how to determine whether the line and plane are parallel?

Question

the line is
x = -5 - 4t
y = 1 - t
z = 3 + 2t

the plane is x + 2y + 3z - 9 = 0

Answer 1

Here's a really rough method.

If the line intersected the plane at any point, then the cordinates of that point will fit the equation for the line and the plane.

So let's sub the x,y,z cords for the line into the plane eqn.
(-5-4t) + 2(1-t) + 3(3+2t) - 9 = 0
the t's cancel off leaving
6 - 9 = 0

My interpretation of that is that the line never intersects the plane. Hence the line is parallel to the plane.
If we had gotten zero on both sides, then we could have concluded that the line was on the plane.

Answer 2

If the line and the plane are parallel then the normal vector to the plane will be perpendicular to the directional vector of the line. So the dot product of the vectors would be zero.

The plane is x + 2y + 3z - 9 = 0

The normal vector n of the plane is:

n = <1, 2, 3>

The directional vector v of the line is:

v = <-4, -1, 2>

Now take the dot product.

n • v = <1, 2, 3> • <-4, -1, 2> = -4 - 2 + 6 = 0

Now we know that the line is either parallel to the plane or lies completely in the plane. Take any point on the line and see if it is also in the plane. If the point is in the plane the entire line is in the plane. If the point is not in the plane, the none of the line is in the plane. The line and plane are parallel.

To pick a point on the line let t = 0. The point is P(-5, 1, 3).

1(-5) + 2*1 + 3*3 - 9 = -3 ≠ 0

The point is not in the plane so the line does not lie in the plane.

The line and plane are parallel.

Answer 3

If the directional vector of the line is orthogonal to the normal vector of the plane, the line and the plane are parallel. The dot product of orthogonal vectors is zero. Let v = directional vector of line n = normal vector of plane v = <3, 2, -1> n = <4, -1, 2> Take the dot product. v • n = <3, 2, -1> • <4, -1, 2> = 12 - 2 - 2 = 8 ≠ 0 The dot product is not zero, so the line and plane are not parallel.

Answer 4

we can substitute x, y, and z to the equation of plane.

- 5 - 4t + 2 - 2t + 9 + 6t - 9 = 0
- 3 = 0

the variable t is now disapppear, so it means that whatever the value of t is, the equation is always correct. so, it can conclude that if we see the location of line x, y, or z on the plane, there are always located on the plane. since the lines are located in the plane, it means that they're parallel to the plane.

Answer 5

putting values of x,y,z
-5-4t+2-2t+9+6t-9=0
-3=0
which is not true
so pt. of line lies on the plane
so they r parallel for all values of t
hence line is || to plane.

Answer 6

this is just a guess ,
(-5-4t) + 2(1-t)+3(3+2t)-9=0
-5-4t+ 2-2t + 9+6t -9=0
-4t-2t+6t-9+9-5=0
-5=0
Hehe, i guess this is now the answer.. . . .

Answer 7

Dr D's method is correct.