1. **State the problem:** Find the equation of the line of intersection between the planes given by $$x + 5y - z = 20$$ and $$3x + 5y + 4z = 23$$. 2. **Recall the formula and approach:** The line of intersection of two planes can be found by solving the system of equations simultaneously. The direction vector of the line is perpendicular to the normal vectors of both planes, so it is the cross product of the normals. 3. **Find the normal vectors:** - For plane 1: normal vector $\mathbf{n_1} = (1, 5, -1)$ - For plane 2: normal vector $\mathbf{n_2} = (3, 5, 4)$ 4. **Find the direction vector $\mathbf{d}$ of the line:** $$\mathbf{d} = \mathbf{n_1} \times \mathbf{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 1 & 5 & -1 \\ 3 & 5 & 4 \end{vmatrix}$$ Calculate the determinant: $$\mathbf{d} = \mathbf{i}(5 \cdot 4 - (-1) \cdot 5) - \mathbf{j}(1 \cdot 4 - (-1) \cdot 3) + \mathbf{k}(1 \cdot 5 - 5 \cdot 3)$$ $$= \mathbf{i}(20 + 5) - \mathbf{j}(4 + 3) + \mathbf{k}(5 - 15)$$ $$= 25\mathbf{i} - 7\mathbf{j} - 10\mathbf{k}$$ So, $$\mathbf{d} = (25, -7, -10)$$ 5. **Find a point on the line:** Solve the system for a convenient variable, for example, set $z = 0$: From plane 1: $$x + 5y = 20$$ From plane 2: $$3x + 5y = 23$$ Subtract plane 1 from plane 2: $$3x + 5y - (x + 5y) = 23 - 20$$ $$2x = 3 \implies x = \frac{3}{2}$$ Substitute back into plane 1: $$\frac{3}{2} + 5y = 20 \implies 5y = 20 - \frac{3}{2} = \frac{40}{2} - \frac{3}{2} = \frac{37}{2}$$ $$y = \frac{37}{10}$$ So the point is: $$\left(\frac{3}{2}, \frac{37}{10}, 0\right)$$ 6. **Write the parametric equations of the line:** Let parameter be $t$, then $$x = \frac{3}{2} + 25t$$ $$y = \frac{37}{10} - 7t$$ $$z = 0 - 10t = -10t$$ **Final answer:** The line of intersection is given by $$\boxed{\begin{cases} x = \frac{3}{2} + 25t \\ y = \frac{37}{10} - 7t \\ z = -10t \end{cases}}$$