Let a⇀=3i^+2j^+xk^ and b⇀=i^-j¯+k^, for some real x. Then a⇀×b⇀=r is possible if :

a¯×b¯=ijk32x1-11=i(2+x)-j(3-x)+k(-5)a¯×b¯=(2+x)2+(3-x)2+25=rr=2x2-2x+38=2(x2-2012x+14-14)+25 =2(x-12)2+492 ≥72

Let a⇀=3i^+2j^+xk^ and b⇀=i^-j¯+k^, for some real x. Then a⇀×b⇀=r is possible if :

a¯×b¯=ijk32x1-11=i(2+x)-j(3-x)+k(-5)a¯×b¯=(2+x)2+(3-x)2+25=rr=2x2-2x+38=2(x2-2012x+14-14)+25 =2(x-12)2+492 ≥72

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$Let \overset{⇀}{a} = 3 \hat{i} + 2 \hat{j} + x \hat{k} and \overset{⇀}{b} = \hat{i} - \bar{j} + \hat{k}, for some real x . Then |\overset{⇀}{a} \times \overset{⇀}{b}| = r is possible if :$

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

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

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

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

answer is B.

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Detailed Solution

$\begin{array}{rcl} \bar{a} \times \bar{b} & = & |\begin{matrix} i & j & k \\ 3 & 2 & x \\ 1 & - 1 & 1 \end{matrix}| = i (2 + x) - j (3 - x) + k (- 5) \\ |\bar{a} \times \bar{b}| & = & \sqrt{{(2 + x)}^{2} + {(3 - x)}^{2} + 25} = r \\ r & = & \sqrt{2 x^{2} - 2 x + 38} = \sqrt{2 (x^{2} - 20 \frac{1}{2} x + \frac{1}{4} - \frac{1}{4}) + 25} \\ = & \sqrt{2 {(x - \frac{1}{2})}^{2} + \frac{49}{2}} \geq \frac{7}{\sqrt{2}} \end{array}$