Resolving a Vector Parallel and Perpendicular to a Plane

Two Components Relative to a Plane

A vector can point partly across a plane and partly along it. For example, a displacement may have a component that moves away from a flat surface and another component that moves along the surface. Resolving the vector into these parts turns a geometric description into a calculation.

The key information is a normal vector to the plane: a nonzero vector perpendicular to every direction that lies in the plane. The component of a vector along this normal is its perpendicular component relative to the plane. Subtracting that component leaves a vector parallel to the plane.

These are vector components, not just lengths. In particular, the perpendicular component can point either in the direction of \(\mathbf{n}\) or in the opposite direction.

The Projection Formulas

The normal vector need not have length one. For a nonzero normal vector \(\mathbf{n}\), the perpendicular component is

The component parallel to the plane is then

Here \(\mathbf{v}_{\perp}\) is parallel to the normal, so it is perpendicular to the plane. The vector \(\mathbf{v}_{\parallel}\) is orthogonal to \(\mathbf{n}\), so it lies in the plane’s direction. The two components add back to the original vector:

If the plane is written as \(ax+by+cz=d\), then \(\mathbf{n}=(a,b,c)\) is a normal vector, provided \((a,b,c)\ne(0,0,0)\). The constant \(d\) determines the plane’s position but does not change its normal direction.

Why the Formula Works

To find the parallel component, look for a vector of the form \(\mathbf{v}-t\mathbf{n}\) that is orthogonal to the normal. The unknown scalar \(t\) is chosen so that the remainder lies along the plane. This condition determines \(t\) directly.

This also proves the decomposition is unique. If \(\mathbf{v}=\mathbf{p}+\mathbf{q}\), where \(\mathbf{p}\) is parallel to the plane and \(\mathbf{q}\) is parallel to \(\mathbf{n}\), then \(\mathbf{q}=s\mathbf{n}\) for some scalar \(s\). Since \(\mathbf{p}=\mathbf{v}-s\mathbf{n}\) must be orthogonal to \(\mathbf{n}\), the condition above forces \(s=(\mathbf{v}\cdot\mathbf{n})/(\mathbf{n}\cdot\mathbf{n})\). Therefore both components are determined uniquely.

Worked Examples

A Common Error: Forgetting the Normal’s Length

A frequent mistake is to multiply the normal by \(\mathbf{v}\cdot\mathbf{n}\) and stop, as if the normal were automatically a unit vector. For the first example, that incorrect calculation would give \(\mathbf{v}_{\perp}=8(2,1,-2)=(16,8,-16)\). But this is not the projection: its dot product with \(\mathbf{v}\) is not the relevant check; instead, verify whether the proposed parallel remainder is orthogonal to \(\mathbf{n}\).

The incorrect remainder would be \((3,4,1)-(16,8,-16)=(-13,-4,17)\), and

So the remainder is not parallel to the plane. The missing denominator is \(\mathbf{n}\cdot\mathbf{n}=9\). Dividing by it gives the correct coefficient \(8/9\), and the resulting remainder has dot product zero with the normal.

Why the Resolution Is Useful

The decomposition separates two geometrically different effects. The perpendicular component measures how much of \(\mathbf{v}\) points across the plane; the parallel component contains what remains along its direction. This makes it possible to describe a vector relative to a surface without changing the vector itself: the two pieces preserve its complete direction and magnitude when recombined.

There are three reliable checks for a calculation:

• \(\mathbf{v}_{\perp}\) must be a scalar multiple of the normal.
• \(\mathbf{v}_{\parallel}\cdot\mathbf{n}\) must equal zero.
• \(\mathbf{v}_{\parallel}+\mathbf{v}_{\perp}\) must equal \(\mathbf{v}\).

If the components are computed correctly, the Pythagorean identity also applies: the squared length of the original vector is the sum of the squared lengths of its perpendicular and parallel components. This follows because their dot product is zero.