Question

Let we have the vectors as $\vec{a}=3\vec{i}+2\vec{j}+x\vec{k}\text{ and }\vec{b}=\vec{i}-\vec{j}+\vec{k}$ for some real X. Then, $\left| \vec{a}\times \vec{b} \right|=r$ is possible if:
\[\begin{align}
  & A.3\sqrt{\dfrac{3}{2}}\text{ }<\text{ }r\text{ }<\text{ }5\sqrt{\dfrac{3}{2}} \\
 & B.0\text{ }<\text{ }r\text{ }<\text{ }\sqrt{\dfrac{3}{2}} \\
 & C.\sqrt{\dfrac{3}{2}}\text{ }<\text{ }r\text{ }<\text{ 3}\sqrt{\dfrac{3}{2}} \\
 & D.r\text{ }\ge \text{ }5\sqrt{\dfrac{3}{2}} \\
\end{align}\]

Answer 1

Hint: Apply cross-multiplication formula in above question and then apply minimum or maximum function rule. For finding minimum or maximum of any given function f (x) = 0, do f'(x) = 0 (i.e. the derivative of function f (x) equals to zero). Here, then we get some value of x. Now, we apply the second derivative test.
Now, once we get the minimum value then introduce the concept of inequality.
Like the given expression will be greater and equal to its minimum value. After doing some minor calculations we will get the final answer.

Complete step-by-step solution:
Now, come to the question, the given vectors are
\[\vec{a}=3\vec{i}+2\vec{j}+x\vec{k}\text{ and }\vec{b}=\vec{i}-\vec{j}+\vec{k}\]
Let us find the cross product as below,
\[\begin{align}
  & \left| \vec{a}\times \vec{b} \right|=\left| \begin{matrix}
   {\vec{i}} & {\vec{j}} & {\vec{k}} \\
   3 & 2 & x \\
   1 & -1 & 1 \\
\end{matrix} \right| \\
 &\Rightarrow \left| \vec{a}\times \vec{b} \right|=\left( 2+x \right)\vec{i}+\left( x-3 \right)\vec{j}+\left( -5 \right)\vec{k} \\
\end{align}\]
Now,
\[\begin{align}
  & \left| \vec{a}\times \vec{b} \right|=\sqrt{{{\left( 2+x \right)}^{2}}+{{\left( x-3 \right)}^{2}}+{{\left( -5 \right)}^{2}}} \\
 &\Rightarrow r=\sqrt{{{\left( 2+x \right)}^{2}}+{{\left( x-3 \right)}^{2}}+25} \\
 & \Rightarrow \sqrt{2\left( {{x}^{2}}-x+19 \right)} \\
\end{align}\]
Now, let \[f\left( x \right)={{x}^{2}}-x+19=0\]
For finding minimum or maximum of f (x) do
\[\begin{align}
  &\Rightarrow f'\left( x \right)=\dfrac{d\left( f\left( x \right) \right)}{dx}=0 \\
 &\Rightarrow 2x-1=0 \\
 &\Rightarrow x=\dfrac{1}{2} \\
 &\Rightarrow f''\left( x \right)=2\text{(+ve)} \\
\end{align}\]
Hence, $x=\dfrac{1}{2}$ for minima.
Here, we have done second derivative test:
i) If $f''\left( a \right)\text{ }>\text{ }0\Rightarrow x=a$ is a point of local minima.
ii) If $f''\left( a \right)\text{ }<\text{ }0\Rightarrow x=a$ is a point of local maxima.
In this way, we can use derivative test for ascertaining maxima/minima.
\[\begin{align}
  & f{{\left( x \right)}_{\text{minimum}}}={{\left( \dfrac{1}{2} \right)}^{2}}-\left( \dfrac{1}{2} \right)+19=\left( \dfrac{75}{4} \right) \\
 &\Rightarrow r\ge \sqrt{2{{\left( {{x}^{2}}-x+19 \right)}_{\text{minimum}}}} \\
 &\Rightarrow r\ge \sqrt{2\times \dfrac{75}{4}} \\
 &\Rightarrow r\ge \sqrt{\dfrac{25\times 3}{2}} \\
 &\Rightarrow r\ge 5\sqrt{\dfrac{3}{2}} \\
\end{align}\]
Therefore, option D $r\ge 5\sqrt{\dfrac{3}{2}}$ is the correct answer.


Note: Before start solving the question, students should focus on the options given. This is one of the crucial habit for getting the idea or approach of the question. Like, in our question, options are in inequality forms then the solution will definitely involve the concept of maxima and minima (90% cases). One other major point is vector product of two vectors (cross product).
If \[\vec{a}={{a}_{1}}\vec{i}+{{a}_{2}}\vec{j}+{{a}_{3}}\vec{k}\text{ and }\vec{b}={{b}_{1}}\vec{i}-{{b}_{2}}\vec{j}+{{b}_{3}}\vec{k}\] then
\[\vec{a}\times \vec{b}=\left| \begin{matrix}
   {\vec{i}} & {\vec{j}} & {\vec{k}} \\
   {{a}_{1}} & {{a}_{2}} & {{a}_{3}} \\
   {{b}_{1}} & {{b}_{2}} & {{b}_{3}} \\
\end{matrix} \right|={{x}_{1}}\vec{i}+{{x}_{2}}\vec{j}+{{x}_{3}}\vec{k}\text{ (let)}\]
Then, \[\left| \vec{a}\times \vec{b} \right|=\sqrt{{{\left( {{x}_{1}} \right)}^{2}}+{{\left( {{x}_{2}} \right)}^{2}}+{{\left( {{x}_{3}} \right)}^{2}}}\]