Find the shortest distance between the lines whose vector equations are r = (i + 2j + 3k) + λ(i - 3j + 2k) and r = 4i + 5j + 6k + μ(2i + 3j + k)

Find the shortest distance between the lines whose vector equations…

Question 19 of 25intermediate🔧 ApplyShort Answer4 marks

Find the shortest distance between the lines whose vector equations are $\vec{r} = (\hat{i} + 2\hat{j} + 3\hat{k}) + \lambda(\hat{i} - 3\hat{j} + 2\hat{k})$ and $\vec{r} = 4\hat{i} + 5\hat{j} + 6\hat{k} + \mu(2\hat{i} + 3\hat{j} + \hat{k})$

Correct Answer

The shortest distance between the given lines is $\frac{3}{\sqrt{19}}$ units (or $\frac{3\sqrt{19}}{19}$ units).

Exercise: EXERCISE 11.2 | Q: 14 | (Page Page 14)

For More Understanding

Explanation

This question requires finding the shortest distance between two skew lines using the formula from Example 10 in the textbook. The student needs to identify the position vectors and direction vectors from both lines, then apply the formula $d = \left| \frac{(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1)}{|\vec{b}_1 \times \vec{b}_2|} \right|$ .

Solution Steps

Step 1: Identify the position vectors and direction vectors from the given equations. For line 1: $\vec{a}_1 = \hat{i} + 2\hat{j} + 3\hat{k}$ and $\vec{b}_1 = \hat{i} - 3\hat{j} + 2\hat{k}$ . For line 2: $\vec{a}_2 = 4\hat{i} + 5\hat{j} + 6\hat{k}$ and $\vec{b}_2 = 2\hat{i} + 3\hat{j} + \hat{k}$ .
Step 2: Calculate $\vec{a}_2 - \vec{a}_1 = (4\hat{i} + 5\hat{j} + 6\hat{k}) - (\hat{i} + 2\hat{j} + 3\hat{k}) = 3\hat{i} + 3\hat{j} + 3\hat{k}$ .
Step 3: Find $\vec{b}_1 \times \vec{b}_2 = \left| \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1 \end{array} \right| = -9\hat{i} + 3\hat{j} + 9\hat{k}$ .
Step 4: Calculate $|\vec{b}_1 \times \vec{b}_2| = \sqrt{81 + 9 + 81} = \sqrt{171} = 3\sqrt{19}$ .
Step 5: Find $(\vec{b}_1 \times \vec{b}_2) \cdot (\vec{a}_2 - \vec{a}_1) = (-9)(3) + (3)(3) + (9)(3) = -27 + 9 + 27 = 9$ .
Step 6: Apply the formula: $d = \left| \frac{9}{3\sqrt{19}} \right| = \frac{3}{\sqrt{19}} = \frac{3\sqrt{19}}{19}$ units.

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