Respuesta :
Answer:
[tex](\frac{1}{2},\frac{-1}{2},\frac{1}{2})[/tex]
Step-by-step explanation:
We have been given the intersection coordinates:
(7,1,-6) and perpendicular to the line x=-1+t ,y=-2+t and z=-1+t
From the condition of perpendicularity we know:
[tex](7,1,-6)(x,y,z)[/tex]
[tex]\Rightarrow 7x+y-6z=0[/tex]
Now, we have given x=-1+t , y=-2+t and z= -1+t
[tex]7(-1+t)+(-2+t)-6(-1+t)=0[/tex]
[tex]-7+7t-2+t+6-6t=0[/tex]
[tex]\Rightarrow -3+2t=0[/tex]
[tex]2t=3[/tex]
[tex]t=\frac{3}{2}[/tex]
Now, substituting t in the given coordinates x,y and z we get:
[tex]x=-1+\frac{3}{2}=\frac{1}{2}[/tex]
[tex]y=-2+\frac{3}{2}=\frac{-1}{2}[/tex]
And [tex]z=-1+\frac{3}{2}=\frac{1}{2}[/tex]
Answer: The answer is [tex]x=-1+2i,~y=-2+2j,~z=-1+2k.[/tex]
Step-by-step explanation: The vector equation of the line can be written as
[tex](x,y,z)=(-1,-2,-1)+t(i+j+k).[/tex]
The general equation of the plane that is perpendicular to this line will be
[tex]x+y+z=c.[/tex]
Since it contains the point (7,1,-6), so
[tex]7+1-6=c\\\\\Rightarrow c=2.[/tex]
Therefore, the plane x+y+z=2 contains the point (7,1,-6) and is perpendicular to the line.
Now, we will substitute the parametric equations of the line into the equation of the plane as follows
[tex]x+y+z=2\\\\\Rightarrow (-1+t)+(-2+t)+(-1+t)=2\\\\\Rightarrow 3t=6\\\\\Rightarrow t=2.[/tex]
So, the parametric equation of the new line is
[tex](x,y,z)=(-1,-2,-1)+2(i+j+k)\\\\\Rightarrow x=-1+2i,~y=-2+2j,~z=-1+2k.[/tex]
