Key Concepts and Formulas

• Vector Cross Product: The cross product of two vectors $\vec{A} = A_x\hat{i} + A_y\hat{j} + A_z\hat{k}$ and $\vec{B} = B_x\hat{i} + B_y\hat{j} + B_z\hat{k}$ is given by: $\vec{A} \times \vec{B} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ A_x & A_y & A_z \\ B_x & B_y & B_z \end{vmatrix}$
• Magnitude of a Vector: The magnitude of a vector $\vec{V} = V_x\hat{i} + V_y\hat{j} + V_z\hat{k}$ is $|\vec{V}| = \sqrt{V_x^2 + V_y^2 + V_z^2}$.
• Quadratic Functions: A quadratic function $f(x) = ax^2 + bx + c$ with $a > 0$ has a minimum value at the vertex $x = -\frac{b}{2a}$. The minimum value is $f\left(-\frac{b}{2a}\right)$.

Step-by-Step Solution

Given vectors are $\vec{a} = 3\hat{i} + 2\hat{j} + x\hat{k}$ and $\vec{b} = \hat{i} - \hat{j} + \hat{k}$. We are asked to find the possible range of values for $r = |\vec{a} \times \vec{b}|$.

Step 1: Calculate the Cross Product $\vec{a} \times \vec{b}$ We compute the cross product using the determinant formula to find the components of the resulting vector. $\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & 2 & x \\ 1 & -1 & 1 \end{vmatrix}$ Expanding the determinant: $\vec{a} \times \vec{b} = \hat{i}((2)(1) - (x)(-1)) - \hat{j}((3)(1) - (x)(1)) + \hat{k}((3)(-1) - (2)(1))$ $\vec{a} \times \vec{b} = \hat{i}(2 + x) - \hat{j}(3 - x) + \hat{k}(-3 - 2)$ $\vec{a} \times \vec{b} = (2+x)\hat{i} + (x-3)\hat{j} - 5\hat{k}$ Explanation: This step yields the vector resulting from the cross product, whose components are functions of $x$.

Step 2: Calculate the Magnitude Squared of the Cross Product Let $r = |\vec{a} \times \vec{b}|$. We will calculate $r^2$ to work with a quadratic expression, which is simpler than dealing with the square root. $r^2 = |\vec{a} \times \vec{b}|^2 = (2+x)^2 + (x-3)^2 + (-5)^2$ Expand the squared terms: $r^2 = (4 + 4x + x^2) + (x^2 - 6x + 9) + 25$ Combine like terms: $r^2 = (x^2 + x^2) + (4x - 6x) + (4 + 9 + 25)$ $r^2 = 2x^2 - 2x + 38$ Explanation: Squaring the magnitude allows us to express it as a quadratic function of $x$, which we can then analyze for its minimum value.

Step 3: Find the Minimum Value of $r^2$ Let $f(x) = r^2 = 2x^2 - 2x + 38$. This is a quadratic function of the form $ax^2 + bx + c$, with $a=2$, $b=-2$, and $c=38$. Since the coefficient of $x^2$ is $a=2 > 0$, the parabola opens upwards, and the function has a minimum value. The minimum occurs at the vertex. The x-coordinate of the vertex is given by $x_v = -\frac{b}{2a}$: $x_v = -\frac{-2}{2(2)} = \frac{2}{4} = \frac{1}{2}$ Now, substitute $x_v = \frac{1}{2}$ into the expression for $r^2$ to find the minimum value of $r^2$: $r^2_{min} = 2\left(\frac{1}{2}\right)^2 - 2\left(\frac{1}{2}\right) + 38$ $r^2_{min} = 2\left(\frac{1}{4}\right) - 1 + 38$ $r^2_{min} = \frac{1}{2} - 1 + 38 = 37.5 = \frac{75}{2}$ Explanation: By finding the minimum of the quadratic $r^2$, we determine the smallest possible value the square of the magnitude can take.

Step 4: Determine the Range of $r$ Since $x$ can be any real number, $r^2$ can take any value greater than or equal to its minimum value. Thus, $r^2 \ge \frac{75}{2}$. Since $r$ represents the magnitude of a vector, $r$ must be non-negative ($r \ge 0$). Taking the square root of the inequality for $r^2$: $r \ge \sqrt{\frac{75}{2}}$ We can simplify the square root: $r \ge \sqrt{\frac{25 \times 3}{2}} = 5\sqrt{\frac{3}{2}}$ Therefore, the possible values for $r$ are $r \ge 5\sqrt{\frac{3}{2}}$.

Common Mistakes & Tips

• Sign Errors in Cross Product: Be very careful with the signs when expanding the determinant for the cross product, especially the $\hat{j}$ component.
• Algebraic Expansion: Ensure accurate expansion and combination of terms when simplifying the expression for $r^2$. Errors here can lead to an incorrect quadratic.
• Minimum vs. Maximum: For a quadratic $ax^2+bx+c$, if $a>0$, there is a minimum. If $a<0$, there is a maximum. Always check the sign of the leading coefficient.
• Non-negativity of Magnitude: Remember that a magnitude is always non-negative. When taking square roots, consider only the positive root.

Summary

We calculated the cross product of the given vectors $\vec{a}$ and $\vec{b}$ to obtain a vector whose components depend on $x$. We then found the square of the magnitude of this cross product, which resulted in a quadratic expression in $x$: $r^2 = 2x^2 - 2x + 38$. Analyzing this quadratic, we found its minimum value to be $\frac{75}{2}$ when $x = \frac{1}{2}$. Since $x$ can be any real number, $r^2$ can take any value greater than or equal to $\frac{75}{2}$. Consequently, the magnitude $r$ must satisfy $r \ge \sqrt{\frac{75}{2}} = 5\sqrt{\frac{3}{2}}$. This range corresponds to option (C).

The final answer is $\boxed{C}$.