AP DSC School Assistant Math Model Test 3

Concept:The shortest distance between two parallel lines is given by $D = \dfrac{|(\vec{a}_2 - \vec{a}_1) \times \vec{b}|}{|\vec{b}|}$, where $\vec{a}_1$ and $\vec{a}_2$ are position vectors of points on the lines, and $\vec{b}$ is the common direction vector.Explanation:Given lines:$\vec{r}_1 = \hat{i} - 2\hat{j} + 3\hat{k} + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$$\vec{r}_2 = 3\hat{i} - 2\hat{j} + \hat{k} + \mu(4\hat{i} + 6\hat{j} + 12\hat{k})$Identify:$\vec{a}_1 = \hat{i} - 2\hat{j} + 3\hat{k}$, $\vec{b}_1 = 2\hat{i} + 3\hat{j} + 6\hat{k}$$\vec{a}_2 = 3\hat{i} - 2\hat{j} + \hat{k}$, $\vec{b}_2 = 4\hat{i} + 6\hat{j} + 12\hat{k}$Observe $\vec{b}_2 = 2\vec{b}_1$, so the lines are parallel. Take the common direction vector $\vec{b} = \vec{b}_1 = 2\hat{i} + 3\hat{j} + 6\hat{k}$.Compute vector between points: $\vec{d} = \vec{a}_2 - \vec{a}_1 = (3-1)\hat{i} + (-2+2)\hat{j} + (1-3)\hat{k} = 2\hat{i} + 0\hat{j} - 2\hat{k}$.Find cross product $\vec{b} \times \vec{d}$:$$\vec{b} \times \vec{d} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 2 & 0 & -2 \end{vmatrix} = (3(-2)-6(0))\hat{i} - (2(-2)-6(2))\hat{j} + (2(0)-3(2))\hat{k} = (-6)\hat{i} - (-4-12)\hat{j} + (-6)\hat{k} = -6\hat{i} + 16\hat{j} - 6\hat{k}.$$Magnitude: $|\vec{b} \times \vec{d}| = \sqrt{(-6)^2 + 16^2 + (-6)^2} = \sqrt{36 + 256 + 36} = \sqrt{328}$.Magnitude of direction vector: $|\vec{b}| = \sqrt{2^2+3^2+6^2} = \sqrt{4+9+36} = \sqrt{49}=7$.Thus distance $D = \dfrac{\sqrt{328}}{7}$.Answer:$\dfrac{\sqrt{328}}{7}$ corresponds to option C.