Find the parametric equations for the line through the point P that is perpendicular to the plane?

Question

a)Find the parametric equations for the line through the point P = (4, -4, 1) that is perpendicular to the plane 3x + 1y - 4z = 1.
t = 0
x=
y=
z=
B)At what point Q does this line intersect the yz-plane?

Answer 1

A) The normal to the plane is (3,1,-4) so the parametic equation of the line is just:
x=4+3t
y=-4+t
z=1-4t

B) the yz plane is when x = 0 so solving for t gives:
0 = 4+3t
t = -4/3
Sub that back in gives:
(0, -16/3, 13/3)

Answer 2

a) Find the parametric equations for the line through the point
P = (4, -4, 1) that is perpendicular to the plane
3x + 1y - 4z = 1.

Since the line is perpendicular to the plane, the directional vector v, of the line is the same as the normal of the plane.

v = <3, 1, -4>

The equation of the line thru P with directional vector v is:

L(t) = P + tv
L(t) = <4, -4, 1> + t<3, 1, -4>
where t is a scalar rangining over the real numbers

Rewrite the equation in parametric form.

L(t):
x = 4 + 3t
y = -4 + t
z = 1 - 4t
_________

b) At what point Q does this line intersect the yz-plane?

The equation of the yz plane is

x = 0

Plug the value of x from the equation of the line L into the equation of the plane and solve for t.

x = 4 + 3t = 0
3t = -4
t = -4/3

Now solve for Q.

x = 4 + 3(-4/3) = 0
y = -4 + t = -4 - 4/3 = -16/3
z = 1 - 4t = 1 - 4(-4/3) = 19/3

Q(0, -16/3, 19/3)

The line intersects the yz plane at Q(0, -16/3, 19/3).
__________

Answer 3

the traditional to the airplane 4x - 4y + 0z = a million is (4, -4, 0) so the line is parallel to (4, -4, 0). the line additionally passes by P = (2, -2, -2) This makes the equation of the line: (x, y, z) = (2 + 4t, -2 - 4t, -2), or x = 2 + 4t y = -2 - 4t z = -2