Orthogonal projection vectors: line and plane formulas with applications

What is an orthogonal projection?

An orthogonal projection is the component of a vector along a specified line or plane, obtained by dropping a perpendicular. Thinking of a shadow can help, provided the light rays are perpendicular to the receiving line or plane.

For a vector $\mathbf a$ and a nonzero direction vector $\mathbf l$, the projection onto the line spanned by $\mathbf l$ is

$$\boxed{\operatorname{proj}_{\mathbf l}\mathbf a=\frac{\mathbf a\cdot\mathbf l}{|\mathbf l|^2}\mathbf l.}$$

The same formula works in two or three dimensions. It is the dimension of the receiving subspace, rather than the dimension of the drawing, that determines which projection formula to use.

Deriving the line-projection formula

Write the projected vector as $\mathbf b=t\mathbf l$. Orthogonality requires the residual $\mathbf a-\mathbf b$ to be perpendicular to $\mathbf l$, so

$$(\mathbf a-t\mathbf l)\cdot\mathbf l=0.$$

Since $\mathbf l\ne\mathbf0$,

$$t=\frac{\mathbf a\cdot\mathbf l}{\mathbf l\cdot\mathbf l}.$$

Substitution gives the formula above. This derivation also proves that it is the unique projection with the stated properties.

The geometric interpretation

If $\theta$ is the angle between nonzero $\mathbf a$ and $\mathbf l$, its signed component in the direction of $\mathbf l$ is

$$|\mathbf a|\cos\theta=\frac{\mathbf a\cdot\mathbf l}{|\mathbf l|}.$$

Multiply that scalar by the unit direction $\mathbf l/|\mathbf l|$ to recover the projection vector. If the dot product is negative, the projection points opposite to $\mathbf l$; its length is still nonnegative.

When $\mathbf e$ is a unit vector, the formula simplifies to $(\mathbf a\cdot\mathbf e)\mathbf e$. The denominator in the general formula is $|\mathbf l|^2$, not a vector square.

Projection onto a plane through the origin

Let $\mathbf n\ne\mathbf0$ be normal to a plane through the origin. Remove the component parallel to the normal:

$$\boxed{\operatorname{proj}_{\mathbf n^\perp}\mathbf a=\mathbf a-\frac{\mathbf a\cdot\mathbf n}{|\mathbf n|^2}\mathbf n.}$$

The resulting vector is in the plane because its dot product with $\mathbf n$ is zero. The removed component is perpendicular to that plane, which establishes the projection property.

For a plane not passing through the origin, translate by a known point on the plane before applying this formula. A point and a free vector are different objects; the translation must not be omitted.

Example 1: a perpendicular foot on a line

Problem. In a triangle $OAB$, suppose $OA=4$ and $\overrightarrow{OA}\cdot\overrightarrow{OB}=3$. Let $P$ be the foot of the perpendicular from $B$ to the line $OA$. Express $\overrightarrow{OP}$ as $k\overrightarrow{OA}$.

The vector $\overrightarrow{OP}$ is the projection of $\overrightarrow{OB}$ onto the direction $\overrightarrow{OA}$. Thus

$$\overrightarrow{OP}=\frac{\overrightarrow{OB}\cdot\overrightarrow{OA}}{|\overrightarrow{OA}|^2}\overrightarrow{OA}=\boxed{\frac3{16}\overrightarrow{OA}.}$$

The length of $OB$ is not needed. The given dot product already determines its component along $OA$.

Example 2: reflection across a plane

Problem. Let $P$ lie on a plane $\pi$ in three-dimensional space, and let $\mathbf n\ne\mathbf0$ be normal to $\pi$. A point $Q$ has reflection $Q'$ across $\pi$. With $\mathbf p=\overrightarrow{OP}$ and $\mathbf q=\overrightarrow{OQ}$, express $\overrightarrow{OQ'}$ in terms of $\mathbf p,\mathbf q,\mathbf n$.

This is the plane-reflection problem discussed in the original article, attributed there to Kwansei Gakuin University's School of Business Administration.

Let $M$ be the midpoint of $QQ'$, which is also the perpendicular foot of $Q$ on $\pi$. The displacement from $Q$ to the plane is the component of $\overrightarrow{QP}=\mathbf p-\mathbf q$ along the normal:

$$\overrightarrow{QM}=\frac{(\mathbf p-\mathbf q)\cdot\mathbf n}{|\mathbf n|^2}\mathbf n.$$

Since $Q'=Q+2\overrightarrow{QM}$,

$$\boxed{\overrightarrow{OQ'}=\mathbf q+2\frac{(\mathbf p-\mathbf q)\cdot\mathbf n}{|\mathbf n|^2}\mathbf n.}$$

Check both defining properties

The displacement $Q'-Q$ is parallel to the normal. Also, the midpoint has position vector

$$\mathbf m=\mathbf q+\frac{(\mathbf p-\mathbf q)\cdot\mathbf n}{|\mathbf n|^2}\mathbf n,$$

and $(\mathbf m-\mathbf p)\cdot\mathbf n=0$, so it lies on the plane. These two checks confirm that the construction is the desired reflection. If $Q$ already lies on the plane, the correction term vanishes and $Q'=Q$.

A practical habit

Before calculating, identify the vector being projected, the receiving direction or plane, and whether a translation is needed. Afterward, check parallelism of the projected component and orthogonality of the residual. These conditions are often quicker to recall than a memorized formula.

Related: vector equations of planes and cross products.