Q79 / 160 · TS EAMCET 2008 · Mathematics
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
- A$-1$
- B$\sqrt{10}+\sqrt{6}$
- C$\sqrt{10}-\sqrt{6}$
- Correct answerD$\sqrt{59}$
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, $\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 1 & -1 \\ 1 & 0 & 3\end{array}\right|$
$\begin{aligned}<br />& =\hat{\mathbf{i}}(3-0)-\hat{\mathbf{j}}(6+1)+\hat{\mathbf{k}}(0-1) \\<br />& =3 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-\hat{\mathbf{k}}<br />\end{aligned}$
$\begin{aligned}<br />& \text {Now, }\left[\begin{array}{lll}<br />\overrightarrow{\mathbf{a}} & \overrightarrow{\mathbf{b}} & \mathbf{c}<br />\end{array}\right]=\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}) \\<br />& =|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}| \cos \theta \\<br />& =1\left(\sqrt{3^2+7^2+1^2}\right) \cos \theta \\<br />& =\sqrt{59} \cos \theta \\<br />& \Rightarrow[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]_{\max }=\sqrt{59} \cdot 1<br />\end{aligned}$
$(\because$ maximum value of $\cos \theta$ is 1)
Hence, maximum value is $\sqrt{59}$.
$\overrightarrow{\mathbf{b}}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{c}}=\hat{\mathbf{i}}+3 \hat{\mathbf{k}}$
Now, $\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}=\left|\begin{array}{ccc}\hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 2 & 1 & -1 \\ 1 & 0 & 3\end{array}\right|$
$\begin{aligned}<br />& =\hat{\mathbf{i}}(3-0)-\hat{\mathbf{j}}(6+1)+\hat{\mathbf{k}}(0-1) \\<br />& =3 \hat{\mathbf{i}}-7 \hat{\mathbf{j}}-\hat{\mathbf{k}}<br />\end{aligned}$
$\begin{aligned}<br />& \text {Now, }\left[\begin{array}{lll}<br />\overrightarrow{\mathbf{a}} & \overrightarrow{\mathbf{b}} & \mathbf{c}<br />\end{array}\right]=\overrightarrow{\mathbf{a}} \cdot(\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}) \\<br />& =|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}} \times \overrightarrow{\mathbf{c}}| \cos \theta \\<br />& =1\left(\sqrt{3^2+7^2+1^2}\right) \cos \theta \\<br />& =\sqrt{59} \cos \theta \\<br />& \Rightarrow[\overrightarrow{\mathbf{a}} \overrightarrow{\mathbf{b}} \overrightarrow{\mathbf{c}}]_{\max }=\sqrt{59} \cdot 1<br />\end{aligned}$
$(\because$ maximum value of $\cos \theta$ is 1)
Hence, maximum value is $\sqrt{59}$.
Hence, the correct answer is option: 'D'.