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.}
Show 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.
The vector equation of the first line is:
\[
\vec{r_1} = (1 - \lambda) \hat{i} + (\lambda - 2) \hat{j} + (3 - 2\lambda) \hat{k}
\]
and the vector equation of the second line is:
\[
\vec{r_2} = (\mu + 1) \hat{i} + (2\mu - 1) \hat{j} - (2\mu + 1) \hat{k}
\]
The lines are skew lines if the direction vectors of the lines are not parallel, i.e., the cross product of the direction vectors is not 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 verify that the lines are skew, we compute the cross product of the 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 not zero, the lines are skew.
The shortest distance between two skew lines is given by:
\[
d = \frac{|\vec{a_2} - \vec{a_1} \cdot (\vec{d_1} \times \vec{d_2})|}{|\vec{d_1} \times \vec{d_2}|}
\]
where \( \vec{a_1} = (1, -2, 3) \) and \( \vec{a_2} = (1, -1, -1) \).
The vector \( \vec{a_2} - \vec{a_1} = (0, 1, -4) \), and we already know that:
\[
\vec{d_1} \times \vec{d_2} = (2, -4, -3)
\]
The shortest distance is then:
\[
d = \frac{|(0, 1, -4) \cdot (2, -4, -3)|}{\sqrt{2^2 + (-4)^2 + (-3)^2}}
\]
\[
= \frac{|0 \times 2 + 1 \times -4 + -4 \times -3|}{\sqrt{4 + 16 + 9}}
\]
\[
= \frac{| -4 + 12 |}{\sqrt{29}} = \frac{}{\sqrt{29}}
\]
Thus, the shortest distance between the lines is \( \frac{}{\sqrt{29}} \).
