To find the shortest distance between two skew lines given by:

$\mathbf{r} = (-\hat{\mathbf{i}} - 2\hat{\mathbf{j}} - 3\hat{\mathbf{k}}) + t(3\hat{\mathbf{i}} - 2\hat{\mathbf{j}} - 2\hat{\mathbf{k}})$

$\mathbf{r} = (7\hat{\mathbf{i}} + 4\hat{\mathbf{k}}) + s(\hat{\mathbf{i}} - 2\hat{\mathbf{j}} + 2\hat{\mathbf{k}})$

follow these steps:

Define the Position Vectors and Direction Vectors:

$\mathbf{a}_1 = -\hat{\mathbf{i}} - 2\hat{\mathbf{j}} - 3\hat{\mathbf{k}}$

$\mathbf{b}_1 = 3\hat{\mathbf{i}} - 2\hat{\mathbf{j}} - 2\hat{\mathbf{k}}$

$\mathbf{a}_2 = 7\hat{\mathbf{i}} + 4\hat{\mathbf{k}}$

$\mathbf{b}_2 = \hat{\mathbf{i}} - 2\hat{\mathbf{j}} + 2\hat{\mathbf{k}}$

Calculate the Vector Connecting Points on the Lines:

$ \mathbf{a}_2 - \mathbf{a}_1 = (7 + 1)\hat{\mathbf{i}} + (0 + 2)\hat{\mathbf{j}} + (4 + 3)\hat{\mathbf{k}} = 8\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 7\hat{\mathbf{k}} $

Find the Cross Product of the Direction Vectors:

$ \mathbf{b}_1 \times \mathbf{b}_2 = \begin{vmatrix} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 3 & -2 & -2 \\ 1 & -2 & 2 \end{vmatrix} $

$ = \hat{\mathbf{i}}((-2) \cdot 2 - (-2) \cdot (-2)) - \hat{\mathbf{j}}(3 \cdot 2 - (-2) \cdot 1) + \hat{\mathbf{k}}(3 \cdot (-2) - (-2) \cdot 1) $

$ = \hat{\mathbf{i}}(-4 - 4) - \hat{\mathbf{j}}(6 + 2) + \hat{\mathbf{k}}(-6 + 2) $

$ = -8\hat{\mathbf{i}} - 8\hat{\mathbf{j}} - 4\hat{\mathbf{k}} $

Calculate the Magnitude of the Cross Product:

$ \left|\mathbf{b}_1 \times \mathbf{b}_2\right| = \sqrt{(-8)^2 + (-8)^2 + (-4)^2} = \sqrt{64 + 64 + 16} = \sqrt{144} = 12 $

Find the Required Distance:

$ \text{Distance} = \frac{|(8\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 7\hat{\mathbf{k}}) \cdot (-8\hat{\mathbf{i}} - 8\hat{\mathbf{j}} - 4\hat{\mathbf{k}})|}{12} $

$ = \frac{|8(-8) + 2(-8) + 7(-4)|}{12} $

$ = \frac{|-64 - 16 - 28|}{12} $

$ = \frac{108}{12} = 9 $

Thus, the shortest distance between these skew lines is 9.

Three Dimensional Geometry

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