Why does the cross product of two direction vectors....

[Specter]

Homework Statement

[/B]
Hopefully this is in the correct section I looked around for others but this seemed like the right one.

Find the scalar, vector, and parametric equations of the plane that passes through the points P(1,0,4), Q(3,1,-6), and R(-2,3,5).

Homework Equations

The Attempt at a Solution

I found the vector and parametric equations but couldn't find the scalar equation. I eventually tried finding the cross product of two of the direction vectors and it gave me the correct answer for the scalar equation, except for D which I found by substituting in a point.

Could anyone explain why the cross product of two direction vectors gives me the scalar equation, or is this one of those things that I should just remember?

 

[Mark44]

Homework Statement

[/B]
Hopefully this is in the correct section I looked around for others but this seemed like the right one.

Find the scalar, vector, and parametric equations of the plane that passes through the points P(1,0,4), Q(3,1,-6), and R(-2,3,5).

Homework Equations

The Attempt at a Solution

I found the vector and parametric equations but couldn't find the scalar equation. I eventually tried finding the cross product of two of the direction vectors and it gave me the correct answer for the scalar equation, except for D which I found by substituting in a point.

Could anyone explain why the cross product of two direction vectors gives me the scalar equation, or is this one of those things that I should just remember?

The cross product doesn't give you the scalar equation -- it gives you the normal to the two direction vectors. So if $\vec u$ and $\vec v$ are the two direction vectors (obtained from your given points), then $\vec u \times \vec v$ will be a vector $\vec N = <A, B, C>$ that is normal to (perpendicular to) the two other vectors. The scalar equation will then be Ax + By + Cz + D = 0.

You can see why this works by taking an arbitrary point in the plane, say S(x, y, z), and one of your given points, say Q. Then the vector SQ has to be perpendicular to $\vec N$, so their dot product will be zero. That will give you an equation that is your scalar equation.

 

[Specter]

The cross product doesn't give you the scalar equation -- it gives you the normal to the two direction vectors. So if $\vec u$ and $\vec v$ are the two direction vectors (obtained from your given points), then $\vec u \times \vec v$ will be a vector $\vec N = <A, B, C>$ that is normal to (perpendicular to) the two other vectors. The scalar equation will then be Ax + By + Cz + D = 0.

You can see why this works by taking an arbitrary point in the plane, say S(x, y, z), and one of your given points, say Q. Then the vector SQ has to be perpendicular to $\vec N$, so their dot product will be zero. That will give you an equation that is your scalar equation.

That makes sense. Thank you!