Question

Find an equation of the plane. the plane through the point (4, 6, 7) and with normal vector 5i + j − k

Answer 1

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 2

Answer:

5x +y -z -19= 0

Step-by-step explanation:

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

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

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

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>.

∴ $x_0=4$, $y_0=6$, $z_0=7$

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:

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