Logic behind normal line in expressing plane

[feryee]

why do we consider normal line in expressing a plane,say in $R^3$ ,of the form $ax + by + cz = d$? What is the logic behind this normal line selection? Plz provide intuitive explanations.Thanks

 

[mfb]

It is a convenient way to express the position and orientation of the plane with a single equation.
There is no ambiguity in the selection apart from an overall scaling of the equation (you can multiply a,b,c,d by a non-zero constant without changing the plane).

 

[fresh_42]

why do we consider normal line in expressing a plane,say in $R^3$ ,of the form $ax + by + cz = d$? What is the logic behind this normal line selection? Plz provide intuitive explanations.Thanks

The right hand rule might be a suitable illustration why it is useful.

 

[feryee]

It is a convenient way to express the position and orientation of the plane with a single equation.
There is no ambiguity in the selection apart from an overall scaling of the equation (you can multiply a,b,c,d by a non-zero constant without changing the plane).

What exactly do you mean by convenient way? I couldn't still get it clearly? would you elaborate more on this!

 

[Mark44]

For a plane in $\mathbb{R}^3$ whose equation is ax + by + cz = d, the vector $\vec{N} = <a, b, c>$ is perpendicular to (or normal to) the plane. The normal gives the orientation of the plane in the three-dimensional space, but that same vector is perpendicular to an infinite family of parallel planes. If you also know a point on the plane, that narrows the choices down to a single plane.

 

[mfb]

What exactly do you mean by convenient way?

It is easy to use it in calculations.

 

[feryee]

For a plane in $\mathbb{R}^3$ whose equation is ax + by + cz = d, the vector $\vec{N} = <a, b, c>$ is perpendicular to (or normal to) the plane. The normal gives the orientation of the plane in the three-dimensional space, but that same vector is perpendicular to an infinite family of parallel planes. If you also know a point on the plane, that narrows the choices down to a single plane.

Thank you very much. That explains quiet well. Can you please tell me how can one claim that this equation actually represent a plane in $\mathbb{R}^3$? If i get your explanations right , the choice of d will determine if we have a single plane or infinite family of parallel plane. right?

 

[Mark44]

Thank you very much. That explains quiet well. Can you please tell me how can one claim that this equation actually represent a plane in $\mathbb{R}^3$?

Think about it in terms of the geometry of the situtation. If you have a vector in $\mathbb{R}^3$, it is perpendicular to an infinite number of planes, all with the same orientation (and therefore parallel). A given value of d identifies one and only one of these planes.

If i get your explanations right , the choice of d will determine if we have a single plane or infinite family of parallel plane. right?

If d is unknown, then yes, we have a family of parallel planes.

For the algebra, let's say we know a vector N = <a, b, c> that is perpendicular to the plane, and a point $P_0(x_0, y_0, z_0)$ that is on the plane. Position the vector so that its tail is at P₀, as in the drawing below.
Take an arbitrary point on the plane P(x, y, z) that is different from P₀ and form the vector $\vec{P_0P} = <x - x_0, y - y_0, z - z_0>$. Since $\vec{P_0P}$ and $\vec{N}$ are perpendicular, their dot product must be zero. IOW, $\vec{P_0P} \cdot \vec{N} = 0$, so $(x - x_0) \cdot a + (y - y_0) \cdot b + (z - z_0) \cdot c = 0$. The equation ax + by + cz = d comes directly from this dot product.

 

[feryee]

Think about it in terms of the geometry of the situtation. If you have a vector in $\mathbb{R}^3$, it is perpendicular to an infinite number of planes, all with the same orientation (and therefore parallel). A given value of d identifies one and only one of these planes.
If d is unknown, then yes, we have a family of parallel planes.

For the algebra, let's say we know a vector N = <a, b, c> that is perpendicular to the plane, and a point $P_0(x_0, y_0, z_0)$ that is on the plane. Position the vector so that its tail is at P₀, as in the drawing below.
Take an arbitrary point on the plane P(x, y, z) that is different from P₀ and form the vector $\vec{P_0P} = <x - x_0, y - y_0, z - z_0>$. Since $\vec{P_0P}$ and $\vec{N}$ are perpendicular, their dot product must be zero. IOW, $\vec{P_0P} \cdot \vec{N} = 0$, so $(x - x_0) \cdot a + (y - y_0) \cdot b + (z - z_0) \cdot c = 0$. The equation ax + by + cz = d comes directly from this dot product.
View attachment 91587

Yeah i see. Really great. But when it comes to plotting the plane by hand, wouldn't this becomes difficult to find all of such a vector on the plane?How can the plotting done in the easy/ handy way(not using software)

 

[Mark44]

Yeah i see. Really great. But when it comes to plotting the plane by hand, wouldn't this becomes difficult to find all of such a vector on the plane?

I'm not sure which vectors you're talking about -- the normal or a vector in the plane.

How can the plotting done in the easy/ handy way(not using software)

The question as you originally asked it didn't have anything to do with graphing a plane. If you have the equation of a plane, it's easy to find three points in the plane, and you can use these points to sketch the plane.

Or, you can find the intersection of the plane in the three coordinate planes.

 

[rs1n]

Yeah i see. Really great. But when it comes to plotting the plane by hand, wouldn't this becomes difficult to find all of such a vector on the plane?How can the plotting done in the easy/ handy way(not using software)

To plot by hand, it is generally much easier to plot where the plane intersects one of the 8 octants. For example, given ax+by+cz=d, set x=0 and you get by+cz=d. This produces a line that you can graph on the yz-plane. Similarly, set y=0 to get ax+cz=d -- a line in the xz-plane. Lastly, ax+by=d (when z=0) gives a line in the xy-plane. These three lines would be the edges of a triangle that lies in the plane ax+by+cz=d.