How do you find parametric equations for the line through P-naught=(3,-1,1) perpendicular to the plane 3x+5y-7z=29?

Question

Cameron Alexander · Accepted Answer

# { (x = 3 + lambda 3), (y = -1+lambda 5), (z = 1 -lambda 7) :}#

Plane #Pi->3x+5y-7z=29# can be written as

#<< vec v , p-p_0 >> = 0#

in which #p = {x,y,z}, p_0 = {x_0,y_0,y_0} # and #vec v# is a normal vector to the plane and #p-p_0# are segments pertaining to the plane.

For #Pi# we have #vec v = {3,5,-7}# so the line perpendicular to #Pi# passing by #q_0 = {3,-1,1}# is

#l->p = q_0 + lambda vec v# or

{ (x = 3 + lambda 3), (y = -1+lambda 5), (z = 1 -lambda 7) :}#

Emma Aldridge · Answer

undefined To find parametric equations for the line through $ P_0 = (3, -1, 1) $ perpendicular to the plane $ 3x + 5y - 7z = 29 $, we first find the normal vector to the plane. Then, we use this normal vector to determine the direction vector of the line. Finally, we use the given point $ P_0 $ to write the parametric equations of the line.

The normal vector to the plane $ 3x + 5y - 7z = 29 $ is $ \mathbf{n} = \langle 3, 5, -7 \rangle $.

The direction vector of the line perpendicular to the plane is the same as the normal vector to the plane, so $ \mathbf{d} = \langle 3, 5, -7 \rangle $.

Thus, the parametric equations of the line are:

$$ x(t) = 3 + 3t $$
$$ y(t) = -1 + 5t $$
$$ z(t) = 1 - 7t $$

where $ t $ is a parameter.