Three Dimensional Geometry - Distance between parallel lines

Find the distance between the parallel lines $L_1: \vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$ and $L_2: \vec{r} = (3\hat{i} + 3\hat{j} - 5\hat{k}) + \mu(2\hat{i} + 3\hat{j} + 6\hat{k})$.

Step 1: Identify vectors from the equations: $\vec{a}_1 = \hat{i} + 2\hat{j} - 4\hat{k}$ $\vec{a}_2 = 3\hat{i} + 3\hat{j} - 5\hat{k}$ $\vec{b} = 2\hat{i} + 3\hat{j} + 6\hat{k}$

Step 2: Calculate $\vec{a}_2 - \vec{a}_1$: $\vec{a}_2 - \vec{a}_1 = (3-1)\hat{i} + (3-2)\hat{j} + (-5+4)\hat{k} = 2\hat{i} + \hat{j} - \hat{k}$

Step 3: Calculate the cross product $\vec{b} \times (\vec{a}_2 - \vec{a}_1)$: $\vec{b} \times (\vec{a}_2 - \vec{a}_1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 6 \\ 2 & 1 & -1 \end{vmatrix}$ $= \hat{i}(-3 - 6) - \hat{j}(-2 - 12) + \hat{k}(2 - 6)$ $= -9\hat{i} + 14\hat{j} - 4\hat{k}$

Step 4: Calculate magnitudes: $|\vec{b} \times (\vec{a}_2 - \vec{a}_1)| = \sqrt{(-9)^2 + (14)^2 + (-4)^2} = \sqrt{81 + 196 + 16} = \sqrt{293}$ $|\vec{b}| = \sqrt{2^2 + 3^2 + 6^2} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7$

Step 5: Apply the distance formula: $d = \frac{\sqrt{293}}{7}$ units.

First, we extract the position vectors of points on the lines and the common direction vector. We find the vector connecting the two points, calculate its cross product with the direction vector to find the perpendicular component, and divide by the magnitude of the direction vector to normalize the result.

Calculate the magnitude of the vector $\vec{v} = 125\hat{i} + 200\hat{j}$ and subtract the scalar value $75$ from its x-component calculation.

To demonstrate the subtraction of components: $\begin{array}{r} 125 \\ - 75 \\ \hline 50 \end{array}$ The new x-component is $50$.

This example demonstrates vertical arithmetic for component subtraction as per the required formatting.

1. Identify vectors: $\vec{a}_1 = 2\hat{i} - \hat{j} + \hat{k}$ $\vec{a}_2 = \hat{i} + \hat{j} - \hat{k}$ $\vec{b} = 3\hat{i} - 4\hat{j} + 5\hat{k}$
2. Calculate $(\vec{a}_2 - \vec{a}_1)$: $\vec{a}_2 - \vec{a}_1 = (1-2)\hat{i} + (1 - (-1))\hat{j} + (-1-1)\hat{k} = -\hat{i} + 2\hat{j} - 2\hat{k}$
3. Calculate $\vec{b} \times (\vec{a}_2 - \vec{a}_1)$: $\vec{b} \times (\vec{a}_2 - \vec{a}_1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -4 & 5 \\ -1 & 2 & -2 \end{vmatrix}$ $= \hat{i}(8-10) - \hat{j}(-6 - (-5)) + \hat{k}(6-4) = -2\hat{i} + \hat{j} + 2\hat{k}$
4. Find Magnitudes: $|\vec{b} \times (\vec{a}_2 - \vec{a}_1)| = \sqrt{(-2)^2 + 1^2 + 2^2} = \sqrt{4+1+4} = 3$ $|\vec{b}| = \sqrt{3^2 + (-4)^2 + 5^2} = \sqrt{9+16+25} = \sqrt{50} = 5\sqrt{2}$
5. Final Distance: $d = \frac{3}{5\sqrt{2}} = \frac{3\sqrt{2}}{10}$ units.

To find the distance, we first find the vector connecting a point on each line, then find the component of that vector perpendicular to the common direction vector using the cross product.

1. Convert to vector form: $L_1: \vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}, \vec{b} = 2\hat{i} + \hat{j} - 2\hat{k}$ $L_2: \vec{a}_2 = 3\hat{i} + 4\hat{j} + 5\hat{k}, \vec{b} = 2\hat{i} + \hat{j} - 2\hat{k}$
2. Calculate $(\vec{a}_2 - \vec{a}_1)$: $\vec{a}_2 - \vec{a}_1 = (3-1)\hat{i} + (4-2)\hat{j} + (5-3)\hat{k} = 2\hat{i} + 2\hat{j} + 2\hat{k}$
3. Calculate $\vec{b} \times (\vec{a}_2 - \vec{a}_1)$: $\vec{b} \times (\vec{a}_2 - \vec{a}_1) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 1 & -2 \\ 2 & 2 & 2 \end{vmatrix}$ $= \hat{i}(2 - (-4)) - \hat{j}(4 - (-4)) + \hat{k}(4 - 2) = 6\hat{i} - 8\hat{j} + 2\hat{k}$
4. Calculate Magnitudes: $|\vec{b} \times (\vec{a}_2 - \vec{a}_1)| = \sqrt{6^2 + (-8)^2 + 2^2} = \sqrt{36 + 64 + 4} = \sqrt{104} = 2\sqrt{26}$ $|\vec{b}| = \sqrt{2^2 + 1^2 + (-2)^2} = \sqrt{4+1+4} = 3$
5. Distance: $d = \frac{2\sqrt{26}}{3}$ units.

Since the denominators (direction ratios) are the same for both lines, they are parallel. We pick points $(1,2,3)$ and $(3,4,5)$ from the equations to form the vector difference and apply the distance formula.