1. **Problem statement:** Find the equations of planes $P_1$ and $P_2$, the angle between them, the vector equation of their line of intersection, and the distance of this line from the origin. 2. **Equation of plane $P_1$:** Plane $P_1$ passes through points $(1,1,1)$, $(1,4,2)$, and $(0,2,0)$. 3. **Find two vectors on $P_1$:** $$\vec{v_1} = (1,4,2) - (1,1,1) = (0,3,1)$$ $$\vec{v_2} = (0,2,0) - (1,1,1) = (-1,1,-1)$$ 4. **Normal vector to $P_1$ is the cross product:** $$\vec{n_1} = \vec{v_1} \times \vec{v_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 3 & 1 \\ -1 & 1 & -1 \end{vmatrix} = (3 \times -1 - 1 \times 1, -(0 \times -1 - 1 \times -1), 0 \times 1 - 3 \times -1) = (-3 -1, -(0 +1), 0 +3) = (-4, -1, 3)$$ 5. **Equation of $P_1$ using point $(1,1,1)$:** $$-4(x-1) -1(y-1) + 3(z-1) = 0$$ Simplify: $$-4x +4 - y +1 + 3z -3 = 0$$ $$-4x - y + 3z + 2 = 0$$ 6. **Equation of plane $P_2$:** $P_2$ is perpendicular to vector $(2,1,1)^T$ and passes through $(0,0,3)$. 7. **Normal vector of $P_2$ is $(2,1,1)$, so equation:** $$2(x-0) + 1(y-0) + 1(z-3) = 0$$ Simplify: $$2x + y + z - 3 = 0$$ 8. **Angle between $P_1$ and $P_2$:** Formula for angle $\theta$ between planes with normals $\vec{n_1}$ and $\vec{n_2}$: $$\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{||\vec{n_1}|| \, ||\vec{n_2}||}$$ Calculate dot product: $$\vec{n_1} \cdot \vec{n_2} = (-4)(2) + (-1)(1) + 3(1) = -8 -1 + 3 = -6$$ Calculate magnitudes: $$||\vec{n_1}|| = \sqrt{(-4)^2 + (-1)^2 + 3^2} = \sqrt{16 + 1 + 9} = \sqrt{26}$$ $$||\vec{n_2}|| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$$ Calculate cosine: $$\cos \theta = \frac{|-6|}{\sqrt{26} \times \sqrt{6}} = \frac{6}{\sqrt{156}} = \frac{6}{2\sqrt{39}} = \frac{3}{\sqrt{39}}$$ Angle: $$\theta = \cos^{-1}\left(\frac{3}{\sqrt{39}}\right)$$ 9. **Vector equation of line of intersection:** Direction vector $\vec{d}$ is cross product of normals: $$\vec{d} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & -1 & 3 \\ 2 & 1 & 1 \end{vmatrix} = (-1)(1) - 3(1), - ( -4(1) - 3(2) ), -4(1) - (-1)(2) = (-1 -3, -(-4 -6), -4 + 2) = (-4, 10, -2)$$ Simplify direction vector: $$\vec{d} = (-2, 5, -1)$$ 10. **Find a point on the line by solving the system:** $$P_1: -4x - y + 3z + 2 = 0$$ $$P_2: 2x + y + z - 3 = 0$$ Add equations to eliminate $y$: $$(-4x - y + 3z + 2) + (2x + y + z - 3) = 0$$ $$-2x + 4z -1 = 0$$ $$2x = 4z -1$$ $$x = 2z - \frac{1}{2}$$ Substitute $x$ into $P_2$: $$2(2z - \frac{1}{2}) + y + z - 3 = 0$$ $$4z -1 + y + z - 3 = 0$$ $$y + 5z - 4 = 0$$ $$y = 4 - 5z$$ Choose $z = 0$ for simplicity: $$x = 2(0) - \frac{1}{2} = -\frac{1}{2}$$ $$y = 4 - 5(0) = 4$$ Point on line: $$\left(-\frac{1}{2}, 4, 0\right)$$ 11. **Vector equation of line:** $$\vec{r} = \left(-\frac{1}{2}, 4, 0\right) + t(-2, 5, -1)$$ 12. **Distance from origin to line:** Formula for distance $d$ from point $\vec{p_0}$ to line through $\vec{a}$ with direction $\vec{d}$: $$d = \frac{|| (\vec{p_0} - \vec{a}) \times \vec{d} ||}{||\vec{d}||}$$ Here, $\vec{p_0} = (0,0,0)$, $\vec{a} = (-\frac{1}{2},4,0)$, $\vec{d} = (-2,5,-1)$. Calculate $\vec{p_0} - \vec{a}$: $$\left(0 + \frac{1}{2}, 0 - 4, 0 - 0\right) = \left(\frac{1}{2}, -4, 0\right)$$ Calculate cross product: $$\left(\frac{1}{2}, -4, 0\right) \times (-2, 5, -1) = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ \frac{1}{2} & -4 & 0 \\ -2 & 5 & -1 \end{vmatrix} = (-4)(-1) - 0(5), - \left( \frac{1}{2}(-1) - 0(-2) \right), \frac{1}{2}(5) - (-4)(-2) = (4 - 0, -(-\frac{1}{2} - 0), \frac{5}{2} - 8) = (4, \frac{1}{2}, -\frac{11}{2})$$ Magnitude of cross product: $$\sqrt{4^2 + \left(\frac{1}{2}\right)^2 + \left(-\frac{11}{2}\right)^2} = \sqrt{16 + \frac{1}{4} + \frac{121}{4}} = \sqrt{16 + 30.5} = \sqrt{46.5}$$ Magnitude of $\vec{d}$: $$||\vec{d}|| = \sqrt{(-2)^2 + 5^2 + (-1)^2} = \sqrt{4 + 25 + 1} = \sqrt{30}$$ Distance: $$d = \frac{\sqrt{46.5}}{\sqrt{30}} = \sqrt{\frac{46.5}{30}} = \sqrt{1.55} \approx 1.245$$