Answer:
[tex]x=1+t\\y=4-t\\z=4+2t[/tex]
Step-by-step explanation:
A vector perpendicular to the plane ax+by+cz+d=0 is of the form [tex](a,b,c)[/tex].
So, a vector perpendicular to the plane x − y + 2z = 7 is [tex](1,-1,2)[/tex].
The parametric equations of a line through the point [tex](x_0,y_0,z_0)[/tex] and parallel to the vector [tex](a,b,c)[/tex] are as follows:
[tex]x=x_0+at\\y=y_0+bt\\z=z_0+ct[/tex]
Put [tex](x_0,y_0,z_0)=(1,4,4)[/tex] and [tex](a,b,c)=(1,-1,2)[/tex]
Therefore,
[tex]x=1+t\\y=4-t\\z=4+2t[/tex]
xy-plane:
Put z = 0 ⇒ t = -2 ⇒x = - 1 , y = 6
So, at point (-1,6,0)
yz-plane:
Put x = 0 ⇒ t = -1 ⇒ y = 5, z =2
So, at point (0,5,2)
xz-plane:
Put y = 0 ⇒ t = 4 ⇒ x = 5, z = 12
So, at point (5,0,12)
