18 The perpendicular distance from the point \( A=(3,-1,0) \) and the straight line \( \vec{r}=(0,4,2)+t(-1,5,0) \) is ... \( \begin{array}{llll}\text { a) } 2.8 & \text { b) } 3.2 & \text { c) } 4.2 & \text { d) } 5.1\end{array} \)

To find the perpendicular distance from the point \( A=(3,-1,0) \) to the line defined by \( \vec{r}=(0,4,2)+t(-1,5,0) \), we can follow this approach: 1. **Direction Vector**: The direction vector of the line is given by \( \vec{d} = (-1, 5, 0) \). 2. **Point on the Line**: A point on the line when \( t = 0 \) is \( P_0 = (0, 4, 2) \). 3. **Vector from Point to Line**: Construct the vector from the point \( A \) to the point \( P_0 \): \[ \vec{AP_0} = P_0 - A = (0 - 3, 4 - (-1), 2 - 0) = (-3, 5, 2). \] 4. **Cross Product**: The perpendicular distance \( d \) from point \( A \) to the line can be found using the formula: \[ d = \frac{|\vec{AP_0} \times \vec{d}|}{|\vec{d}|}. \] Calculate the cross product \( \vec{AP_0} \times \vec{d} \): \[ \vec{AP_0} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 5 & 2 \\ -1 & 5 & 0 \end{vmatrix} = \hat{i}(5 \cdot 0 - 2 \cdot 5) - \hat{j}(-3 \cdot 0 - 2 \cdot -1) + \hat{k}(-3 \cdot 5 - 5 \cdot -1). \] Simplifying this determinant gives: \[ \hat{i}(0 - 10) - \hat{j}(0 + 2) + \hat{k}(-15 + 5) = -10 \hat{i} - 2 \hat{j} - 10 \hat{k}. \] 5. **Magnitude of Cross Product**: \[ |\vec{AP_0} \times \vec{d}| = \sqrt{(-10)^2 + (-2)^2 + (-10)^2} = \sqrt{100 + 4 + 100} = \sqrt{204} = 2\sqrt{51}. \] 6. **Magnitude of Direction Vector**: \[ |\vec{d}| = \sqrt{(-1)^2 + 5^2 + 0^2} = \sqrt{1 + 25} = \sqrt{26}. \] 7. **Distance Calculation**: \[ d = \frac{|\vec{AP_0} \times \vec{d}|}{|\vec{d}|} = \frac{2\sqrt{51}}{\sqrt{26}}. \] 8. **Numerical Approximation**: \[ \sqrt{51} \approx 7.14 \quad \text{and} \quad \sqrt{26} \approx 5.1 \implies d \approx \frac{2 \cdot 7.14}{5.1} \approx \frac{14.28}{5.1} \approx 2.8. \] Thus, the closest option is: The perpendicular distance is \( 2.8 \). The correct answer is \( \text{a) } 2.8 \).