Subjects algebra

Parametric Line Daf849

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1. **State the problem:** Find the parametric equations of the line of intersection of the two planes given by: $$2x - y + 2z - 12 = 0$$ $$2x - 4y - z + 6 = 0$$ 2. **Recall the method:** The intersection of two planes is a line. To find its parametric equations, we: - Find the direction vector of the line by taking the cross product of the normals of the planes. - Find a particular point on the line by solving the system. 3. **Find the normal vectors:** - For plane 1: $\vec{n_1} = (2, -1, 2)$ - For plane 2: $\vec{n_2} = (2, -4, -1)$ 4. **Find the direction vector $\vec{d}$ of the line:** $$\vec{d} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 2 \\ 2 & -4 & -1 \end{vmatrix}$$ Calculate the determinant: $$\vec{d} = \mathbf{i}((-1)(-1) - 2(-4)) - \mathbf{j}(2(-1) - 2(2)) + \mathbf{k}(2(-4) - (-1)(2))$$ $$= \mathbf{i}(1 + 8) - \mathbf{j}(-2 - 4) + \mathbf{k}(-8 + 2)$$ $$= 9\mathbf{i} + 6\mathbf{j} - 6\mathbf{k}$$ 5. **Simplify the direction vector:** $$\vec{d} = (9, 6, -6) = 3(3, 2, -2)$$ We can use the simplified direction vector: $$\vec{d} = (3, 2, -2)$$ 6. **Find a particular point on the line:** Solve the system: $$2x - y + 2z = 12$$ $$2x - 4y - z = -6$$ Let $z = t$ (parameter). Then: From the first equation: $$2x - y + 2t = 12 \implies 2x - y = 12 - 2t$$ From the second equation: $$2x - 4y - t = -6 \implies 2x - 4y = t - 6$$ 7. **Subtract the first from the second to eliminate $2x$:** $$ (2x - 4y) - (2x - y) = (t - 6) - (12 - 2t)$$ $$ \cancel{2x} - 4y - \cancel{2x} + y = t - 6 - 12 + 2t$$ $$ -3y = 3t - 18$$ $$ y = \frac{18 - 3t}{3} = 6 - t$$ 8. **Substitute $y$ back to find $x$:** $$2x - y = 12 - 2t$$ $$2x - (6 - t) = 12 - 2t$$ $$2x - 6 + t = 12 - 2t$$ $$2x = 12 - 2t + 6 - t = 18 - 3t$$ $$x = \frac{18 - 3t}{2} = 9 - \frac{3t}{2}$$ 9. **Write parametric equations:** $$x = 9 - \frac{3}{2}t$$ $$y = 6 - t$$ $$z = t$$ 10. **Final answer:** The parametric equations of the line of intersection are: $$x = 9 - \frac{3}{2}t, \quad y = 6 - t, \quad z = t$$ where $t$ is a parameter.