Let \vec{a}=3\hat{i}+2\hat{j}+x\hat{k} 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 : 


 

  • Option 1)

    \sqrt{\frac{3}{2}}<r\leq 3\sqrt{\frac{3}2{}}

  • Option 2)

    r\geq 5\sqrt{\frac{3}{2}}

  • Option 3)

    0<r\leq \sqrt{\frac{3}{2}}

     

  • Option 4)

    3\sqrt{\frac{3}{2}}<r<5\sqrt{\frac{3}{2}}

 

Answers (1)

\vec{a}=3i+2j+xk

\vec{b}=i-j+k

\left | \vec{a}\times \vec{b} \right |=\begin{vmatrix} i &j &k \\ 3 &2 &x \\ 1 &-1 &1 \end{vmatrix}

               =i\left ( 2-x \right )-j(3-x)+k(-3-2)

                =\left | (2+x)i+j(x-3)+(-5)k \right |

        r= \sqrt{(x-2)^{2}+(x-3)^{2}+25}

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

      r= \sqrt{2x^{2}-2x+38}

\frac{\mathrm{d} r}{\mathrm{d} x}=0=\frac{1}{2}\frac{4x-2}{\sqrt{2x^{2}-2x+38}}=0

x=\frac{1}{2}

\frac{\mathrm{d} ^{2}r}{\mathrm{d} x^{2}}>0 hence r has minimum at x=\frac{1}{2}

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

r=\sqrt{0.5-1+38}

r=\sqrt{37.5}

\Rightarrow r\geqslant 5\sqrt{\frac{3}{2}}


Option 1)

\sqrt{\frac{3}{2}}<r\leq 3\sqrt{\frac{3}2{}}

Option 2)

r\geq 5\sqrt{\frac{3}{2}}

Option 3)

0<r\leq \sqrt{\frac{3}{2}}

 

Option 4)

3\sqrt{\frac{3}{2}}<r<5\sqrt{\frac{3}{2}}

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