Уравнението на правата $l_ {AB} \begin{cases}A(2,1,1) \\ B(4, -2, 1) \end{cases} \rightarrow l_{AB}: r(t)=(B-A)\cdot{}t+A \Leftrightarrow \\ r(t)= \left((4,-2,1)-(2,1,1) \right)\cdot{}t+(2,1,1)=(2,-3,0)t+(2,1,1) \Rightarrow l_{AB}:\begin{cases} x=2t+2 \\ y=-3t+1 \\ z=0t+1\end{cases} \\[12pt]$ Уравнението на равнината $\alpha_{CDE} \begin{cases} C(0,1,0)\\ D(2,2,0) \\ E(1,0,1) \end{cases} \\[6pt] \vec{CD}=D-C=(2,2,0)-(0,1,0)=(2, 1, 0) \\[6pt] \vec{CE}=E-C=(1, 0, 1)-(0, 1, 0)=(1, -1, 1) \\ \vec{CD}\times{}\vec{CE}=\begin{vmatrix} i&j&k\\2&1&0\\1&-1&1 \end{vmatrix}=(-1)^{1+1}\cdot{}i\cdot{}\begin{vmatrix} 1&0\\-1&1\end{vmatrix}+(-1)^{1+2}\cdot{}j\cdot{}\begin{vmatrix}2&0\\1&1 \end{vmatrix}+(-1)^{1+3}\cdot{}k\cdot{}\begin{vmatrix} 2&1\\1&-1\end{vmatrix} \\[6pt] \vec{CD}\times{}\vec{CE}=(1\cdot{}1-0\cdot{}(-1))i-(2\cdot{}1-0\cdot{}1)j+(2\cdot{}(-1)-1\cdot{}1)k \\[6pt] \vec{CD}\times{}\vec{CE}=1i-2j-3k=\underbrace{\begin{pmatrix}1\\-2\\-3 \end{pmatrix}}_{n}\cdot{}\begin{pmatrix} i&j&k \end{pmatrix} \\ \alpha_{CDE}: n\cdot{}\left(\begin{pmatrix}x&y&z\end{pmatrix}-C\right)=0 \Rightarrow \quad \begin{pmatrix}1\\-2\\-3 \end{pmatrix}\cdot{}\begin{pmatrix} x-0&y-1&z-0 \end{pmatrix} =0 \\ \quad 1(x-0)-2(y-1)-3(z-0)=0 \\ \quad x-0-2y+2-3z+0=0 \\ \boxed{\quad \alpha_{CDE}: \quad x-2y-3z+2=0 \quad } \\[12pt] l_{AB}\cap{}\alpha_{CDE}=H(x_{H}, y_{H}, z_{H}) \\[6pt] H\in{}l_{AB} \Rightarrow x_{H}=2t_{H}+2, \quad y_{H}=-3t_{H}+1, \quad z_{H}=1 \\[6pt] H \in \alpha_{CDE} \Rightarrow \quad x_{H}-2y_{H}-3z_{H}+2=0 \quad \Leftrightarrow \quad 2t_{H}+2-2(-3t_{H}+1)-3\cdot{}1+2=0 \\[6pt] \quad 2t_{H}+2+6t_{H}-2-1=0 \\[6pt] \quad 8t_{H}=1 \\[6pt] \quad t_{H}=\dfrac{1}{8} \\ x_{H}=2t_{H}+2=2\cdot{}\dfrac{1}{8}+2=2\frac{1}{4}, \quad y_{H}=-3t_{H}+1=-3\cdot{}\dfrac{1}{8}+1=\frac{5}{8}, \quad z_{H}=1 $ $$ H\begin{pmatrix} 2\dfrac{1}{4},&\dfrac{5}{8},&1\end{pmatrix} $$
Проверете сметките за изчислителни грешки или грешки при пренасянията.

- Screenshot 2024-12-26 133357.png (74.68 KiB) Прегледано 159 пъти
[tex]\color{lightseagreen}\text{''Който никога не е правил грешка, никога не е опитвал нещо ново.''} \\
\hspace{21em}\text{(Алберт Айнщайн)}[/tex]