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# 4.8: Planes in Rⁿ

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## Outcomes

1. Find the vector and scalar equations of a plane.

Much like the above discussion with lines, vectors can be used to determine planes in $$\mathbb{R}^n$$. Given a vector $$\vec{n}$$ in $$\mathbb{R}^n$$ and a point $$P_0$$, it is possible to find a unique plane which contains $$P_0$$ and is perpendicular to the given vector.

## Definition $$\PageIndex{1}$$: Normal Vector

Let $$\vec{n}$$ be a nonzero vector in $$\mathbb{R}^n$$. Then $$\vec{n}$$ is called a normal vector to a plane if and only if $\vec{n} \bullet \vec{v} = 0\nonumber$ for every vector $$\vec{v}$$ in the plane.

In other words, we say that $$\vec{n}$$ is orthogonal (perpendicular) to every vector in the plane.

Consider now a plane with normal vector given by $$\vec{n}$$, and containing a point $$P_0$$. Notice that this plane is unique. If $$P$$ is an arbitrary point on this plane, then by definition the normal vector is orthogonal to the vector between $$P_0$$ and $$P$$. Letting $$\overrightarrow{0P}$$ and $$\overrightarrow{0P_0}$$ be the position vectors of points $$P$$ and $$P_0$$ respectively, it follows that $\vec{n} \bullet (\overrightarrow{0P} - \overrightarrow{0P_0}) = 0\nonumber$ or $\vec{n} \bullet \overrightarrow{P_0P} = 0\nonumber$

The first of these equations gives the vector equation of the plane.

## Definition $$\PageIndex{2}$$: Vector Equation of a Plane

Let $$\vec{n}$$ be the normal vector for a plane which contains a point $$P_0$$. If $$P$$ is an arbitrary point on this plane, then the vector equation of the plane is given by $\vec{n} \bullet (\overrightarrow{0P} - \overrightarrow{0P_0}) = 0\nonumber$

Notice that this equation can be used to determine if a point $$P$$ is contained in a certain plane.

## Example $$\PageIndex{1}$$: A Point in a Plane

Let $$\vec{n} = \left[ \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right]$$ be the normal vector for a plane which contains the point $$P_0 = \left( 2, 1, 4 \right)$$. Determine if the point $$P = \left( 5, 4, 1 \right)$$ is contained in this plane.

Solution

By Definition $$\PageIndex{2}$$, $$P$$ is a point in the plane if it satisfies the equation $\vec{n} \bullet (\overrightarrow{0P} - \overrightarrow{0P_0}) = 0\nonumber$

Given the above $$\vec{n}$$, $$P_0$$, and $$P$$, this equation becomes \begin{aligned} \left[ \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right] \bullet \left( \left[ \begin{array}{r} 5 \\ 4 \\ 1 \end{array} \right] - \left[ \begin{array}{r} 2 \\ 1 \\ 4 \end{array} \right] \right) &= \left[ \begin{array}{r} 1 \\ 2 \\ 3 \end{array} \right] \bullet \left( \left[ \begin{array}{r} 3 \\ 3 \\ -3 \end{array} \right] \right) \\ &= 3 + 6 - 9 = 0\end{aligned}

Therefore $$P = ( 5, 4, 1)$$ is contained in the plane.

Suppose $$\vec{n} = \left[ \begin{array}{c} a \\ b \\ c \end{array} \right]$$, $$P = \left( x,y,z\right)$$ and $$P_0 = (x_0, y_0, z_0 )$$.

Then \begin{aligned} \vec{n} \bullet (\overrightarrow{0P} - \overrightarrow{0P_0}) &= 0 \\ \left[ \begin{array}{c} a \\ b \\ c \end{array} \right] \bullet \left( \left[ \begin{array}{c} x \\ y \\ z \end{array} \right] - \left[ \begin{array}{c} x_0 \\ y_0 \\ z_0 \end{array} \right] \right) &= 0 \\ \left[ \begin{array}{c} a \\ b \\ c \end{array} \right] \bullet \left[ \begin{array}{c} x - x_0 \\ y - y_0 \\ z - z_0 \end{array} \right] &= 0 \\ a(x - x_0) + b (y - y_0) + c (z-z_0) &= 0 \end{aligned}

We can also write this equation as $ax + by + cz = ax_0 + by_0 + cz_0\nonumber$

Notice that since $$P_0$$ is given, $$ax_0+by_0+cz_0$$ is a known scalar, which we can call $$d$$. This equation becomes $ax + by + cz = d\nonumber$

## Definition $$\PageIndex{3}$$: Scalar Equation of a Plane

Let $$\vec{n} = \left[ \begin{array}{c} a \\ b \\ c \end{array} \right]$$ be the normal vector for a plane which contains the point $$P_0 = (x_0, y_0, z_0)$$.Then if $$P=(x,y,z)$$ is an arbitrary point on the plane, the scalar equation of the plane is given by $ax + by + cz = d\nonumber$ where $$a,b,c,d \in \mathbb{R}$$ and $$d = ax_0 + by_0 + cz_0$$.

Consider the following equation.

## Example $$\PageIndex{2}$$: Finding the Equation of a Plane

Find an equation of the plane containing $$P_0 = (3, -2, 5)$$ and orthogonal to $$\vec{n} = \left[ \begin{array}{r} -2 \\ 4 \\ 1 \end{array} \right]$$.

Solution

The above vector $$\vec{n}$$ is the normal vector for this plane. Using Definition $$\PageIndex{2}$$, we can determine the vector equation for this plane. \begin{aligned} \vec{n} \bullet (\overrightarrow{0P} - \overrightarrow{0P_0}) &= 0 \\ \left[ \begin{array}{r} -2 \\ 4 \\ 1 \end{array} \right] \bullet \left(\left[ \begin{array}{c} x \\ y \\ z \end{array} \right] - \left[ \begin{array}{r} 3 \\ -2 \\ 5 \end{array} \right] \right) &= 0 \\ \left[ \begin{array}{r} -2 \\ 4 \\ 1 \end{array} \right] \bullet \left[ \begin{array}{c} x - 3 \\ y + 2 \\ z - 5 \end{array} \right] &= 0 \end{aligned}

Using Definition $$\PageIndex{3}$$, we can determine the scalar equation of the plane. $-2x + 4y + 1z = -2(3) + 4(-2) + 1(5) = -9\nonumber$

Hence, the vector equation of the plane is $\left[ \begin{array}{r} -2 \\ 4 \\ 1 \end{array} \right] \bullet \left[ \begin{array}{c} x - 3 \\ y + 2 \\ z - 5 \end{array} \right] = 0\nonumber$ and the scalar equation is $-2x + 4y + 1z = -9\nonumber$

Suppose a point $$P$$ is not contained in a given plane. We are then interested in the shortest distance from that point $$P$$ to the given plane. Consider the following example.

## Example $$\PageIndex{3}$$: Shortest Distance From a Point to a Plane

Find the shortest distance from the point $$P = (3,2,3)$$ to the plane given by
$$2x + y + 2z = 2$$, and find the point $$Q$$ on the plane that is closest to $$P$$.

Solution

Pick an arbitrary point $$P_0$$ on the plane. Then, it follows that $\overrightarrow{QP} = proj_{\vec{n}}\overrightarrow{P_0P}\nonumber$ and $$\| \overrightarrow{QP} \|$$ is the shortest distance from $$P$$ to the plane. Further, the vector $$\overrightarrow{0Q} = \overrightarrow{0P} - \overrightarrow{QP}$$ gives the necessary point $$Q$$.

From the above scalar equation, we have that $$\vec{n} = \left[ \begin{array}{c} 2 \\ 1 \\ 2 \end{array} \right]$$. Now, choose $$P_0 = (1, 0, 0)$$ so that $$\vec{n} \bullet \overrightarrow{0P} = 2 = d$$. Then, $$\overrightarrow{P_0P} = \left[ \begin{array}{c} 3 \\ 2 \\ 3 \end{array} \right] - \left[ \begin{array}{c} 1 \\ 0 \\ 0 \end{array} \right] = \left[ \begin{array}{c} 2 \\ 2 \\ 3 \end{array} \right]$$.

Next, compute $$\overrightarrow{QP} = proj_{\vec{n}}\overrightarrow{P_0P}$$. \begin{aligned} \overrightarrow{QP} &= proj_{\vec{n}}\overrightarrow{P_0P} \\ &= \left( \frac{ \overrightarrow{P_0P} \bullet \vec{n}}{\| \vec{n} \| ^2}\right)\vec{n} \\ &= \frac{12}{9} \left[ \begin{array}{r} 2 \\ 1 \\ 2 \end{array} \right] \\ &= \frac{4}{3} \left[ \begin{array}{r} 2 \\ 1 \\ 2 \end{array} \right] \end{aligned}

Then, $$\| \overrightarrow{QP} \| = 4$$ so the shortest distance from $$P$$ to the plane is $$4$$.

Next, to find the point $$Q$$ on the plane which is closest to $$P$$ we have \begin{aligned} \overrightarrow{0Q} &= \overrightarrow{0P} - \overrightarrow{QP} \\ &= \left[ \begin{array}{r} 3 \\ 2 \\ 3 \end{array} \right] - \frac{4}{3} \left[ \begin{array}{r} 2 \\ 1 \\ 2 \end{array} \right] \\ &= \frac{1}{3} \left[ \begin{array}{r} 1 \\ 2 \\ 1 \end{array} \right]\end{aligned}

Therefore, $$Q = (\frac{1}{3}, \frac{2}{3}, \frac{1}{3} )$$.

This page titled 4.8: Planes in Rⁿ is shared under a CC BY license and was authored, remixed, and/or curated by Ken Kuttler (Lyryx) .