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2D planes in 3-space

vector equation of a plane

A 2D plane can be described by a single point on its surface together with two non-collinear vectors parallel to its surface. Let ${\bf a}$ be the position vector of $A$, a point on the plane, and ${\bf b}$ and ${\bf c}$ be two such vectors, then the vector equation of the plane is given by $$ {\bf r} = {\bf a} + \lambda {\bf b} + \mu {\bf c} $$

Cartesian equation of a plane

For a 2D plane in 3-space, we can determine its Cartesian equation using the formula $$ {\bf r}\cdot{\bf n} = {\bf a}\cdot{\bf n} $$ where ${\bf n}$ is a vector normal to the plane and ${\bf a}$ is the position vector of a point on the plane.

vector equation of a plane 1

Determine there vector equation of a plane passing through points $A(1,4,2)$, $B(3,8,-1)$, $C(2,5,9)$.
solution - press button to display

Let us first find the vectors $OA,\;AB,\;AC$

$$ OA = \left(\begin{array}{c}1 \\ 4 \\ 2 \end{array}\right),\; AB = \left(\begin{array}{c}2 \\ 4 \\ -3 \end{array}\right),\; AC=\left(\begin{array}{c}1 \\ 1 \\ 7 \end{array}\right) $$

The it follows that the plane $\pi$ has equation $$ \pi : {\bf r} = \left(\begin{array}{c}1 \\ 4 \\ 2 \end{array}\right) +\lambda\left(\begin{array}{c}2 \\ 4 \\ -3 \end{array}\right) +\mu \left(\begin{array}{c}1 \\ 1 \\ 7 \end{array}\right) $$

Note that this equation is in no way uniquely defined, we can, for example, change the vector OA for the position vector of any other point in the plane, or indeed change the non collinear vectors for any others parallel to the plane!

Cartesian equation of a plane 1

Find both the vector equation and Cartesian equation of the plane passing through the points $A(1,2,3),\; B(2,5,7),\;C(3,-2,2)$

solution - press button to display

Let us first find the vectors $OA$, $AB$ and $AC$ $$ OA = \left(\begin{array}{c}1 \\ 2 \\ 3 \end{array}\right),\;AB = \left(\begin{array}{c}1 \\ 3 \\ 4 \end{array}\right),\; AC = \left(\begin{array}{c}2 \\ -4 \\ -1 \end{array}\right) $$

The vector equation is $$ {\bf r} = \left(\begin{array}{c}1 \\ 2 \\ 3 \end{array}\right) + \lambda \left(\begin{array}{c}1 \\ 3 \\ 4 \end{array}\right) + \mu \left(\begin{array}{c}2 \\ -4 \\ -1 \end{array}\right) $$

The Cartesian equation is given by $$ {\bf r}\cdot {\bf n} = {\bf a}\cdot{\bf n} $$ where ${\bf a}$ is the position vector of a point on the plane and ${\bf n}$ is a vector normal to the plane.

The easiest way to calculate a normal vector is to do the cross product of the two internal vectors.

$$ {\bf n} = \left|\begin{array}{ccc}i & j & k \\ 1 & 3 & 4 \\ 2 & -4 & -1 \end{array} \right| = 13{\bf i} + 9{\bf j} - 10{\bf k} $$

 

Applying ${\bf r}\cdot {\bf n} = {\bf a}\cdot{\bf n} $ yields $$ 13x + 9y -10z = 13 + 18 - 30 = 1 $$

Hence the plane has Cartesian equation $13x + 9y - 10z = 1$

intersection of two planes (Cartesian method)

Determine the vector equation of the line of intersection of the planes $$ \begin{align} \pi_1\; &: \; x+y+z = 4 \\ \pi_2 \; &: \; x - 2y +3z = 2 \end{align} $$

solution - press button to display

To construct a vector equation of a line, we require the position vector of a single point on the line, ${\bf a}$ together with a vector parallel to the line, ${\bf v}$.

In the case of the two planes $\pi_1$ and $\pi_2$, we know that the line is contained within each plane and so the line must be perpendicular to the normal vector of each plane. It follows that the direction vector ${\bf v}$ of the line is given by $$ {\bf v} = {\bf n_1} \times {\bf n_2} $$ where ${\bf n_i}$ is the normal vector to plane $\pi_i$

the vector ${\bf v}$ is given by $$ {\bf v} = \left|\begin{array}{ccc}i & j & k \\ 1 & 1 & 1 \\ 1 & -2 & 3 \end{array}\right| = 5{\bf i} -2{\bf j} -3{\bf k} $$

There are a number of ways to determine a single point on the plane, in this case, we will eliminate a variable, and then determine a solution of the reduced system of equations $$\pi_1 - \pi_2 \; : -3y - 2z = 2$$ Let $z = -1$, then $y = 0\; \Rightarrow x = 5$, so our vector ${\bf a}$ is $$ {\bf a} = 5{\bf i} + 0{\bf j} - 1{\bf k} $$

The equation of the line is therefore $$ {\bf r} = \left(\begin{array}{c}5 \\ 0 \\ -1\end{array}\right) + \lambda\left(\begin{array}{c}5 \\ -2 \\ -3\end{array}\right) $$

intersection of two planes (vector method)

Determine the vector equation of the line of intersection of the two planes, $\pi_1$ and $\pi_2$, defined below

$$ \begin{align} \pi_1 &: {\bf r} = \left(\begin{array}{c}1 \\ 0 \\ 1\end{array}\right) + \lambda\left(\begin{array}{c}1 \\ 2 \\ 3\end{array}\right) + \mu\left(\begin{array}{c}-1 \\ -1 \\ 2\end{array}\right)\\ \pi_2 &: {\bf r} = \left(\begin{array}{c}1 \\ 1 \\ 2\end{array}\right) + s\left(\begin{array}{c}2 \\ 1 \\ -1\end{array}\right) +t\left(\begin{array}{c}1 \\ -3 \\ 1\end{array}\right)\\ \end{align} $$

solution - press button to display

Setting the two equations equal to each other yields a system of equations in four unknowns. 

$$ \left(\begin{array}{c}1 \\ 0 \\ 1\end{array}\right) + \lambda\left(\begin{array}{c}1 \\ 2 \\ 3\end{array}\right) + \mu\left(\begin{array}{c}-1 \\ -1 \\ 2\end{array}\right) = \left(\begin{array}{c}1 \\ 1 \\ 2\end{array}\right) + s\left(\begin{array}{c}2 \\ 1 \\ -1\end{array}\right) +t\left(\begin{array}{c}1 \\ -3 \\ 1\end{array}\right) $$

We can rearrange this to yield a matrix equation, which can then be reduced to solve.

Let us rewrite our equation as $M{\bf x} = {\bf v}$

$$ \left(\begin{array}{cccc}1 & 1 & -2 & -1\\ 2 & -1 & -1 & 3 \\ 3 & 2 & 1 & -1\end{array}\right)\left(\begin{array}{c}\lambda \\ \mu \\ s \\ t\end{array}\right) = \left(\begin{array}{c}0 \\ 1\\ 1\end{array}\right) $$

Expressing this as an augmented matrix, we can then perform row operations, to reduce the matrix

$$ \left(\begin{array}{cccc|c} 1 & 1 & -2 & -1 & 0 \\ 2 & -1 & -1 & 3 & 1 \\ 3 & 2 & 1 & -1 & 1 \end{array}\right) $$ Performing $r'_2 = r_2 - 2r_1$ and $r'_3 = r_3 - 3r_1$ gives $$ \left(\begin{array}{cccc|c} 1 & 1 & -2 & -1 & 0 \\ 0 & -3 & 3 & 5 & 1 \\ 0 & -1 & 7 & 2 & 1 \end{array}\right) $$

Finally, performing $r''_3 = r'_3 + r'_2$ gives $$ \left(\begin{array}{cccc|c} 1 & 1 & -2 & -1 & 0 \\ 0 & -3 & 3 & 5 & 1 \\ 0 & 0 & -18 & -1 & -2 \end{array}\right) $$

solving the equations to determine $s$ in terms of $t$ is sufficient to determine the equation of the line of intersection.

The variable $t$ is a free variable, so let us say $t=\alpha$ and now try to write $s$ in terms of $\alpha$

From row three of our final augmented matrix, we get $$-18s - \alpha = -2 \Rightarrow s = \frac{1}{18}(2-\alpha)$$

The equation of the line of intersection is therefore

$$ {\bf r} =\left(\begin{array}{c}1 \\ 1 \\ 2\end{array}\right) + \frac{2-\alpha}{18}\left(\begin{array}{c}2 \\ 1 \\ -1 \end{array}\right) + \alpha\left(\begin{array}{c}1 \\ -3 \\ 1\end{array}\right) $$

 

This can now be rewritten as

$${\bf r} = \frac{1}{9}\left(\begin{array}{c}11 \\ 10 \\ 17 \end{array}\right) + \alpha \left(\begin{array}{c}16 \\ -55 \\ 19\end{array}\right)$$

Intersection of three planes

Determine the unique point of intersection of the three planes, $\pi_1,\pi_2,\pi_3$ defined by the equations below. $$ \begin{align} \pi_1 &: x + y + z = -2 \\ \pi_2 &: x+2y +3z = 6\\ \pi_3 &: x +y + 2z = 4 \end{align} $$
solution - press button to display

This is perhaps the most straightforward type of plane-intersection questions. We can rewrite these equations as a matrix equation:

$$ \left(\begin{array}{ccc} 1 & 1 & 1 \\ 1 & 2 & 3 \\ 1 & 1 & 2\end{array}\right)\left(\begin{array}{c}x \\ y \\ z\end{array}\right) = \left(\begin{array}{c}-2 \\ 6 \\ 4\end{array}\right) $$

Using row operations, this reduces to the following

$$ \left(\begin{array}{ccc} 1 & 1 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 1\end{array}\right)\left(\begin{array}{c}x \\ y \\ z\end{array}\right) = \left(\begin{array}{c}-2 \\ 8 \\ 6\end{array}\right) $$

We can now either continue with row operations to reduce the matrix to the identity or we can use back substitution to determine the values of x,y,z from this reduced system

We can see from the bottom row that $z = 6$, We also have $y + 2z = 8 \Rightarrow y = -4$. Finally, we have $x + y + z = -2$ which gives $x = -4$

The point, $P$, of intersection of the three planes is therefore $$ P=(-4,-4,6) $$

Alternatively for this question, we could simply find the inverse of the $3\times 3$ matrix and determine the solution by matrix multiplication.