Let $\overrightarrow{\mathbf{a}}$ be a unit vector, $\overrightarrow{\mathbf{b}}=2…

Let $\overrightarrow{\mathbf{a}}$ be a unit vector, $\overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}+3 \hat{\mathbf{k}}$. Then, maximum value of $[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]$ is
  1. $-1$
  2. $\sqrt{10}+\sqrt{6}$
  3. $\sqrt{10}-\sqrt{6}$
  4. $\sqrt{59}$

Solution

Given that, $ \overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}+3 \hat{\mathbf{k}} $ Now, $\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 1 & -1 \\ 1 & 0 & 3\end{array}\right|$ $ \begin{aligned} & =\hat{\mathbf{i}}(3-0)-\hat{\mathbf{j}}(6+1)+\hat{\mathbf{k}}(0-1) \\ & =3 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-\hat{\mathbf{k}} \end{aligned} $ $ \begin{aligned} & \text { Now, }\left[\begin{array}{lll} \overrightarrow{\mathbf{a}} & \overrightarrow{\mathbf{b}} & \mathbf{c} \end{array}\right]=\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}) \\ & =|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}| \cos \theta \\ & =1\left(\sqrt{3^2+7^2+1^2}\right) \cos \theta \\ & =\sqrt{59} \cos \theta \\ & \Rightarrow[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]_{\max }=\sqrt{59} \cdot 1 \\ & \end{aligned} $ $(\because$ maximum value of $\cos \theta$ is 1 ) Hence, maximum value is $\sqrt{59}$

Asked in: AP EAMCET 2008

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