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The equation of a plane in the three-dimensional space is determined with the help of a normal vector and any point on the plane which is known. A vector is a physical quantity that not only has a magnitude but also has a direction. The line which is perpendicular to the plane is called the normal vector of the plane. In a three dimensional Cartesian system, position vector defines the position of any point on the plane with reference to the origin. This article is curated to explain the general equation and vector equation of a plane in the 3D Cartesian system.

Equation of a plane can be defined by the following methods:

Using the equation with three non-collinear points.

One non-coincident line and two parallel lines.

The normal vector and the point on the plane.

Two lines intersecting on a plane.

A straight line and a point on the plane.

There can be an infinite number of planes that have the same normal vector, but only one plane has a specific point in it which remains perpendicular to it throughout.

Let us consider the following 3D representation to understand the normal vector of a plane, the normal vector formula and the vector equation of a plane.

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Here we can see that the plane is passing through the fixed point P1(x1, y1, z1) having a position vector \[\vec{r_{1}}\]. The vector normal to the plane is \[\vec{N}\]. Suppose there is an arbitrary point P(x, y, z) lying on this plane, the position vector of the point is \[\vec{r}\]. The equation of the line in vector form for P1P can be represented as:

\[\vec{P_{1} P} = \vec{r} - \vec{r_{1}}\]

The above vector lies on the plane. Since \[\vec{N}\] is orthogonal to the plane, it is also said to be orthogonal to any vector that lies in the plane. Thus, the dot product or the scalar product of the normal vector \[\vec{N}\] and the vector on the plane drawn from P1 to P i.e. \[\vec{P_{1} P}\] is always zero.

\[\vec{P_{1} P} \cdot \vec{N} = 0\]

Thus, the vector equation of the plane is:

\[(\vec{r} - \vec{r_{1}}). \vec{N} = 0\]

Here, the vector form of the normal vector formula is \[\vec{N} = A \widehat{i} + B \widehat{j} + C \widehat{k}\]. Here A, B, and C are the direction cosines of the unit vectorÂ parallelÂ to the normal vector to the planet. The vectors \[\vec{r}\] and \[\vec{r_{1}}\] can be defined by the following:

\[\vec{r} = x \widehat{i} + y \widehat{j} + z \widehat{k}\]

\[\vec{r_{1}} = x_{1} \widehat{i} + y_{1} \widehat{j} + z_{1} \widehat{k}\]

Substituting the vector equation with the above values we get:

\[[(x \widehat{i} + y \widehat{j} + z \widehat{k}) - (x_{1} \widehat{i} + y_{1} \widehat{i} + z_{1} \widehat{k})] \cdot (A \widehat{i} + B \widehat{k} + C \widehat{k}) = 0\]

\[[(x - x_{1}) \widehat{i} + (y - y_{1}) \widehat{j} + (z - z_{1}) \widehat{k}] \cdot (A \widehat{i} + B \widehat{j} + C \widehat{k}) = 0\]

Therefore,

\[A( x - x_{1}) + B (y - y_{1}) + C(z - z_{1}) = 0\]

Thus, the general equation of a plane is :

Ax + By + Cz + D = 0

Where, \[D = -(Ax_{1} + By_{1} + Cz_{1})\]

When the plane is supposed to pass through the origin O, then x = 0, y = 0 and z = 0. Thus substituting the values of x, y and z in the above equation we get:

A.0 + B.0 + C.0 + D = 0

Therefore, D = 0

Thus, when we have an equation of a line that says D = 0, that means the plane passes through the origin.

When writing the equation of a plane in a normal form derived from the vector equation we get:

\[\vec{r} \cdot \widehat{N} = d\]

where \[\vec{r}\] is the position vector of an arbitrary point lying on the plane and N is a unit normal vector parallel to the normal that joins the origin to the plane. D here is the perpendicularÂ distanceÂ of the plane from the origin.

In the given figure, distance from point A having coordinates (x0, y0, z0) to the plane is positive as it is on the opposite side to side of the origin with respect to the plane.

(image will be uploaded soon)

The length of the normal segment form the origin to point A through the plane is written as p + d. Here p is the length of the perpendicular from origin to plane and d is the perpendicular distance from point A to the plane.

Thus the distance formula is:

\[x_{0} cos \alpha + y_{0} cos \beta + z_{0} cos \gamma = p + d\]

Thus distance of point A to the plane is:

\[d = x_{0} cos \alpha + y_{0} cos \beta + z_{0} cos \gamma - p\]

Now from the normal form where, \[\vec{N} = \widehat{i} + B \widehat{j} + C \widehat{k}\], the unit vector of the normal vector is:

\[N_{0} = \frac{\vec{N}}{|\vec{N}|} = \frac{A \widehat{i} + B \widehat{j} + C \widehat{k}}{\pm \sqrt{A^{2} + B^{2} + C^{2}}} = cos \alpha \widehat{i} + cos \beta \widehat{j} + cos \gamma \widehat{k}\]

Here,

\[cos \alpha = \frac{A}{\pm \sqrt{A^{2} + B^{2} + C^{2}}}, cos \beta = \frac{B}{\pm \sqrt{A^{2} + B^{2} + C^{2}}} \text{ and } cos \gamma = \frac{C}{\pm \sqrt{A^{2} + B^{2} + C^{2}}}\]

Thus, in order to convert the equation of a plane in the normal form to Hessianâ€™s normal form, we have to divide the general form of the equation by \[\pm \sqrt{A^{2} + B^{2} + C^{2}}\]. The sign of the square root should be opposite to that of D where D â‰ 0.

\[p = \frac{-D}{\pm \sqrt{A^{2} + B^{2} + C^{2}}}\]

Just as the equation of a line in three dimensionsÂ is notÂ unique, it is similar for the equation of a plane due to choosing different points or vectors.

FAQ (Frequently Asked Questions)

Q1. The Graph of a Plane is in the First Octant. Find the Equation of the Plane.

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Ans: Let us first look at the graph and determine the points of intersection of the plane in the x, y and z axes. Before doing that let us take into account the general form of the equation of the plane:

Ax + By + Cz + D = 0

The plane intersects x-axis when y = 0 and z = 0. Thus, in such a case = -D/A .

Similarly when the plane intersects y-axis then y= -D/B.

In the same fashion when the plane intersects z-axis then z= -D/C.

When we look at the graph provided with the question, we can say:

4= -D/A

5= -D/B

3= -D/C

Thus, the equation of the plane is general form stands at:

x/4 + y/5 + z/3 = 1

Q2. Find an Equation for the Plane Passing Through the PointÂ P(2, -1, 1) and Parallel to the Plane:

2x + 3y - zÂ =Â 5.

Ans: Parallel planes share common normals. Thus, any plane that is parallel to 2x + 3y - zÂ =Â 5Â has a normal N^{â†’} 2 iĚ‚ + 3 jĚ‚ - kĚ‚. Thus we can say that:

N^{âź¶}Â =Â âź¨2,3,-1âź©

Now the point P(2,-1,1), lies on the plane parallel to the plane 2x + 3y - zÂ =Â 5. So we can say that:

r_{1}^{âź¶} = 2 iĚ‚ + 3 jĚ‚ - kĚ‚

r_{1}^{âź¶}Â =Â âź¨2, -1, 1âź©

Now letÂ Q (x,y,x) be an arbitrary point on the parallel plane. Then the vector of P will be:

r_{1}^{âź¶} = xiĚ‚ + yjĚ‚ + zkĚ‚

On writing the vector form of PQ we get

P^{âź¶}Q = âź¨(x - 2), (y + 1), (z - 1)âź©

Thus the equation of the plane will be:

(r^{âź¶} - r1) . N^{âź¶} = 0

âź¨x - 2, y + 1, z - 1âź©.âź¨2, 3, -1âź© = 0

2x - 2 + 3y + 1 - (z - 1) = 0

2x + 3y - z = 0