Here you will learn how to find intersection of a line and a plane with examples.
Let’s begin –
Intersection of a Line and a Plane
Let the equation of a line be \(x – x_1\over l\) = \(y – y_1\over m\) = \(z – z_1\over n\) and that of a plane be ax + by + cz + d = 0.
The coordinates of any point on the line \(x – x_1\over l\) = \(y – y_1\over m\) = \(z – z_1\over n\) is given
\(x – x_1\over l\) = \(y – y_1\over m\) = \(z – z_1\over n\) = r (say)
or, \((x_1 + lr, y_1 + mr, z + nr)\) …………(i)
If it lies on the plane ax + by + cz + d = 0, then
\(a(x_1 + lr)\) + \(b(y_1 + mr)\) + \(c(z_1 + nr)\) + d = 0
\(\implies\) \((ax_1 + by_1 + cz_1 + d)\) + r(al + bm + cn) = 0
\(\implies\) r = -\((ax_1 + by_1 + cz_1 + d)\over al + bm + cn\)
Substituting the value of r in (i), we obtain the coordinates of the required point of intersection.
In order to find the coordinates of the point of intersection of a line and a plane, we may use the following algorithm,
Algorithm :
1). Write the coordinates of any point on the line in terms of some parameters r (say).
2). Substitute these coordinates in the equation of the plane to obtain the value of r.
3). Put the value of r in the coordinates of the point in step 1.
Example : Find the coordinates of the point where the line through the points A(3, 4, 1) and B(5, 1, 6) crosses the XY-plane.
Solution : The equation of the line passing through A and B is
\(x – 3\over 5 – 3\) = \(y – 4\over 1 – 4\) = \(z – 1\over 6 – 1\)
or, \(x – 3\over 2\) = \(y – 4\over -3\) = \(z – 1\over 5\)
The coordinates of any point on this line are given by
\(x – 3\over 2\) = \(y – 4\over -3\) = \(z – 1\over 5\) = \(\lambda\)
\(\implies\) x = \(2\lambda + 3\), y = \(-3\lambda + 4\), z = \(5\lambda + 1\)
So, \((2\lambda + 3, -3\lambda + 4, 5\lambda + 1)\) are coordinates of any point on the line passing through A and B. If it lies on XY-plane i.e z = 0.Then.
\(5\lambda + 1\) = 0 \(\implies\) \(\lambda\) = -\(1\over 5\)
Thus, the coordinates of the required point are (13/5, 23/5, 0).
