If your given the plane $- x + 6 y + 5 z = 21$ $-x+6y+5z=21$ . How do you find the normal vector for that plane and a point on that plane?

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Answer 1 · 2016-08-31T16:44:01

$\to n = (- 1, 6, 5)$ $vec n = (-1,6,5)$
$p_{0} = (0, 0, \frac{21}{5})$ $p_0 =(0,0,21/5)$

A plane can be represented as

$Pi-> << p-p_0, vec n >> = 0$

where

$p = (x,y,z) in Pi$ is a generic point
$p_0 = (x_0,y_0,z_0) in Pi$ is a given point
and $vec n = (n_x,n_y,n_z)$ is a vector normal to $Pi$

Developping

$n_x(x-x_0)+n_y(y-y_0)+n_z(z-z_0) =$
$x n_x + y n_y + z n_z =-(x_0n_x+y_0n_y+z_0n_z)$

by comparisson we have

$n_x = -1$
$n_y = 6$
$n_z =5$

and

$-(-x_0+6y_0+5z_0) = 21$

choosing $x_0=y_0=0$ we obtain

$z_0 = 21/5$

Finally

$vec n = (-1,6,5)$
$p_0 =(0,0,21/5)$

If your given the plane -x+6y+5z=21−x+6y+5z=21-x+6y+5z=21. How do you find the normal vector for that plane and a point on that plane?