How do we find the intersection point of a line and a plane? Take a look at the graph below. When a line intersects with a plane, there will be either just one intersection point or infinite number of intersection points.When the line does not fall on the plane (refer to the red line), there is only one intersection point; when the line falls on the plane (refer to the blue line), all the points of the line intersect with the plane, i.e. there are infinite intersection points.
Find Intersection Point AlgebraicallyTo find the intersection point algebraically, let's look at the equations of the plane and line:
1)
| Plane
| :
| Ax + By + Cz = D (please take note the A, B, C and D here are NOT the same as the points in the graphs in this page.
|
2)
| Line
| :
| Vector Equation
| r(t) = r0 + tv
| | Parametric Equations | x = x0 + tv1
y = y0 + tv2
z = z0 + tv3
|
|
We can substitute the variables
x,
y and
z in the plane equation with the parametric equations as below to find the value of
t.
+B(y_0+tv_2)+C(z_0+tv_3)&space;&=D&space;\\&space;Ax_0+By_0+Cz_0+t(Av_1+Bv_2+Cv_3)&space;&=&space;D\\&space;t(Av_1+Bv_2+Cv_3)&space;&=&space;D-(Ax_0+By_0+Cz_0)&space;\\&space;t&space;&=&space;\frac{D-(Ax_0+By_0+Cz_0)}{Av_1+Bv_2+Cv_3}&space;\end{align*} "\begin{align*} Ax+By+Cz &= D\\ A(x_0+tv_1)+B(y_0+tv_2)+C(z_0+tv_3) &=D \\ Ax_0+By_0+Cz_0+t(Av_1+Bv_2+Cv_3) &= D\\ t(Av_1+Bv_2+Cv_3) &= D-(Ax_0+By_0+Cz_0) \\ t &= \frac{D-(Ax_0+By_0+Cz_0)}{Av_1+Bv_2+Cv_3} \end{align*}")
You don't have to memorize the value of
t at all! You just need to apply the concept above and carry on with your steps to solve the problem. After getting the value of
t, we use the value of
t to find the values of
x,
y and
z, in which (
x,
y,
z) is the coordinate of the intersection point.
The graph below shows a point and a plane. Move point A to move the line, C or D to move the plane. Point B is the intersection point with its coordinate given at the bottom of the page.
Betty, Created with GeoGebra and Thomas Calculus, Pearson.
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