6.0 Intersection of a line and a plane

Let us assume the equation of line in cartesian form be $$\frac{{x - {x_1}}}{l} = \frac{{y - {y_1}}}{m} = \frac{{z - {z_1}}}{n} = r({\text{constant)}}$$ and the equation of plane in cartesian form be $$ax+by+cz+d=0$$

We can write the co-ordinates of any intersection point $P$ of line and plane from the equation of line i.e.,

$$x = rl + {x_1};\quad y = rm + {y_1};\quad z = rn + {z_1}$$ This point must satisfy the equation of plane. Therefore, $$\begin{equation} \begin{aligned} a(rl + {x_1}) + b\left( {rm + {y_1}} \right) + c\left( {rn + {z_1}} \right) + d = 0 \\ \left( {a{x_1} + b{y_1} + c{z_1} + d} \right) + r\left( {al + bm + cn} \right) = 0 \\ r = - \frac{{a{x_1} + b{y_1} + c{z_1} + d}}{{al + bm + cn}} \\\end{aligned} \end{equation} $$ Substituting the value of $r$, we get the co-ordinates of point of intersection of line and plane i.e., $P$.