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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)

S solutionqc

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

$\vec{b}=i-j+k$

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

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

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

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

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

$r= \sqrt{2x^{2}-2x+38}$

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

$x=\frac{1}{2}$

$\frac{\mathrm{d} ^{2}r}{\mathrm{d} x^{2}}>0$ 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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