Question
If $\vec a = 2\hat i + 2\hat j + 3\hat k,\vec b = - \hat i + 2\hat j + \hat k$ and $\vec c = 3\hat i + \hat j$ are such that $\vec a + \lambda \vec b$ is perpendicular to $\vec c$, then find the value of $\lambda $
Now $\vec a + \lambda \vec b = 2\hat i + 2\hat j + 3\hat k + \lambda \left( { - \hat i + 2\hat j + \hat k} \right)$=
$ = 2\hat i + 2\hat j + 3\hat k - \lambda \hat i + 2\lambda \hat j + \lambda \hat k$
$\Rightarrow \vec a + \lambda \vec b = \left( {2 - \lambda } \right)\hat i + \left( {2 + 2\lambda } \right)\hat j + \left( {3 + \lambda } \right)\hat k$
Again, $\vec c = 3\hat i + \hat j = 3\hat i + \hat j + 0\hat k$
Since, $\left( {\vec a + \lambda \vec b} \right)$ is perpendicular to $\vec c$ therefore, $\left( {\vec a + \lambda \vec b} \right).\vec c = 0$
$\Rightarrow 6 - 3\lambda + 2 + 2\lambda = 0 \Rightarrow - \lambda + 8 = 0$
$ \Rightarrow - \lambda = - 8 \Rightarrow \lambda = 8$
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