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Part of three-dimensional coordinate geometry, a plane is a flat, 2D surface that stretches put infinitely far. A plane of the equation is a 2D analogue of a point having (zero dimensions), a line (one dimension), and three-dimensional space. A plane in 3 D space contains the following equation.

ax + by + cz + d=0,

in which a minimum of one of the numbers a, b, and c should be non-zero. A plane in a 3-dimensional coordinate space is identified by a point and a vector that is perpendicular to the plane.

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An equation plane in 3D coordinate space is identified by a point and a vector which is perpendicular to the plane. Moreover, suppose that P =(x,y,zâ€‹) be any point in the plane, and r and râ‚€ be the position vectors of points P and P0â€‹, respectively. Now, if we let \[\overrightarrow{n}\] = a,b,c , then since \[\overrightarrow{P0P}\] is perpendicular to \[\overrightarrow {n}\],we have;

\[\overrightarrow {P0â€‹P}\]â€‹â‹…nâ€‹= (râˆ’râ‚€) â‹…n

=(xâˆ’xâ‚€â€‹,yâˆ’yâ‚€â€‹,zâˆ’zâ‚€â€‹)â‹…(a,b,c)

=a(xâˆ’xâ‚€â€‹)+b(yâˆ’yâ‚€â€‹)+c(zâˆ’zâ‚€â€‹)

=0.â€‹

We can also express the above equation of the plane as:

Ax + by + cz + d = 0

In which d= - (axâ‚€ + byâ‚€ + czâ‚€)

This does not quite work if one of a, b, c is equivalent to zero. In such a case, that vector is parallel to one of the coordinate planes. Letâ€™s say c = 0, then the vector is parallel to the xy-plane and the equation of the needed plane is a(x-xâ‚€) + b(y-yâ‚€) = 0 which is of course a straight line in the xy plane and z is unrestricted. Likewise, arguments apply if two of a, b, c are zero.

Another way to imagine the vector equation of a plane in 3d is as a flattened parallelepiped. A flattened parallelepiped, composed of three vectors and has a volume equals to 0 Thus, refer to the image shown for the equation of scalar triple product we can use in order to calculate this volume:

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Now, Suppose that the endpoints of a vector are ( x, y, z )(x,y,z) and (xâ‚€â€‹,yâ‚€â€‹,zâ‚€â€‹) and the components of vector are âŸ¨a,b,câŸ©. Then by taking the dot product, we obtain the equation of a plane, which is as below:

0 = a(x-xâ‚€) + b(y-yâ‚€) + c (z-zâ‚€)

Example:

If a plane is crossing across the point A= (1, 3, 2) and has a normal vector \[\overrightarrow{n}\] = (3, 2, 5). Then, find out the equation of the plane?

Solution:

The equation of the plane that crosses through A= (1,3,2) and consist of a normal vector \[\overrightarrow{n}\] = (3,2,5) is

3(x-1) + 2(y-3) + 5(z-2) = 0

3xâˆ’3+2yâˆ’6+5zâˆ’10 = 0

3x+2y+5zâˆ’19 = 0

Example:

In the case where the plane is crossing through the point A= (5, 6, 2) and has a normal vector \[\overrightarrow{n}\] = (-1, 3,-7). Then, find out the equation of the plane?

Solution:

The equation of the plane that passes through the point A=(5,6,2) and has normal vector \[\overrightarrow{n}\] = (-1,3,-7) is:

âˆ’1(x-5) + 3(y-6) -7(z-2) = 0

âˆ’x + 5 + 3y â€“ 18 âˆ’ 7z + 14 = 0

âˆ’x + 3y âˆ’ 7z + 1 = 0

When we are given three points on a plane, we can determine the equation of the plane by solving simultaneous equations.

Suppose that ax + by + cz + d = 0 be the equation of a plane on which there are the three points: A=(1,0,2), B=(2,1,1) and C=(-1,2,1). Then, the equation of the plane is acknowledged as follows:

We already have the equation of the plane with the four unknown constants:

ax + by + cz +d = 0Â â€¦.(1)

We also obtain the following three equations by substituting the coordinates of A, B, and C into equation (1)

aâ‹…1 + bâ‹…0 + câ‹…2 + d = 0

aâ‹…2 + bâ‹…1 + câ‹…1 + d = 0

aâ‹…(-1) + bâ‹…2 +câ‹…1+d=0

Which provides b=3a, c=4a, d=-9aÂ â€¦(2)

Substituting equation (2) into (1), we have

ax + 3ay + 4az -9a = 0

x + 3y + 4z - 9 =0

therefore, the equation of the plane crossing across the three points A=(1,0,2), B=(2,1,1), and C=(-1,2,1) is

x + 3y + 4z - 9 =0

Considering this method, we can identify the equation of a plane if we know 3 points.

FAQ (Frequently Asked Questions)

Q1. What is Meant by a Normal Vector?

Answer: Letâ€™s start by assuming that we are familiar with a point which is on the plane, Pâ‚€=(xâ‚€, yâ‚€, zâ‚€). Letâ€™s also suppose that we consist of an orthogonal vector (perpendicular) to the plane, â†’n=âŸ¨a, b, câŸ©. This vector is known as the normal vector. Now, assuming that P=(x,y,z) is any point in the plane. Ultimately, because we are going to be working with vectors in the starting and weâ€™ll assume thatâ†’râ‚€ and â†’r be the position vectors for P and Pâ‚€ respectively.

Observe that we added in the vector â†’râˆ’â†’râ‚€ that will lie completely in the plane. Moreover, notice that we have put the normal vector on the plane, but there is no reason to anticipate this to be the case. It is possible that the normal vector does not touch the plane in any way.Â

Q2. How to Identify if the Given Equation Plane is Orthogonal, Parallel, or Neither?

Answer: Letâ€™s say, if the plane is given by âˆ’x+2z=10 and the line given by â†’r=âŸ¨5, 2âˆ’t, 10+4tâŸ© are orthogonal, parallel, or neither. To determine, we would pick off a vector that is normal to the plane. This is â†’n=âŸ¨âˆ’1, 0, 2âŸ©. We can also obtain a vector that is parallel to the line as well as an equation of plane parallel to two vectors. This is v=âŸ¨0, âˆ’1, 4âŸ©v=âŸ¨0, âˆ’1, 4âŸ©.

Now, if these two vectors are parallel then the line and the plane will be orthogonal. If â†’nâ†’ and â†’vâ†’ are parallel, then â†’vâ†’ is orthogonal to the plane, but â†’vâ†’ is also parallel to the line. Thus, if the 2 vectors are parallel, the plane and line will be orthogonal.