Find the distance between the skew lines \( \vec{r}(t)=\langle-1,4,-5\rangle t+\langle 6,0,-6\rangle \) anc \( \vec{p}(s)=\langle 0,2,5\rangle s+\langle-4,-3,-7\rangle \).

To find the distance between the skew lines represented by the vector equations \( \vec{r}(t) = \langle -1, 4, -5 \rangle t + \langle 6, 0, -6 \rangle \) and \( \vec{p}(s) = \langle 0, 2, 5 \rangle s + \langle -4, -3, -7 \rangle \), we can follow these steps: ### Step 1: Determine Direction Vectors The direction vector of the first line is \( \vec{d_1} = \langle -1, 4, -5 \rangle \) and the direction vector of the second line is \( \vec{d_2} = \langle 0, 2, 5 \rangle \). ### Step 2: Find the Vector Between Two Points on Each Line We can obtain two points, one from each line. Let's take \( t = 0 \) for the first line: \[ \vec{r}(0) = \langle 6, 0, -6 \rangle \] And take \( s = 0 \) for the second line: \[ \vec{p}(0) = \langle -4, -3, -7 \rangle \] Let's denote these points as \( \vec{A} = \langle 6, 0, -6 \rangle \) and \( \vec{B} = \langle -4, -3, -7 \rangle \). ### Step 3: Construct the Vector Between Points A and B The vector between points \( A \) and \( B \) is given by: \[ \vec{AB} = \vec{B} - \vec{A} = \langle -4 - 6, -3 - 0, -7 + 6 \rangle = \langle -10, -3, -1 \rangle \] ### Step 4: Use the Cross Product to Find the Direction of Minimum Distance The cross product of the direction vectors \( \vec{d_1} \) and \( \vec{d_2} \) gives a vector that is perpendicular to both direction vectors. Calculating the cross product: \[ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 4 & -5 \\ 0 & 2 & 5 \end{vmatrix} = \hat{i}(4 \cdot 5 - (-5) \cdot 2) - \hat{j}(-1 \cdot 5 - (-5) \cdot 0) + \hat{k}(-1 \cdot 2 - 4 \cdot 0) \] Calculating we get: \[ = \hat{i}(20 + 10) - \hat{j}(-5) - \hat{k}(2) = \hat{i} \cdot 30 + 5 \hat{j} - 2 \hat{k} = \langle 30, 5, -2 \rangle \] ### Step 5: Magnitude of the Cross Product The magnitude of the cross product \( \lVert \vec{d_1} \times \vec{d_2} \rVert \) can be calculated as: \[ \lVert \langle 30, 5, -2 \rangle \rVert = \sqrt{30^2 + 5^2 + (-2)^2} = \sqrt{900 + 25 + 4} = \sqrt{929} \] ### Step 6: Using the Formula for the Distance Between Skew Lines Finally, we can find the distance \( D \) between the two skew lines using the formula: \[ D = \frac{|\vec{AB} \cdot (\vec{d_1} \times \vec{d_2})|}{\lVert \vec{d_1} \times \vec{d_2} \rVert} \] Calculating \( \vec{AB} \cdot (\vec{d_1} \times \vec{d_2}) \): \[ \vec{AB} \cdot \langle 30, 5, -2 \rangle = \langle -10, -3, -1 \rangle \cdot \langle 30, 5, -2 \rangle = -10 \cdot 30 + (-3