Home»Math Guides»Equations for Lines and Planes (Example 3)
Lines and Planes – Creating a Plane Equation from 3 Points (Example 3)
We’ve shown how to solve problems where you find a line that’s parallel or perpendicular to another line that goes through a particular point, but what about if you’re only given points?
Example question: Find the plane going through the points (0,1,1), (1,0,1) and (1,1,0).
Let’s call the points listed above A, B and C, respectively.
To get an equation for a plane we need a norm, and to get a norm we need to find a vector perpendicular to vectors on the plane.
The vectors on the plane are two vectors made from these three points, and their cross product will be a vector perpendicular to both of them.
In other words, make vectors from these points and find their cross product.
Let’s make AB and CB vectors. To make vectors from points, subtract the points from each other.
But be careful with notation, AB is the vector made from subtracting A points from B points. BA is the vector made from subtracting B points from A points.
\[\mathop {AB}\limits^ \to = \left\langle {1 – 0,0 – 1,1 – 1} \right\rangle \]
\[\mathop {AB}\limits^ \to = \left\langle {1, – 1,0} \right\rangle \]
\[\mathop {CB}\limits^ \to = \left\langle {1 – 1,0 – 1,1 – 0} \right\rangle \]
\[\mathop {CB}\limits^ \to = \left\langle {0, – 1,1} \right\rangle \]
And find their cross product.
Let’s draw the two vectors as a matrix, I will find \(\mathop {CB}\limits^ \to \times \mathop {AB}\limits^ \to \)
\[\begin{array}{l}\mathop {CB}\limits^ \to \times \mathop {AB}\limits^ \to \\\begin{array}{*{20}{c}}i&j&k\\0&{ – 1}&1\\1&{ – 1}&0\end{array}\end{array}\]
\[\mathop {CB}\limits^ \to \times \mathop {AB}\limits^ \to = \left\langle {( – 1)(0) – (1)( – 1), – [(0)(0) – (1)(1)],(0)( – 1) – ( – 1)(1)} \right\rangle \]
\[\mathop {CB}\limits^ \to \times \mathop {AB}\limits^ \to = \left\langle {1,1,1} \right\rangle \]
So the norm of our plane is \(\left\langle {1,1,1} \right\rangle \).
The plane will have the equation:
\[1x – {x_o} + 1y – {y_o} + 1z – {z_o} = 0\]
We got this by multiplying x, y and z by the norm’s values, which were 1, 1 and 1, respectively.
We need to substitute in x\(_O\), y\(_O\), and z\(_O\) values, which we can use from any of the points given.
Let’s substitute the point (0,1,1).
\[1x – 0 + 1y – 1 + 1z – 1 = 0\]
\[x + y + z = 2\]
This is the equation of a plane that goes through the points.
There are other ways to find a cross product, so this plane isn’t necessarily the only plane that fulfills the conditions.
Try another topic, such as finding partial derivatives of multi-variable functions!