Equation of a line in space

In this article, we will be discussing the class 12 mathematics topic equation of a line in space with solved examples. 

Equation of a straight line in space 

• The equation of a line can be determined if- 

1. the line has a given direction and passes through a given point. 

2. the line passes through two given points in space. 

Equation of a line through a given point and parallel to a given vector b

Vector Form 

For a given point A, with respect to the origin O in the rectangular coordinate system, let a be the position vector of the point. 

Let us suppose that there is a line l which passes through the given point A and is parallel to the given vector b.  And let us also assume that r is the random vector of any arbitrary point D on the line. 

Therefore, AD will be parallel to the given vector b such that AD = b, and is a real number. 

However, we know that AD = OD – OA 

Hence, b = r – a 

Therefore, for any value of , the following equation will help determine the position vector of the point D on the line. The vector equation of a line through a given point and parallel to give vector  b is 

r = a + b… (1) 

Cartesian Form 

We will now derive the cartesian form of the equation of a through a given point and parallel to a given vector with the help of the equation derived from the vector form. 

Let us assume that the coordinates of a given point A is (x1, y1, z1) and that a, b, c are the direction ratios of that line. Let us also assume that the coordinates of any given point P are (x, y, z). 

r = x i + y j + z k 

a = x1 i + y1 j + z1 k

b = a i + b j + c k

Substitute these values in equation, we get 

r = a + b

x i + y j + z k = x1 i + y1 j + z1 k + (a i + b j + c k)

x i + y j + z k = x1 i + y1 j + z1 k + a i + b j + c k 

x i + y j + z k = (x1 + a) i + (y1 + b) j + (z1 + c) k

After equating the coefficients of i, j, and k, we get 

x =  x1 + λ a

y = y1+ λ b

z = z1 + λ c

The aforementioned equations are called parametric equations of a line. After we remove the parameter from the above equations, we get 

x- x1a = y- y1b = z- z1c

Equation of a line passing through two given points 

Vector Form 

A (x1, y1, z1) and B (x2, y2, z3) are two points on a line and a and b are the position vector of those points respectively. 

Let r be the position vector of any arbitrary point P (x, y, z) on the line. 

P is on the line only if 

r – a = λ(b – a)

Therefore, r = a + λ (b– a), R… (1)

Cartesian Form 

r = x i + y j + z k 

a = x1 i + y1 j + z1 k

b = x2 i + y2 j + z2 k

By substituting these values in equation 1, we get 

r = a + λ (b– a)

x i + y j + z k = x1 i + y1 j + z1 k + (x2 i + y2 j + z2 k – x1 i – y1 j – z1 k)

x i + y j + z k = x1 i + y1 j + z1 k + {(x2 – x1) i + (y2 – y1) j + (z2 – z1) k }

After equating the coefficients of i, j, and k, we get

X = x1 + (x2 – x1)

Y = y1 + (y2 – y1) 

Z = z1 + (z2 – z1) 

After eliminating the , we obtain the equation of a line in 3d space in cartesian form 

x- x1 (x2 – x1) = y-y1 (y2 – y1)=z-z1 (z2 – z1)  

Lines and planes in space problems 

Problem 1- Find the equation of a line in space which is passing through the points A (-10, 0, 20) and B (30, 40, 60). 

Solution- Let  a and  b be the two position vectors of points A (-10, 0, 20) and B (30, 40, 60).

Then a = -10 i + 20 k

b = 30 i + 40 j + 60 k

b – a = 40 i + 40 j + 40 k

If r is the position vector, then the vector equation of the line is 

r = a + λ (b– a)

r = -10 i + 20 k + (40 i + 40 j + 40 k)

Problem 2- Find the vector and Cartesian equations of the line through the point (15, 20, -14) and which is parallel to the vector 30 i + 20 j – 60 k. 

Solution-a = 15 i + 20 j – 14 k

 b = 30 i + 20 j – 60 k

Vector equation of the line is

 r = 15 i + 20 j – 14 k + (30 i + 20 j – 60 k)

Cartesian equation of the line is 

x i + y j + z k = 15 i + 20 j – 14 k + (30 i + 20 j – 60 k)

x i + y j + z k = (15 + 30 ) i + (20 + 20 ) j – (14 + 60 ) k

After eliminating the , we get 

x-1530 = y-2020 = Z+14-60 

Conclusion 

So,in this article we learnt that there are two ways to represent a line in 3-D.First is the vector form which is of type r = a + b  where a  is the position vector of a point on the line and b  is the direction of line.Second ways is the cartesian form x- x1a = y- y1b = z- z1c   where a,b,c are direction ratios and x1,y1,z1 are coordinates of a point on line.