The unit vector which are perpendicular to both the vectors \(\rm \hat{i}-2\hat{j}+3\hat{k} \)  and \(\rm \hat{2i}+\hat{3j}-\hat{k} \) are , 

  1. \(\rm \frac{1}{\sqrt{3}}\left ( -\hat{i}+\hat{j}+\hat{k} \right )\)
  2. \(\rm \frac{1}{\sqrt{3}}\left ( \hat{i}+\hat{j}+\hat{k} \right )\)
  3. \(\rm \frac{1}{\sqrt{3}}\left ( -\hat{i}+\hat{j}-\hat{k} \right )\)
  4. \(\rm \frac{-1}{\sqrt{3}}\left ( \hat{i}+\hat{j}+\hat{k} \right )\)

Answer (Detailed Solution Below)

Option 1 : \(\rm \frac{1}{\sqrt{3}}\left ( -\hat{i}+\hat{j}+\hat{k} \right )\)
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Detailed Solution

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Concept:

If \(\rm\overrightarrow{c}\) perpendicular to both the vectors \(\rm\overrightarrow{a}\) and  \(\rm\overrightarrow{b}\) then \(\rm \vec c = \vec a \times \vec b\)

Unit vector, \(\rm \hat{c}= \frac{\overrightarrow{c}}{\left | \overrightarrow{c} \right |}\)  

Calculation :

Here the given vectors are, \(\overrightarrow{a}=\rm \hat{i}-2\hat{j}+3\hat{k}\)  and \(\overrightarrow{b}=\rm 2\hat{i}+3\hat{j}-\hat{k}\) , 

Let, vector \(\rm\overrightarrow{c}\) are perpendicular to both the vectors, 

So,  \(\rm \overrightarrow{c}= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k}\\ 1 & -2 & 3\\ 2& 3 & -1 \end{vmatrix}\) 

⇒ \(\rm \overrightarrow{c}= \hat{i}(2-9)-\hat{j}(-1-6)+\hat{k}(3+4)\) 

⇒ \(\rm \overrightarrow{c}= -7\hat{i}+7\hat{j}+7\hat{k}\) 

∴ \(\rm \left | \overrightarrow{c} \right | = \sqrt{(-7)^{2}+7^{2}+7^{2}}\) = \(7\sqrt{3}\) 

Hence the unit vector perpendicular to both the vectors are, 

\(\rm \hat{c}= \frac{\overrightarrow{c}}{\left | \overrightarrow{c} \right |}\) = \(\rm\frac{-7\hat{i}+7\hat{j}+7\hat{k}}{7\sqrt{3}}\) 

⇒ \(\rm \hat{c}= \frac{1}{\sqrt{3}}\left ( -\hat{i}+\hat{j}+\hat{k} \right )\) 

The correct option is 1. 

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