A Geometric Projection Problem-Solving Workshop

Choosing the Vector That Matches the Geometry

A projection problem is often less about complicated arithmetic than about choosing the right direction vector. For a line, use a vector running along the line. For a rectangular box, represent each diagonal by its coordinate changes along the edges. For a parallel-perpendicular decomposition, use the direction to which the vector is being compared.

Once the direction is chosen, the dot product identifies how much of a vector points along it. The remaining part is perpendicular. This tutorial uses that idea to solve several geometric problems and to check each answer in more than one way.

Projection Formulas for Geometric Problems

Suppose a line passes through a point \(A\) and has nonzero direction vector \(\mathbf{d}\). A point \(X\) on that line can be written as \(X=A+t\mathbf{d}\), where \(t\) is a real number. Given a point \(P\), the vector from \(A\) to \(P\) is \(\mathbf{w}=P-A\).

The component of \(\mathbf{w}\) parallel to \(\mathbf{d}\) is its vector projection onto \(\mathbf{d}\). The formula applies when \(\mathbf{d}\ne\mathbf{0}\), so that \(\mathbf{d}\cdot\mathbf{d}\ne0\):

Consequently, the foot \(H\) of the perpendicular from \(P\) to the line, and the distance from \(P\) to that line, are

The same formula gives a parallel-perpendicular decomposition of any vector \(\mathbf{v}\): choose a nonzero direction vector \(\mathbf{d}\), set \(\mathbf{v}_{\parallel}=\operatorname{proj}_{\mathbf{d}}\mathbf{v}\), and set \(\mathbf{v}_{\perp}=\mathbf{v}-\mathbf{v}_{\parallel}\). The choice of \(\mathbf{d}\) determines what “parallel” means in the problem.

Why the Perpendicular-Foot Formula Works

Any candidate point on the line has the form \(A+t\mathbf{d}\). The vector from that candidate to \(P\) is \(\mathbf{w}-t\mathbf{d}\). At a perpendicular foot, this error vector must have zero dot product with the line direction. Thus \(t\) must satisfy

Expanding the dot product gives \(\mathbf{w}\cdot\mathbf{d}-t(\mathbf{d}\cdot\mathbf{d})=0\). Since \(\mathbf{d}\ne\mathbf{0}\), divide by \(\mathbf{d}\cdot\mathbf{d}\) to obtain \(t=(\mathbf{w}\cdot\mathbf{d})/(\mathbf{d}\cdot\mathbf{d})\). Substitution into \(A+t\mathbf{d}\) gives the stated foot formula. The value of \(t\) is unique, so there is exactly one such foot on the line.

Worked Example: Foot and Distance to a Line

Box Diagonals: Let the Edges Set the Coordinates

For a rectangular box, the three edge directions meeting at a vertex are mutually perpendicular. If their lengths are \(a\), \(b\), and \(c\), coordinates aligned with those edges represent a body diagonal by a vector with components \((a,b,c)\). Another body diagonal may have one or more signs reversed, depending on its direction.

To find an angle, compute the dot product of direction vectors and divide by the product of their lengths. For directed vectors \(\mathbf{u}\) and \(\mathbf{v}\), the angle \(\theta\) between them satisfies \(\cos\theta=(\mathbf{u}\cdot\mathbf{v})/(\|\mathbf{u}\|\|\mathbf{v}\|)\). Reversing a direction changes the sign of the dot product. Thus the angle between two chosen directed vectors may be obtuse even though the smaller angle between the geometric lines is acute.

Choosing a Direction for a Parallel-Perpendicular Split

For a vector \(\mathbf{v}\) and nonzero direction vector \(\mathbf{d}\), the projection gives the part parallel to \(\mathbf{d}\); subtracting it from \(\mathbf{v}\) gives the perpendicular part. These components must add to the original vector. The perpendicular component must also have dot product zero with \(\mathbf{d}\).

Because the components are perpendicular, their squared lengths add. This provides an additional independent check: \(\|\mathbf{v}\|^2=\|\mathbf{v}_{\parallel}\|^2+\|\mathbf{v}_{\perp}\|^2\).

A Common Mistake: Forgetting the Denominator

A frequent error is to treat \((\mathbf{w}\cdot\mathbf{d})\mathbf{d}\) as the projection onto a direction vector that is not a unit vector. The formula without a denominator works only when the direction vector has length \(1\).

For example, in the preceding decomposition, an incorrect calculation would give \(\mathbf{v}_{\parallel}=11(1,2,2)=(11,22,22)\). Its dot product with \(\mathbf{d}\) is \(11+44+44=99\), but the component along \(\mathbf{d}\) should have coefficient \(11/9\), since \(\mathbf{d}\cdot\mathbf{d}=9\). The wrong vector is not even a component of \((3,4,0)\) that could be completed by subtracting a perpendicular remainder: the remainder would be \((-8,-18,-22)\), whose dot product with \(\mathbf{d}\) is \(-8-36-44=-88\ne0\). Including the denominator produces the correct projection and makes the remainder perpendicular.

A Reliable Problem-Solving Routine

Before calculating, state what the chosen direction represents. Then use the formula that fits the question and verify the defining geometric conditions. This prevents a correct dot product from being attached to the wrong vector or interpreted as the wrong quantity.