1. **State the problem:** Find the vector equation of the tangent line to the curve formed by the intersection of the cylinders defined by the equations $$x^2 + y^2 = 25$$ and $$y^2 + z^2 = 20$$ at the point $(3,4,2)$. 2. **Understand the problem:** The curve of intersection lies on both surfaces. The tangent line at a point on the curve is in the direction of the cross product of the gradients of the two surfaces at that point. 3. **Find gradients:** - For the first cylinder, define $$F(x,y,z) = x^2 + y^2 - 25 = 0$$. - For the second cylinder, define $$G(x,y,z) = y^2 + z^2 - 20 = 0$$. Calculate gradients: $$\nabla F = \left(\frac{\partial F}{\partial x}, \frac{\partial F}{\partial y}, \frac{\partial F}{\partial z}\right) = (2x, 2y, 0)$$ $$\nabla G = \left(\frac{\partial G}{\partial x}, \frac{\partial G}{\partial y}, \frac{\partial G}{\partial z}\right) = (0, 2y, 2z)$$ 4. **Evaluate gradients at point $(3,4,2)$:** $$\nabla F(3,4,2) = (2\times3, 2\times4, 0) = (6, 8, 0)$$ $$\nabla G(3,4,2) = (0, 2\times4, 2\times2) = (0, 8, 4)$$ 5. **Find direction vector of tangent line:** The direction vector $$\mathbf{v}$$ is the cross product: $$\mathbf{v} = \nabla F \times \nabla G = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 6 & 8 & 0 \\ 0 & 8 & 4 \end{vmatrix}$$ Calculate determinant: $$\mathbf{v} = (8 \times 4 - 0 \times 8)\mathbf{i} - (6 \times 4 - 0 \times 0)\mathbf{j} + (6 \times 8 - 8 \times 0)\mathbf{k} = (32)\mathbf{i} - (24)\mathbf{j} + (48)\mathbf{k} = (32, -24, 48)$$ 6. **Simplify direction vector:** Divide by 8: $$\mathbf{v} = (\cancel{32}^{4}, \cancel{-24}^{-3}, \cancel{48}^{6}) = (4, -3, 6)$$ 7. **Write vector equation of tangent line:** Using point $(3,4,2)$ and direction vector $(4,-3,6)$: $$\mathbf{r}(t) = (3,4,2) + t(4,-3,6)$$ **Final answer:** $$\boxed{\mathbf{r}(t) = \langle 3 + 4t, 4 - 3t, 2 + 6t \rangle}$$