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 $P_1$:** Given points: $A(1,1,1)$, $B(1,4,2)$, $C(0,2,0)$. - Find vectors $\vec{AB} = B - A = (0,3,1)$ and $\vec{AC} = C - A = (-1,1,-1)$. - The normal vector $\vec{n_1}$ to $P_1$ is $\vec{AB} \times \vec{AC}$: $$\vec{n_1} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 0 & 3 & 1 \\ -1 & 1 & -1 \end{vmatrix} = (3 \times -1 - 1 \times 1)\mathbf{i} - (0 \times -1 - 1 \times -1)\mathbf{j} + (0 \times 1 - 3 \times -1)\mathbf{k} = (-3 -1)\mathbf{i} - (0 +1)\mathbf{j} + (0 +3)\mathbf{k} = (-4, -1, 3)$$ - Equation of $P_1$ using point $A$: $$-4(x-1) -1(y-1) + 3(z-1) = 0$$ Simplify: $$-4x +4 - y +1 + 3z -3 = 0 \Rightarrow -4x - y + 3z + 2 = 0$$ 3. **Equation of Plane $P_2$:** - $P_2$ is perpendicular to vector $\vec{n_2} = (2,1,1)$ and passes through $(0,0,3)$. - Equation: $$2(x-0) + 1(y-0) + 1(z-3) = 0 \Rightarrow 2x + y + z - 3 = 0$$ 4. **Angle Between $P_1$ and $P_2$:** - Angle $\theta$ between planes is angle between their normals: $$\cos \theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{||\vec{n_1}|| \cdot ||\vec{n_2}||}$$ - Compute dot product: $$\vec{n_1} \cdot \vec{n_2} = (-4)(2) + (-1)(1) + 3(1) = -8 -1 + 3 = -6$$ - Norms: $$||\vec{n_1}|| = \sqrt{(-4)^2 + (-1)^2 + 3^2} = \sqrt{16 + 1 + 9} = \sqrt{26}$$ $$||\vec{n_2}|| = \sqrt{2^2 + 1^2 + 1^2} = \sqrt{4 + 1 + 1} = \sqrt{6}$$ - Calculate angle: $$\cos \theta = \frac{6}{\sqrt{26} \times \sqrt{6}} = \frac{6}{\sqrt{156}} = \frac{6}{12.49} \approx 0.48$$ $$\theta = \cos^{-1}(0.48) \approx 61.93^\circ$$ 5. **Vector Equation of Line of Intersection:** - Direction vector $\vec{d}$ is cross product of normals: $$\vec{d} = \vec{n_1} \times \vec{n_2} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & -1 & 3 \\ 2 & 1 & 1 \end{vmatrix} = (-1)(1) - 3(1)\mathbf{i} - (-4)(1) - 3(2)\mathbf{j} + (-4)(1) - (-1)(2)\mathbf{k} = (-1 -3)\mathbf{i} - (-4 -6)\mathbf{j} + (-4 + 2)\mathbf{k} = (-4, 10, -2)$$ - Simplify direction vector by dividing by 2: $$\vec{d} = (-2, 5, -1)$$ - Find a point on the line by solving the system: $$\begin{cases} -4x - y + 3z + 2 = 0 \\ 2x + y + z - 3 = 0 \end{cases}$$ - Add equations: $$(-4x - y + 3z + 2) + (2x + y + z - 3) = -2x + 4z -1 = 0 \Rightarrow 2x = 4z -1 \Rightarrow x = 2z - \frac{1}{2}$$ - Substitute $x$ into second equation: $$2(2z - \frac{1}{2}) + y + z - 3 = 0 \Rightarrow 4z -1 + y + z - 3 = 0 \Rightarrow y + 5z -4 = 0 \Rightarrow y = 4 - 5z$$ - Let $z = 0$ for simplicity: $$x = 2(0) - \frac{1}{2} = -\frac{1}{2}, \quad y = 4 - 0 = 4, \quad z = 0$$ - Point on line: $(-\frac{1}{2}, 4, 0)$ - Vector equation: $$\vec{r} = \left(-\frac{1}{2}, 4, 0\right) + t(-2, 5, -1)$$ 6. **Distance of Line from Origin:** - Distance $d$ from origin to line with point $\vec{p}$ and direction $\vec{d}$: $$d = \frac{||\vec{p} \times \vec{d}||}{||\vec{d}||}$$ - Compute cross product: $$\vec{p} = \left(-\frac{1}{2}, 4, 0\right), \quad \vec{d} = (-2, 5, -1)$$ $$\vec{p} \times \vec{d} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -\frac{1}{2} & 4 & 0 \\ -2 & 5 & -1 \end{vmatrix} = (4 \times -1 - 0 \times 5)\mathbf{i} - \left(-\frac{1}{2} \times -1 - 0 \times -2\right)\mathbf{j} + \left(-\frac{1}{2} \times 5 - 4 \times -2\right)\mathbf{k} = (-4)\mathbf{i} - \left(\frac{1}{2}\right)\mathbf{j} + \left(-\frac{5}{2} + 8\right)\mathbf{k} = (-4, -0.5, 5.5)$$ - Norms: $$||\vec{p} \times \vec{d}|| = \sqrt{(-4)^2 + (-0.5)^2 + 5.5^2} = \sqrt{16 + 0.25 + 30.25} = \sqrt{46.5} \approx 6.82$$ $$||\vec{d}|| = \sqrt{(-2)^2 + 5^2 + (-1)^2} = \sqrt{4 + 25 + 1} = \sqrt{30} \approx 5.48$$ - Distance: $$d = \frac{6.82}{5.48} \approx 1.24$$ **Final answers:** - $P_1$: $-4x - y + 3z + 2 = 0$ - $P_2$: $2x + y + z - 3 = 0$ - Angle between planes: $\approx 61.93^\circ$ - Line of intersection: $\vec{r} = \left(-\frac{1}{2}, 4, 0\right) + t(-2, 5, -1)$ - Distance from origin to line: $\approx 1.24$