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Find the distance between the lines:$$\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$$$$\vec{r} = (3\hat{i} + 3\hat{j} - 5\hat{k}) + \mu(4\hat{i} + 6\hat{j} + 12\hat{k})$$
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Q: Find the distance between the lines:$$\vec{r} = (\hat{i} + 2\hat{j} - 4\hat{k}) + \lambda(2\hat{i} + 3\hat{j} + 6\hat{k})$$$$\vec{r} = (3\hat{i} + 3\hat{j} - 5\hat{k}) + \mu(4\hat{i} + 6\hat{j} + 12\hat{k})$$

Question Analysis & Solution

Step-by-Step Solution

Step 1: Check if the lines are parallel.

The direction vectors of the lines are b1 = 2î + 3ĵ + 6k and b2 = 4î + 6ĵ + 12k.

Notice that b2 = 2 * b1, which means the lines are parallel.

Step 2: Find a point on each line.

Point A on the first line: (1, 2, -4)

Point B on the second line: (3, 3, -5)

Step 3: Find the vector connecting the two points.

a = AB = (3-1)î + (3-2)ĵ + (-5-(-4))k = 2î + ĵ - k

Step 4: Calculate the cross product of the direction vector and the vector connecting the points.

b1 x a = (2î + 3ĵ + 6k) x (2î + ĵ - k) =
î(3*(-1) - 6*1) - ĵ(2*(-1) - 6*2) + k(2*1 - 3*2) = -9î + 14ĵ - 4k

Step 5: Find the magnitude of the cross product.

|b1 x a| = √((-9)^2 + (14)^2 + (-4)^2) = √(81 + 196 + 16) = √293

Step 6: Find the magnitude of the direction vector.

|b1| = √(2^2 + 3^2 + 6^2) = √(4 + 9 + 36) = √49 = 7

Step 7: Calculate the distance between the parallel lines.

Distance = |b1 x a| / |b1| = √293 / 7

Correct Answer: √293 / 7

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