Respuesta :
Answer:
[tex]22x-8y+28z=-144[/tex]
Step-by-step explanation:
Equations of lines,
x + y - z = 2,
2x - y + 3z = 2
If z = 0,
Equations would be,
x + y = 2
2x - y = 2
Adding these equations, we get, 3x = 4 ⇒ x = [tex]\frac{4}{3}[/tex]
[tex]\implies \frac{4}{3} + y = 2\implies y = 2 - \frac{4}{3} = \frac{6-4}{3} = \frac{2}{3}[/tex]
Thus, solution would be ( 4/3, 2/3, 0)
Now, if x = 0,
Equation are,
y - z = 2,
-y + 3z = 2
Adding equations, We get, 2z = 4 ⇒ z = 2,
⇒ y - 2 = 2 ⇒ y = 2 + 2 = 4,
Thus, solution would be ( 0, 4, 2 )
Let a be the vector from (-2, 2, 3) to (4/3, 2/3, 0)
⇒ a = (4/3 + 2)i + (2/3 -2)j + (0-3)k = [tex]\frac{10}{3}i-\frac{4}{3}j-3k[/tex]
Similarly,
Let b be the vector from (-2, 2, 3) to (0, 4, 2),
⇒ b = (0+2)i + (4-2)j + (2-3)k ⇒ b = 2i + 2j - k
So, the normal of the plane is,
[tex]n=a\times b = \begin{vmatrix}i & j & k \\ \frac{10}{3} & -\frac{4}{3} & -3 \\ 2 & 2 & -1\end{vmatrix}=\frac{22}{3}i-\frac{8}{3}j+\frac{28}{3}k[/tex]
∵ General equation of a plane,
[tex]n.<x-x_0, y-y_0, z-z_0>=0[/tex]
Hence, the equation of the given plane is,
[tex]<\frac{22}{3}, -\frac{8}{3}, \frac{28}{3}><x+2, y-2, z+3>[/tex]
[tex]\frac{22}{3}(x+2) -\frac{8}{3}(y-2)+\frac{28}{3}(z+3)=0[/tex]
[tex]22(x+2)-8(y-2)+28(z+3)=0[/tex]
[tex]22x+44-8y+16+28z+84=0[/tex]
[tex]\implies 22x-8y+28z=-144[/tex]
