Question

Verify that lines given by \( \vec{r} = (1 - \lambda) \hat{i + (\lambda - 2) \hat{j} + (3 - 2\lambda) \hat{k} \) and \( \vec{r} = (\mu + 1) \hat{i} + (2\mu - 1) \hat{j} - (2\mu + 1) \hat{k} \) are skew lines. Hence, find shortest distance between the lines.}

Hint:

When finding the shortest distance between skew lines, remember to use the formula: \[ d = \frac{|\vec{a_2} - \vec{a_1} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|} \] This formula involves calculating the cross product of the direction vectors and using the vector connecting the points on the two lines.

Answer 1

The vector equation for the first line is: \[ \vec{r_1} = (1 - \lambda) \hat{i} + (\lambda - 2) \hat{j} + (3 - 2\lambda) \hat{k} \] and for the second line is: \[ \vec{r_2} = (\mu + 1) \hat{i} + (2\mu - 1) \hat{j} - (2\mu + 1) \hat{k} \] Two lines are skew if their direction vectors are not parallel, meaning their cross product is non-zero. The direction vector of the first line is: \[ \vec{d_1} = \frac{d\vec{r_1}}{d\lambda} = -\hat{i} + \hat{j} - 2\hat{k} \] and the direction vector of the second line is: \[ \vec{d_2} = \frac{d\vec{r_2}}{d\mu} = \hat{i} + 2\hat{j} - 2\hat{k} \] To confirm the lines are skew, we compute the cross product of their direction vectors: \[ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 1 & -2 \\ 1 & 2 & -2 \end{vmatrix} \] \[ \vec{d_1} \times \vec{d_2} = \hat{i}(1 \times -2 - 2 \times -2) - \hat{j}(-1 \times -2 - 1 \times -2) + \hat{k}(-1 \times 2 - 1 \times 1) \] \[ = \hat{i}(-2 + 4) - \hat{j}(2 + 2) + \hat{k}(-2 - 1) \] \[ = 2\hat{i} - 4\hat{j} - 3\hat{k} \] Since the cross product is non-zero, the lines are skew. The shortest distance between two skew lines is calculated using the formula: \[ d = \frac{|\vec{a_2} - \vec{a_1} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|} \] where \( \vec{a_1} = (1, -2, 3) \) and \( \vec{a_2} = (1, -1, -1) \). The vector difference is \( \vec{a_2} - \vec{a_1} = (0, 1, -4) \), and the cross product is \( \vec{d_1} \times \vec{d_2} = (2, -4, -3) \). The shortest distance is therefore: \[ d = \frac{|(0, 1, -4) \cdot (2, -4, -3)|}{\sqrt{2^2 + (-4)^2 + (-3)^2}} \] \[ = \frac{|0 \times 2 + 1 \times -4 + -4 \times -3|}{\sqrt{4 + 16 + 9}} \] \[ = \frac{| -4 + 12 |}{\sqrt{29}} = \frac{8}{\sqrt{29}} \] The shortest distance between the lines is \( \frac{8}{\sqrt{29}} \).