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JEE Main 2026 Preparation: Question Papers, Solutions, Mock Tests & Strategy Unacademy » JEE Study Material » Mathematics » Intersection of a Straight Line and a Plane

Intersection of a Straight Line and a Plane

The point that satisfies both equations of a line and a point is the intersection of a straight line and a plane.

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The intersection of a straight line and plane focuses on the relation between line and plane equations in a three-dimensional coordinate system. The intersection of a straight line and a plane aids in understanding equation intersections in R2 to R3.

In a three dimensional space, an intersection can be:

  • An empty set 
  • A point 
  • A line 

A point is the meeting of a line and a plane.

P (xo, yo, zo) satisfies the equation of the line and the plane in R3. 

What is a Line and a Plane? 

In mathematics, a line is a straight one-dimensional figure. It is often defined as the shortest distance between two points, and it has no curves. A plane, on the other hand, is a two-dimensional figure. It is defined as a flat surface that extends indefinitely and has two linear independent vectors. 

Equation of a Straight Line and Intercept Form 

The equation of a straight line is:
ax+by+c= 0


Here: 

a, b and c are constants
x and y are variables

Equation of intercept form with x intercept as a and y intercept as b is: 

xa + yb =1 

 

What are the Possibilities when a Line and a Plane Intersect ?

When the line lies on a plane, there will be three possible intersections: 

  • If the line is in a plane, there are an infinite number of intersections between the line and the plane. 
  • If the line is parallel to the plane, the line and plane will not intersect. 
  • If the line intersects the plane once, the line and the plane intersect.

Lines that are Parallel and Plane 

A plane can be expressed as the set of points p (in a vector notation). If the line is parallel to the plane, the normal vector and plane are also at a right angle to the direction vector v of the line. 

Product of n.v = 0 

This confirms that the two vectors are perpendicular to each other and that there is no interaction.

Intersecting Lines and Planes 

The interaction of a line and a plane will have a common meeting point. 

The parametric equation of a line is 

 x=( x0 + at , y= y0+ bt, z= z0 +ct )

The scalar equation for a plane: 

ax + by + cz + d = 0

When a straight line and plane cross at one point:

a ( x0 +at ) + b ( y0 + bt ) + c ( z0 + ct) + d = 0

In the equation above, the parameter “t” is found.

Components to Find the Line and Plane Intersection

These are the steps to find the line and plane intersection:

  • First, write down the calculation of the line in the variable form: 

x=( x0 + at , y= y0+ bt, z= z0 +ct )

 

  • Write the calculation of the plane in its scalar form:

Ax + By + Cz + D = 0

  • Rewrite the scalar equation of the plane with x, y, and z to the variable calculation.

     

  • Substitute t in the variable calculation. This way x, y, and z components are established.

     

  • The single-variable equation thus helps us solve t.

Examples of Straight Line and Point Questions 

A line intersects with a plane in the following calculation, let us find the meeting point.

2x + y -2 z = 4

x= 1+ t

y= 1+ 2t

z= t 

We now rewrite the scalar equation of the plane, using the parametric forms:

2x +y – 2z = 4 

2 ( 1+t ) + ( 4+ 2t) – 2 (t ) = 4

Let us now solve to get the value of t:

2+ 2t + 4 + 2t- 2t = 4

2t + 6 = 4

2t =  -2

t= -1

Using t= -1 , and the parametric equation:

x= 1 + ( -1 )

= 0

y= 4 + 2 ( -1 )

=2. 

Hence x, y, z =  (0, 2, -1). 

The intersection of the straight line and plane is 0, 2, -1. 

Conclusion

By solving straight line and a plane questions, we have ample information to understand all about this section of analytic geometry.  Apart from mathematics, the intersection of a straight line and a plane finds its uses in the ray tracing method in computer graphics.

It is also used in vision-based 3D reconstruction, like video game programming, by making use of the algorithm of the intersection of a straight line with a polyhedron. 

faq

Frequently Asked Questions

Get answers to the most common queries related to the JEE Examination Preparation.

What is the intersection of a straight line and a plane?

The intersection of a straight line and plane focuses on the relation b...Read full

If C is the centroid of the triangle having vertices (3,−1), (1, 3) and (2, 4), let P be the point of intersection of the lines x + 3y − 1 = 0 and 3x − y + 1 = 0. The line passing through the points C and P also passes through which points?

Coordinates of C =  3+1+23...Read full

A right angle is drawn from point (1, 0, 3) with a line passing through (α, 7, 1) is (5/3, 7/3, 17/3). What will be the value of ‘a’?

Given points= P (1, 0, 3)  ...Read full

Consider the plane P = 2x+ y -4z= 4.

(a) What are the points of intersection of P with the line:              x=t  , y= 2+ 3t  , z= t...Read full

The intersection of a straight line and plane focuses on the relation between line and plane equations in a three-dimensional coordinate system.

Coordinates of C = 

3+1+23, -1+3+43

 = (2, 2)

Point of intersection of two lines: 

x + 3y – 1 = 0 and 

3x – y + 1 = 0 is A(-1/5, 2/5)

Slope, m = 8/11

Now, the equation of line CA is:

 y – 2 = (8/11)(x – 2)

8x – 11y + 6 = 0

Point (-9, -6) lies on CP

Hence, the points are ( -9, 6 )

Given points=

P (1, 0, 3) 

Q (5/3, 7/3, 17/3)

Line L direction ratios=

(α – 5/3, 7 – 7/3, 1 – 17/3) = ( (3α – 5)/3, 14/3, -14/3 )

PQ direction ratios = (-2/3, -7/3, -8/3)

 

According to the question, line L is at a right angle to PQ.

 

So, 

3a-53×−23+143×73+−143×83=0

=-6α +10 – 98 + 112 = 0

= 6α = 24

= α = 4

Hence, α = 4

 

(a) What are the points of intersection of P with the line:

             x=t  , y= 2+ 3t  , z= t

To find the points of intersection of P with the line, substitute the formula for x, y and z into the equation for P and solve to find t.
2t+( 2+ 3t) – 4t = 4

t= 2. 

Hence, using t=2 to find the point of intersection
x, y, z = 2, 8, 2.

(b) Find all the points of intersection of P with the line:

x= 1+t,  y= 4+ 2t, z=t. 

 

Substitution of the values results in the following equation: 

2( 1+t ) + (4+ 2t)  – 4 (t) =4

6= 4

The value of t does not satisfy the equation. 

Therefore there are no points of intersection.

 

Let the vertices of a triangle ABC be: 

A =(1, 0)

B =(6, 2)

C =(3/2, 6). 

The triangles APC, APB and BPC have the same areas with a point P in the centre. What is the length of the line PQ?

𝑄 = (-7/6, -1/3).

 

Point P = centroid

Coordinates =

 ( 1+6+323,1+5+23 )

 = (176, 83 )

Point Q coordinates are : (-76, -13)
Therefore, formula of distance:

PQ = √[42+32] = 5.

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