Respuesta :
Answer:
The equation of the line is given by:
[tex]x = 1 + t[/tex]
[tex]y = 1 - 2t[/tex]
[tex]z = 1 + t[/tex]
Step-by-step explanation:
Parametrizing the equation of the line in function of t, the equation of the line is given by:
[tex]x = x_0 + at[/tex]
[tex]y = y_0 + bt[/tex]
[tex]z = z_0 + ct[/tex]
In which we have an initial point [tex](x_0,y_0,z_0)[/tex] and [tex](a,b,c)[/tex] is a vector parallel to the line.
Line passing through the point (1, 1, 1)
This means that [tex]x_0 = y_0 = z_0 = 1[/tex]. So
[tex]x = 1 + at[/tex]
[tex]y = 1 + bt[/tex]
[tex]z = 1 + ct[/tex]
Perpendicular to the plane containing the points (1, 0, 0), (2, 1, 1) and (1, 1, 2).
From this, we get two vectors:
[tex](2-1,1-0,1-0) = (1,1,1)[/tex]
[tex](1-2,1-1,2-1) = (-1,0,1)[/tex]
The parallel vector is given by the determinant of the following matrix:
[tex]\left[\begin{array}{ccc}i&j&k\\1&1&1\\-1&0&1\end{array}\right][/tex]
Which is:
[tex]D = i*1*1 + j*1*-1 + k*1*0 - k*1*(-1) - j*1*1 -i*1*0[/tex]
[tex]D = i - j + k - j[/tex]
[tex]D = i - 2j + k[/tex]
So the vector is (1,-2,1), and the equation of the line is:
[tex]x = 1 + t[/tex]
[tex]y = 1 - 2t[/tex]
[tex]z = 1 + t[/tex]
