Question

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 \(|\vec{a} \times \vec{b}| = r\) is possible if :

• A. \(0 \lt r \lt \sqrt{\frac{3}{2}}\)
• B. \(3\sqrt{\frac{3}{2}} \lt r \lt 5\sqrt{\frac{3}{2}}\)
• C. \(r \ge 5\sqrt{\frac{3}{2}}\)
• D. \(\sqrt{\frac{3}{2}} \lt r \le 3\sqrt{\frac{3}{2}}\)

Hint:

Expand the cross product, write \(|\vec{a}\times\vec{b}|^2\) as a quadratic in \(x\) and find its minimum value.

Answer 1

The Correct Option is C

\(\vec{a}\times\vec{b} = \begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&2&x\\1&-1&1\end{vmatrix} = (2+x)\hat{i} + (x-3)\hat{j} - 5\hat{k}\)
\(r^2 = (x+2)^2 + (x-3)^2 + 25 = 2x^2 - 2x + 38 = 2\left(x - \frac{1}{2}\right)^2 + \frac{75}{2}\)
Since \(x\) is real, the minimum of \(r^2\) is \(\frac{75}{2}\) (at \(x = \frac12\)) and \(r^2\) can be any larger value.
So \(r \ge \sqrt{\frac{75}{2}} = 5\sqrt{\frac{3}{2}}\).
Answer: (3) \(r \ge 5\sqrt{\frac{3}{2}}\)