 + (2 + 2\lambda) = 0$
$ 6 - 3\lambda + 2 + 2\lambda = 0$
$ 8 - \lambda = 0$
$ 8 = \lambda$
answered Feb 16 by suganyadevi.p
 

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