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#### Raerin

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- Oct 7, 2013

- 46

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)?

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- Thread starter
- #1

- Oct 7, 2013

- 46

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)?

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- Mar 5, 2012

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A plane has 2 independent direction vectors.

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)?

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

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?

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- Oct 7, 2013

- 46

So the second direction vector in the parallel plane just need to have z=0 and be non-colinear with the first vector?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 isnota 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?

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

z + 7 = 0?

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- Mar 5, 2012

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

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

z + 7 = 0?

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- #5

- Oct 7, 2013

- 46

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=0Yep.

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).

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

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- Mar 5, 2012

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The method is correct yes.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?

- Jan 29, 2012

- 1,151

The "direction vector" of what? You use "direction vectors" to find the equaton of aWhat 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).

And here youWould a direction vector of the parallel line be (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?If so the final equation is p=(4,2,-7)+s(1,1,0) + t(2,2,0)?