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If two points are taken on a surface, then a surface is known as a plane when a straight line joining these two points lies completely inside the surface. A Plane can be determined if one of the three things are present. First, the plane’s equation in a normal formal, second, it passes through a point and also is perpendicular to a particular direction, third it runs through three non Collinear points when given.
Also Read: NCERT Solutions For Class 12 Mathematics Chapter 10 Vector Algebra
Key Terms: Three-dimensional Plane, Vectors, Magnitude, Dimensions, One-dimensional Geometry, Angle Equation, Distance
Plane Definition
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A plane is two-dimensional in nature and a flat surface that prolongs infinitely far in two dimensions. Simply, in geometry, a plane is defined as a surface that comprises all the straight lines which join any two points lying on the surface.
Plane
Read More: Coplanar Vector
General Equation of a Plane
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The first degree’s general equation in x,y,z represents a plane. Therefore, a general equation of a plane is represented as ax + by + cz + d = 0. In the cartesian equation of a plane, the coefficients of x,y, and z are the direction ratios of the normal to plane.
When the plane passes through a fixed point which has coordinates (x1, y1, z1) , its equation is given as
a(x – x1) + b (y — y1) + c (z — z1) = 0
Read More: Coordinate Geometry
Normal Form of the Equation of Plane
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A plane which is at the distance p from the origin and has direction cosines of the normal that are from the origin to the plane as l,m,n, is represented by the equation
lx + my + nz = p
The perpendicular N from the origin to the plane has coordinates of its foot as
(1p, mp, np)
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Intercept Form of Equation of a Plane
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The intercept form of the equation of a plane, where a,b, and c are intercepts that are on the respective X, Y, and Z-axes, is represented as
x / a + y / b + z / c = 1
To find out the x-intercept, we put y = 0 and z = 0 in the plane’s equation. Therefore, the value of x-intercept can be found out. In a similar way, the intercept for all other axes can be found.
In three-dimensional space, any two planes can be related to each other in three ways, which are-
- The two planes can intersect each other.
- The two planes can be similar/identical.
- The two planes can be parallel to each other.
Read More: Addition of Vectors
Properties of a Plane
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- If two separate planes are perpendicular to the same line, then by default, they should be parallel to each other.
- A line on a plane could either be parallel to the plane, could be intersecting the plane at a singular point, or is itself existing inside the plane.
- If we have two lines that are different from each other and are perpendicular to the same plane, then they must be parallel to each other too.
- In a three dimensional space, if there are two planes that are different, then they are either intersecting in a line or are parallel to each other.
Read More: Optimization
Sample Questions
Ques. What is a plane? (2 marks)
Ans. In geometry, a plane is defined as a surface that comprises all the straight lines which join any two points lying on the surface. It is two dimensional in nature and is known as a flat surface that prolongs infinitely far in two dimensions.
Ques. How can the value of a y-intercept be found? (3 marks)
Ans. The intercept form of the equation of a plane is depicted as
x / a + y / b + z / c = 1
Here, a,b, and c are the intercepts on the X, Y, and Z axes respectively. To find the y-intercept, the values of x and z are made 0. Hence, the value of the y-intercept can be found.
Ques. What is the equation of a plane passing through a fixed point having coordinates (x1, y1, z1)? (2 marks)
Ans. When the plane passes through a fixed point which has coordinates (x1, y1, z1) , its equation is given as
a(x – x1) + b (y — y1) + c (z — z1) = 0
Ques. Find the vector and Cartesian equation of the plane passing through the points (2, 2, -1), (3, 4, 2) and (7, 0, 6). Also find the vector equation of a plane passing through (4, 3, 1) and parallel to the plane obtained above. (3 marks)
Ans. Let’s take equation for the plane passing through A (2, 2, -1) as
a(x-2) + b(y-2) + c(z+1) = 0 …(i)
We know that B (3, 4, 2) falls on the plane …(i)
a(3-2) + b(4-2) + c(2+1) = 0
a + 2b + 3c =0 ... (ii)
When C (7, 0, 6) falls on the plane (i)
a(7 - 2) + b(0 - 2) + c(6+1) = 0
5a - 2b + 7c = 0 …(iii)
Therefore we can obtain from (ii) and (iii)
Ques. Find the vector equation of the plane that contains the lines 
and the point (-1, 3, – 4). Also, find the length of the perpendicular drawn from the point (2, 1, 4) to the plane, thus obtained. (3 marks)
Ans. We have 
as the equation of the line provided.
The point that the plane pass through = (-1, 3, -4)
Therefore the equation of the plane: a(x+1) + b(y-3) + c(z+4) = 0 …(i)
Since (1,1) falls on the plane,
Therefore 2a - 2b + 4c = 0 …(ii)
Also (1, 2, -1) falls on the same plane
Therefore 2a - b + 3c = 0 …(iii)
When we solve equation (ii) and (iii) we obtain:
Ques. Find the vector and cartesian equations of the plane passing through the points having position vectors 
Write the equation of a plane passing through a point (2, 3, 7) and parallel to the plane obtained above. Hence, find the distance between the two parallel planes. (3 marks)
Ans. Let's consider A, B, C as the points with position,
Ques. Find the distance of the point (-1, -5, -10) from the point of intersection of the line 
(3 marks)
Ans.
Ques. Find the equation of the plane through the line of intersection of 
and perpendicular to the plane 
. Hence find whether the plane thus obtained contains the line x – 1 = 2y – 4 = 3z – 12. (3 marks)
Ans.
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