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
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
