1. **Problem statement:** Given points $A_1(3,5,4)$, $A_2(5,8,3)$, $A_3(1,2,-2)$, and $A_4(-1,0,2)$, find: a) Equation of plane $A_1A_2A_3$. b) Equation of line $A_1A_2$. c) Equation of line $A_4M$ perpendicular to plane $A_1A_2A_3$. d) Equation of line $A_3N$ parallel to line $A_1A_3$. e) Equation of plane through $A_4$ perpendicular to line $A_1A_2$. f) Sine of angle between line $A_1A_4$ and plane $A_1A_2A_3$. g) Cosine of angle between planes $XOY$ and $A_1A_2A_3$. h) Projection of point $A_4$ on plane $A_1A_2A_3$. k) Equation of plane through $A_2$ parallel to vectors $A_1A_4$ and $A_3A_4$. --- 2. **Formulas and rules:** - Plane through points $P,Q,R$: normal vector $\vec{n} = \overrightarrow{PQ} \times \overrightarrow{PR}$, plane equation $n_x(x-x_0)+n_y(y-y_0)+n_z(z-z_0)=0$. - Line through $P$ with direction $\vec{d}$: parametric $x=x_0+td_x$, $y=y_0+td_y$, $z=z_0+td_z$. - Line perpendicular to plane has direction vector equal to plane's normal. - Angle between line and plane: if $\theta$ is angle between line and plane, then $\sin\theta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}||\vec{n}|}$. - Angle between planes: cosine of angle between normals. - Projection of point $P$ on plane with normal $\vec{n}$ and point $Q$ on plane: $P' = P - \frac{(\overrightarrow{QP} \cdot \vec{n})}{|\vec{n}|^2} \vec{n}$. - Plane through point $P$ parallel to vectors $\vec{v}$ and $\vec{w}$ has normal $\vec{n} = \vec{v} \times \vec{w}$. --- 3. **Calculations:** - Vectors: $\overrightarrow{A_1A_2} = (5-3,8-5,3-4) = (2,3,-1)$ $\overrightarrow{A_1A_3} = (1-3,2-5,-2-4) = (-2,-3,-6)$ $\overrightarrow{A_1A_4} = (-1-3,0-5,2-4) = (-4,-5,-2)$ $\overrightarrow{A_3A_4} = (-1-1,0-2,2+2) = (-2,-2,4)$ - a) Plane $A_1A_2A_3$ normal: $\vec{n} = \overrightarrow{A_1A_2} \times \overrightarrow{A_1A_3} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 3 & -1 \\ -2 & -3 & -6 \end{vmatrix} = (3 \cdot (-6) - (-1)(-3), - (2 \cdot (-6) - (-1)(-2)), 2 \cdot (-3) - 3 \cdot (-2)) = (-18 -3, -(-12 -2), -6 +6) = (-21,14,0)$ Simplify normal vector dividing by 7: $\vec{n} = (-3,2,0)$ Plane equation: $-3(x-3) + 2(y-5) + 0(z-4) = 0 \Rightarrow -3x + 9 + 2y - 10 = 0 \Rightarrow -3x + 2y -1 = 0$ - b) Line $A_1A_2$ parametric: $x=3+2t$, $y=5+3t$, $z=4 - t$ - c) Line $A_4M$ perpendicular to plane $A_1A_2A_3$ has direction $\vec{n} = (-3,2,0)$ and passes through $A_4$: $x = -1 - 3s$, $y = 0 + 2s$, $z = 2 + 0s = 2$ - d) Line $A_3N$ parallel to $A_1A_3$ direction $\overrightarrow{A_1A_3} = (-2,-3,-6)$ passing through $A_3$: $x=1 - 2u$, $y=2 - 3u$, $z = -2 - 6u$ - e) Plane through $A_4$ perpendicular to line $A_1A_2$ (direction $\overrightarrow{A_1A_2} = (2,3,-1)$): Normal vector $\vec{n} = (2,3,-1)$ Equation: $2(x+1) + 3(y-0) -1(z-2) = 0 \Rightarrow 2x + 2 + 3y - z + 2 = 0 \Rightarrow 2x + 3y - z + 4 = 0$ - f) Sine of angle between line $A_1A_4$ and plane $A_1A_2A_3$: $\vec{d} = \overrightarrow{A_1A_4} = (-4,-5,-2)$, $\vec{n} = (-3,2,0)$ $|\vec{d}| = \sqrt{(-4)^2 + (-5)^2 + (-2)^2} = \sqrt{16 + 25 + 4} = \sqrt{45}$ $|\vec{n}| = \sqrt{(-3)^2 + 2^2 + 0^2} = \sqrt{9 + 4} = \sqrt{13}$ Dot product: $\vec{d} \cdot \vec{n} = (-4)(-3) + (-5)(2) + (-2)(0) = 12 - 10 + 0 = 2$ Sine of angle $\theta$ between line and plane: $\sin \theta = \frac{|\vec{d} \cdot \vec{n}|}{|\vec{d}||\vec{n}|} = \frac{2}{\sqrt{45} \sqrt{13}} = \frac{2}{\sqrt{585}}$ - g) Cosine of angle between planes $XOY$ (normal $\vec{k} = (0,0,1)$) and $A_1A_2A_3$ (normal $\vec{n} = (-3,2,0)$): $\cos \phi = \frac{|\vec{n} \cdot \vec{k}|}{|\vec{n}||\vec{k}|} = \frac{|0|}{\sqrt{13} \cdot 1} = 0$ - h) Projection of $A_4$ on plane $A_1A_2A_3$: Vector $\overrightarrow{A_1A_4} = (-4,-5,-2)$ Projection formula: $P' = A_4 - \frac{(\overrightarrow{A_1A_4} \cdot \vec{n})}{|\vec{n}|^2} \vec{n}$ Dot product: $\overrightarrow{A_1A_4} \cdot \vec{n} = 2$ (from f) $|\vec{n}|^2 = 13$ So: $P' = (-1,0,2) - \frac{2}{13}(-3,2,0) = \left(-1 + \frac{6}{13}, 0 - \frac{4}{13}, 2 - 0\right) = \left(-\frac{7}{13}, -\frac{4}{13}, 2\right)$ - k) Plane through $A_2(5,8,3)$ parallel to vectors $\overrightarrow{A_1A_4} = (-4,-5,-2)$ and $\overrightarrow{A_3A_4} = (-2,-2,4)$: Normal vector: $\vec{n} = \overrightarrow{A_1A_4} \times \overrightarrow{A_3A_4} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ -4 & -5 & -2 \\ -2 & -2 & 4 \end{vmatrix} = (-5 \cdot 4 - (-2)(-2), -(-4 \cdot 4 - (-2)(-2)), -4 \cdot (-2) - (-5)(-2)) = (-20 -4, -(-16 -4), 8 -10) = (-24, 20, -2)$ Simplify dividing by 2: $\vec{n} = (-12,10,-1)$ Plane equation: $-12(x-5) + 10(y-8) -1(z-3) = 0 \Rightarrow -12x + 60 + 10y - 80 - z + 3 = 0 \Rightarrow -12x + 10y - z -17 = 0$ --- **Final answers:** a) Plane $A_1A_2A_3$: $-3x + 2y - 1 = 0$ b) Line $A_1A_2$: $x=3+2t$, $y=5+3t$, $z=4 - t$ c) Line $A_4M$: $x = -1 - 3s$, $y = 2s$, $z=2$ d) Line $A_3N$: $x=1 - 2u$, $y=2 - 3u$, $z=-2 - 6u$ e) Plane through $A_4$ perpendicular to $A_1A_2$: $2x + 3y - z + 4 = 0$ f) $\sin \theta = \frac{2}{\sqrt{585}}$ g) $\cos \phi = 0$ h) Projection of $A_4$ on plane $A_1A_2A_3$: $\left(-\frac{7}{13}, -\frac{4}{13}, 2\right)$ k) Plane through $A_2$ parallel to $\overrightarrow{A_1A_4}$ and $\overrightarrow{A_3A_4}$: $-12x + 10y - z - 17 = 0$