1. **Problem Statement:** We have a satellite trajectory given by the vector equation $$\vec{r}(t) = \langle 1, 2, 0 \rangle + t \langle 2, 1, 2 \rangle$$ and a ground station at point $$S(4,4,10)$$. (a) Convert the vector equation to symmetric form and explain the difficulty of symmetric form in loops. (b) Calculate the minimum distance from the ground station to the satellite's path using the point-to-line distance formula. 2. **Part (a): Symmetric Form of the Trajectory** The vector equation is: $$\vec{r}(t) = \langle x_0, y_0, z_0 \rangle + t \langle a, b, c \rangle = \langle 1, 2, 0 \rangle + t \langle 2, 1, 2 \rangle$$ The parametric equations are: $$x = 1 + 2t$$ $$y = 2 + t$$ $$z = 0 + 2t$$ To write the symmetric form, solve each for $t$: $$t = \frac{x - 1}{2} = y - 2 = \frac{z}{2}$$ So the symmetric form is: $$\frac{x - 1}{2} = y - 2 = \frac{z}{2}$$ **Explanation:** Symmetric equations can be harder to implement inside a loop because they require checking equality of multiple expressions, which can be undefined or cause division by zero if any denominator is zero. Parametric (vector) form uses a single parameter $t$ which is easier to iterate over in a loop. 3. **Part (b): Minimum Distance to the Line** Given: - Point on line at $t=0$: $$P = (1, 2, 0)$$ - Direction vector: $$\vec{v} = \langle 2, 1, 2 \rangle$$ - Ground station point: $$S = (4, 4, 10)$$ The vector from $P$ to $S$ is: $$\vec{PS} = \langle 4 - 1, 4 - 2, 10 - 0 \rangle = \langle 3, 2, 10 \rangle$$ The formula for distance is: $$d = \frac{|\vec{PS} \times \vec{v}|}{|\vec{v}|}$$ Calculate the cross product: $$\vec{PS} \times \vec{v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 3 & 2 & 10 \\ 2 & 1 & 2 \end{vmatrix} = \mathbf{i}(2 \cdot 2 - 10 \cdot 1) - \mathbf{j}(3 \cdot 2 - 10 \cdot 2) + \mathbf{k}(3 \cdot 1 - 2 \cdot 2)$$ $$= \mathbf{i}(4 - 10) - \mathbf{j}(6 - 20) + \mathbf{k}(3 - 4) = \langle -6, 14, -1 \rangle$$ Magnitude of cross product: $$|\vec{PS} \times \vec{v}| = \sqrt{(-6)^2 + 14^2 + (-1)^2} = \sqrt{36 + 196 + 1} = \sqrt{233}$$ Magnitude of direction vector: $$|\vec{v}| = \sqrt{2^2 + 1^2 + 2^2} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$$ Distance: $$d = \frac{\sqrt{233}}{3}$$ **Final answers:** (a) Symmetric form: $$\frac{x - 1}{2} = y - 2 = \frac{z}{2}$$ (b) Minimum distance: $$\frac{\sqrt{233}}{3}$$