# [SOLVED]Plane parallel to the xy plane?

#### Raerin

##### Member
What is the equation of a plane that passes through the point (4,2,-7) and is parallel to the xy plane?

I know that on the xy plane z = 0, so the direction vector is (1,1,0). Would a direction vector of the parallel line be (2,2,0)?

If so the final equation is p=(4,2,-7)+s(1,1,0) + t(2,2,0)?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
What is the equation of a plane that passes through the point (4,2,-7) and is parallel to the xy plane?

I know that on the xy plane z = 0, so the direction vector is (1,1,0). Would a direction vector of the parallel line be (2,2,0)?

If so the final equation is p=(4,2,-7)+s(1,1,0) + t(2,2,0)?
A plane has 2 independent direction vectors.
I'm afraid (1,1,0) and (2,2,0) are not independent.
That's because one is a multiple of the other.
Proper independent vectors would for instance be (1,0,0) and (0,1,0).
They both have z-coordinate 0, and one is not a multiple of the other.

Anyway, the question asks for an equation.
That would be an equation of the form ax+by+cz=d.

Since you know that the plane is supposed to be parallel to the xy plane, what would its general form be?

#### Raerin

##### Member
A plane has 2 independent direction vectors.
I'm afraid (1,1,0) and (2,2,0) are not independent.
That because one is a multiple of the other.
Proper independent vectors would for instance be (1,0,0) and (0,1,0).
They both have z-coordinate 0, and one is not a multiple of the other.

Anyway, the question asks for an equation.
That would be an equation of the form ax+by+cz=d.

Since you know that the plane is supposed to be parallel to the xy plane, what would its general form be?
So the second direction vector in the parallel plane just need to have z=0 and be non-colinear with the first vector?

For (1,0,0) the equation would be:

z + 7 = 0?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
So the second direction vector in the parallel plane just need to have z=0 and be non-colinear with the first vector?
Yep.

For (1,0,0) the equation would be:

z + 7 = 0?
Not sure how you tie (1,0,0) in, but yes, that is the equation of the plane parallel to the xy plane which contains (4,2,-7). #### Raerin

##### Member
Yep.

Not sure how you tie (1,0,0) in, but yes, that is the equation of the plane parallel to the xy plane which contains (4,2,-7). I found the cross product between (1,1,0) and (1,0,0). The result was (0,0,1) and I just subbed in (4, 2, -7) to z=0 to find the d value of the equation ax+by+cz+d=0

This method is correct, right? Or was my answer just coincidentally right?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
I found the cross product between (1,1,0) and (1,0,0). The result was (0,0,1) and I just subbed in (4, 2, -7) to z=0 to find the d value of the equation ax+by+cz+d=0

This method is correct, right? Or was my answer just coincidentally right?
The method is correct yes.

#### HallsofIvy

##### Well-known member
MHB Math Helper
What is the equation of a plane that passes through the point (4,2,-7) and is parallel to the xy plane?

I know that on the xy plane z = 0, so the direction vector is (1,1,0).
The "direction vector" of what? You use "direction vectors" to find the equaton of a line, not a plane. For a plane, you want to use the normal/b] vector.

Would a direction vector of the parallel line be (2,2,0)?
And here you say "parallel line" rather than "plane". You appear to be confusing the two.

If so the final equation is p=(4,2,-7)+s(1,1,0) + t(2,2,0)?
If the xy- plane, where every point has the form (x, y, 0) has equation z= 0, isn't it obvious that the equation of a plane parallel to that must be "z= constant" for some constant?