Question:

Step 1: Relate vector magnitudes using the cross product. Given the vector relation $\mathbf{c} \times \mathbf{a} = \mathbf{b}$, we can take the magnitude of both sides. The magnitude of the cross product of two vectors is given by $|\mathbf{u} \times \mathbf{v}| = |\mathbf{u}||\mathbf{v}|\sin\theta$, where $\theta$ is the angle between the vectors. Let $\alpha$ be the angle between vectors $\mathbf{c}$ and $\mathbf{a}$. $$|\mathbf{c} \times \mathbf{a}| = |\mathbf{b}|$$ $$|\mathbf{c}||\mathbf{a}|\sin\alpha = |\mathbf{b}|$$ Given $|\mathbf{a}| = 2$ and $|\mathbf{b}| = 3$, we substitute these values: $$|\mathbf{c}|(2)\sin\alpha = 3$$ Step 2: Express $|\mathbf{c}|$ in terms of $\alpha$. From the relation obtained in Step 1, we can isolate $|\mathbf{c}|$: $$|\mathbf{c}| = \frac{3}{2\sin\alpha} = \frac{3}{2}\csc\alpha$$ Step 3: Express the squared magnitude of the difference vector. We need to find the least value of $|\mathbf{c} - \mathbf{a}|$. It is often easier to minimize the square of the magnitude. The formula for the squared magnitude of the difference of two vectors is: $$|\mathbf{c} - \mathbf{a}|^2 = |\mathbf{c}|^2 - 2(\mathbf{c} \cdot \mathbf{a}) + |\mathbf{a}|^2$$ Step 4: Substitute expressions to form an equation in terms of $\alpha$. We substitute the expression for $|\mathbf{c}|$ from Step 2 and the given value $|\mathbf{a}| = 2$ into the equation from Step 3. We also use the definition of the dot product $\mathbf{c} \cdot \mathbf{a} = |\mathbf{c}||\mathbf{a}|\cos\alpha$. $$|\mathbf{c} - \mathbf{a}|^2 = \left(\frac{3}{2}\csc\alpha\right)^2 - 2\left(\frac{3}{2}\csc\alpha\right)(2)\cos\alpha + (2)^2$$ $$|\mathbf{c} - \mathbf{a}|^2 = \frac{9}{4}\csc^2\alpha - 6\csc\alpha\cos\alpha + 4$$ Recall that $\csc\alpha\cos\alpha = \frac{1}{\sin\alpha}\cos\alpha = \frac{\cos\alpha}{\sin\alpha} = \cot\alpha$. $$|\mathbf{c} - \mathbf{a}|^2 = \frac{9}{4}\csc^2\alpha - 6\cot\alpha + 4$$ Step 5: Rewrite the expression in terms of $\cot\alpha$ and identify the quadratic form. Using the trigonometric identity $\csc^2\alpha = 1 + \cot^2\alpha$, we rewrite the expression for $|\mathbf{c} - \mathbf{a}|^2$ entirely in terms of $\cot\alpha$: $$|\mathbf{c} - \mathbf{a}|^2 = \frac{9}{4}(1 + \cot^2\alpha) - 6\cot\alpha + 4$$ $$|\mathbf{c} - \mathbf{a}|^2 = \frac{9}{4} + \frac{9}{4}\cot^2\alpha - 6\cot\alpha + 4$$ $$|\mathbf{c} - \mathbf{a}|^2 = \frac{9}{4}\cot^2\alpha - 6\cot\alpha + \frac{25}{4}$$ Let $x = \cot\alpha$. The expression is a quadratic in $x$: $f(x) = \frac{9}{4}x^2 - 6x + \frac{25}{4}$. Step 6: Determine the value of $\cot\alpha$ for the least value of $|\mathbf{c} - \mathbf{a}|$. The expression $f(x) = \frac{9}{4}x^2 - 6x + \frac{25}{4}$ represents a parabola opening upwards (since the coefficient of $x^2$ is positive, $\frac{9}{4} > 0$). Its minimum value occurs at the vertex. For a quadratic $Ax^2 + Bx + C$, the x-coordinate of the vertex is given by $x = -\frac{B}{2A}$. Here, $A = \frac{9}{4}$ and $B = -6$. $$x = \cot\alpha = -\frac{-6}{2\left(\frac{9}{4}\right)} = \frac{6}{\frac{9}{2}} = \frac{12}{9} = \frac{4}{3}$$ So, $|\mathbf{c} - \mathbf{a}|$ attains its least value when $\cot\alpha = \frac{4}{3}$. Step 7: Calculate $\csc\alpha$ from the obtained $\cot\alpha$. Using the identity $\csc^2\alpha = 1 + \cot^2\alpha$: $$\csc^2\alpha = 1 + \left(\frac{4}{3}\right)^2 = 1 + \frac{16}{9} = \frac{9+16}{9} = \frac{25}{9}$$ Since $\alpha$ is the angle between two vectors, $0 \leq \alpha \leq \pi$, so $\sin\alpha \geq 0$ and thus $\csc\alpha \geq 0$. $$\csc\alpha = \sqrt{\frac{25}{9}} = \frac{5}{3}$$ Step 8: Calculate the value of $|\mathbf{c}|$. Substitute the value of $\csc\alpha$ back into the expression for $|\mathbf{c}|$ from Step 2: $$|\mathbf{c}| = \frac{3}{2}\csc\alpha = \frac{3}{2} \cdot \frac{5}{3} = \frac{5}{2}$$ Step 9: Conclude the final answer. When $|\mathbf{c} - \mathbf{a}|$ attains its least value, the value of $|\mathbf{c}|$ is $\frac{5}{2}$. The final answer is $\frac{5}{2}$, which corresponds to Option 3.