Respuesta :

The coefficients of x, y, z in the equation of the plane are the same as the normal vector components.
  5x +y -z = 5*4 +6 -7

The equation of the plane is ...
  5x +y -z = 19

Answer:

5x +y -z -19= 0

Step-by-step explanation:

The equation of a plane in the point-normal form is given as:

[tex]\boxed{a(x-x_0)+b(y-y_0)+c(z-z_0)=0}[/tex] , where

  • [tex]P_0(x_0,y_0,z_0)[/tex] is a point in which the plane passes through
  • the normal vector is [tex]< a,b,c >[/tex]

Given that the plane passes through the point (4, 6, 7), we can find the positional vector of this point, which is <4, 6, 7>.

∴ [tex]x_0=4[/tex], [tex]y_0=6[/tex], [tex]z_0=7[/tex]

Normal vector= 5i +j -k

In vector notation, the normal vector can be written as <5, 1, -1>.

∴ a= 5, b= 1, c= -1

Substitute these values into the equation:

5(x -4) +1(y -6) -(z -7)= 0

Expand:

5x -20 +y -6 -z +7= 0

Simplify:

5x +y -z -19= 0

Additional:

For a similar question on equation of plane, do check out the following!

  • https://brainly.com/question/17402797
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