Point of Intersection of Two lines in 3d

Here you will learn how to find point of intersection of two lines in 3d for both vector and cartesian form with example.

Let’s begin –

Point of Intersection of Two lines in 3d

(a) Cartesian Form

Algorithm :

Let the two lines be 

\(x – x_1\over a_1\) = \(y – y_1\over b_1\) = \(z – z_1\over c_1\)                            ……(i)

and \(x – x_2\over a_2\) = \(y – y_2\over b_2\) = \(z – z_2\over c_2\)                     ………(ii)

1). Write the coordinates of general point on (i) and (ii). The coordinates of general points on (i) and (ii) are given by

\(x – x_1\over a_1\) = \(y – y_1\over b_1\) = \(z – z_1\over c_1\) = \(\lambda\)  and  \(x – x_2\over a_2\) = \(y – y_2\over b_2\) = \(z – z_2\over c_2\) = \(\mu\) respectively,

i.e. (\(a_1\lambda + x_1\), \(b_1\lambda + y_1\), \(c_1\lambda + z_1\)) and (\(a_2\mu + x_2\), \(b_2\mu + y_2\), \(c_2\mu + z_2\))

2). If all the line (i) and (ii) intersect, then they have a common point.

\(\therefore\) \(a_1\lambda + x_1\) = \(a_2\mu + x_2\), \(b_1\lambda + y_1\) = \(b_2\mu + y_2\) and \(c_1\lambda + z_1\) = \(c_2\mu + z_2\)

3). Solve any two two of the equations in \(\lambda\) and \(\mu\) obtained in step 2. If the values of \(\lambda\) and \(\mu\) satisfy the third equation, then the lines (i) and (ii) intersect. Otherwise they do not intersect.

4). To obtain the coordinates of the point of intersection, substitute the value of \(\lambda\) (or \(\mu\)) in the coordinates of general point(s) obtained in step 1.

(b) Vector Form

Algorithm :

Let the two lines be 

\(\vec{r}\) = (\(a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\)) + \(\lambda\) (\(b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\))                        ……(i)

and \(\vec{r}\) = (\({a_1}’ \hat{i} + {a_2}’ \hat{j} + {a_3}’ \hat{k}\)) + \(\mu\) (\({b_1}’\hat{i} + {b_2}’\hat{j} + {b_3}’\hat{k}\))                   ………(ii)

1). If all the line (i) and (ii) intersect, then they have a common point.

\(\therefore\) (\(a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\)) + \(\lambda\) (\(b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\)) = (\({a_1}’ \hat{i} + {a_2}’ \hat{j} + {a_3}’ \hat{k}\)) + \(\mu\) (\({b_1}’\hat{i} + {b_2}’\hat{j} + {b_3}’\hat{k}\)) 

\(\implies\) \(a_1 + \lambda b_1\) = \({a_1}’ + \mu {b_1}’\), \(a_2 + \lambda b_2\) = \({a_2}’ + \mu {b_2}’\) and \(a_3 + \lambda b_3\) = \({a_3}’ + \mu {b_3}’\)

2). Solve any two two of the equations in \(\lambda\) and \(\mu\) obtained in step 2. If the values of \(\lambda\) and \(\mu\) satisfy the third equation, then the lines (i) and (ii) intersect. Otherwise they do not intersect.

3). To obtain the position vector of the point of intersection, substitute the value of \(\lambda\) (or \(\mu\)) in (i) and (ii).

Example : Show that the line \(x – 1\over 2\) = \(y – 2\over 3\) = \(z – 3\over 4\) and \(x – 4\over 5\) = \(y – 1\over 2\) = z intersect. Finf their point of intersection.

Solution : The coordinates of any point on first line are given by

\(x – 1\over 2\) = \(y – 2\over 3\) = \(z – 3\over 4\) = \(\lambda\)

or, x = \(2\lambda + 1\), y = \(3\lambda + 2\) and z = \(4\lambda + 3\)

The coordinates of any point on second line are given by

\(x – 4\over 5\) = \(y – 1\over 2\) = z = \(\mu\) 

or, x = \(5\mu + 4\), y = \(2\mu + 1\), z = \(\mu\)

If the lines intersect, then they have a common point. So, for some values of \(\lambda\) and \(\mu\), we must have,

\(2\lambda + 1\) = \(5\mu + 4\), \(3\lambda + 2\) = \(2\mu + 1\) and \(4\lambda + 3\) = \(\mu\)

Solving first two of these equations, we get: \(\lambda\) = -1 and \(\mu\) = -1.

Clearly, \(\lambda\) = -1 and \(\mu\) = -1 satisfy the third equation. So, the given lines intersect.

Putting \(\lambda\) in (\(2\lambda + 1\), \(3\lambda + 2\), \(4\lambda + 3\)), the coordinates of the required point of intersection are (-1, -1, -1).

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