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Let \(\vec a = 3\hat i + 2\hat j + x\hat k{\rm{\;and\;}}\vec b = \hat i - \hat j + \hat k\), for some real x. Then \(\left| {\vec a \times \vec b} \right| = r\) is possible if:
1. \(\sqrt {\frac{3}{2}} < r \le 3\sqrt {\frac{3}{2}}\)
2. \(r \ge 5\sqrt {\frac{3}{2}}\)
3. \(0 < r \le \sqrt {\frac{3}{2}}\)
4. \(3\sqrt {\frac{3}{2}} < r < 5\sqrt {\frac{3}{2}}\)

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Best answer
Correct Answer - Option 2 : \(r \ge 5\sqrt {\frac{3}{2}}\)

Given, \(\vec a = 3\hat i + 2\hat j + x\hat k{\rm{\;and\;}}\vec b = \hat i - \hat j + \hat k\)

Now, \(\vec a \times \vec b = \left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ 3&2&x\\ 1&{ - 1}&1 \end{array}} \right|\)

\(= \left( {2 + x} \right)\hat i + \left( {x - 3} \right)\hat j - 5\hat k\) 

\(\left| {\vec a \times \vec b} \right| = \sqrt {{{(2 + x)}^2} + {{(x - 3)}^2} + {{( - 5)}^2}} = r\) 

\(\Rightarrow r = \sqrt {4 + {x^2} + 4x + {x^2} + 9 - 6x + 25}\) 

\(= \sqrt {2{x^2} - 2x + 38}\) 

Add \(\frac{1}{2}{\rm{\;and\;}} - \frac{1}{2}\) in above equation, we get

\(\Rightarrow r = \sqrt {2{x^2} - 2x + \frac{1}{2} + 38 - \frac{1}{2}}\) 

\(= \sqrt {2\left( {{x^2} - x + \frac{1}{4}} \right) + 38 - \frac{1}{2}}\) 

\(= \sqrt {2{{\left( {x - \frac{1}{2}} \right)}^2} + \frac{{75}}{2}}\) 

\(\Rightarrow r \ge \sqrt {\frac{{75}}{2}}\) 

\(\Rightarrow r \ge \sqrt {\frac{{25 \times 3}}{2}}\) 

Hence, \(r \ge 5\sqrt {\frac{3}{2}}\)

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