Respuesta :

LammettHash

Replace x, y, and z in the plane's equation with their appropriate parametric equations, then solve for t, then evaluate x, y, and z at this t to find the point P.

[tex]x+y-z=-5[/tex]

[tex](1+t)+2t-(-3t)=-5[/tex]

[tex]6t+1=-5[/tex]

[tex]6t=-6[/tex]

[tex]t=-1[/tex]

[tex]\implies x=0,y=-2,z=-3[/tex]

So the point P is (0, -2, -3).

adioabiola

Here, we are required to find the point, P where the lines:

  • x = 1 + t
  • x = 1 + ty = 2t
  • x = 1 + ty = 2tz = -3t

intersects the plane x + y - z = -5.

The point, P where the lines intersects the plane is given by;

he point, P where the lines intersects the plane is given by;P (0, -2, 3)

Therefore, to find this point, we have to substitute for x, y and z in the equation of the plane.

Therefore, we have;

  • 1 + t + 2t -(-3t) = -5.
  • therefore 6t = - 6
  • and ultimately t = -1

To get the coordinates of the point P where the lines intersect the plane; we can then find x, y and z by substituting t into each of their equations.

Therefore;

  • x = 1 + (-1) = 0
  • x = 1 + (-1) = 0y = 2(-1) = -2
  • x = 1 + (-1) = 0y = 2(-1) = -2z = -3(-1) = 3

Therefore, the point, P where the lines intersects the plane is given by;

  • P (0, -2, 3)

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