The Projection Problem for a Line
A line through the origin in \(\mathbb{R}^n\) can be described by all scalar multiples of one nonzero vector. Given another vector \(\mathbf{b}\), we want to identify the vector on that line that fits \(\mathbf{b}\)'s component along the line. The defining geometric condition is that the difference between \(\mathbf{b}\) and its projection is perpendicular to the line. This condition leads directly to a formula, and also gives a simple way to check a calculation.
Throughout this tutorial, vectors are in \(\mathbb{R}^n\) with the standard dot product \(\mathbf{x}\cdot\mathbf{y}=\mathbf{x}^{\top}\mathbf{y}\). Two vectors are perpendicular, or orthogonal, when their dot product is zero. For a nonzero vector \(\mathbf{u}\), the line it spans is \(L=\operatorname{span}\{\mathbf{u}\}=\{t\mathbf{u}:t\in\mathbb{R}\}\). The requirement \(\mathbf{u}\ne\mathbf{0}\) matters: if \(\mathbf{u}=\mathbf{0}\), its span is just \(\{\mathbf{0}\}\), not a line.
To find \(\mathbf{p}\), write it as \(t\mathbf{u}\), since it must lie in \(L\). Every vector in the line is a multiple \(s\mathbf{u}\). The residual \( \mathbf{b}-t\mathbf{u}\) is perpendicular to every such multiple exactly when it is perpendicular to \(\mathbf{u}\): taking \(s=1\) gives the necessary condition, and multiplying a zero dot product by any \(s\) gives the sufficient condition. Thus the defining condition reduces to one scalar equation.
Because \(\mathbf{u}\ne\mathbf{0}\), at least one of its coordinates is nonzero, so the sum of the squares of its coordinates, \(\mathbf{u}\cdot\mathbf{u}\), is positive. We may divide by it to find \(t\).
Proof. Any candidate in \(L\) has the form \(t\mathbf{u}\). By the definition of projection, its residual must be perpendicular to every vector in \(L\). This is equivalent to \((\mathbf{b}-t\mathbf{u})\cdot\mathbf{u}=0\), since \(\mathbf{u}\in L\) and each vector in \(L\) is a multiple of \(\mathbf{u}\). Expanding gives \(\mathbf{b}\cdot\mathbf{u}-t(\mathbf{u}\cdot\mathbf{u})=0\). Since \(\mathbf{u}\ne\mathbf{0}\), we have \(\mathbf{u}\cdot\mathbf{u}>0\), so the equation has exactly one solution, \(t=(\mathbf{b}\cdot\mathbf{u})/(\mathbf{u}\cdot\mathbf{u})\). The resulting \(t\mathbf{u}\) lies in \(L\) and its residual has zero dot product with \(\mathbf{u}\), hence with every multiple of \(\mathbf{u}\). It therefore satisfies the definition and is the unique projection. If \(\mathbf{b}=\mathbf{0}\), the same equation gives \(t=0\), so the formula returns \(\mathbf{0}\). \(\square\)
Calculating and Checking a Projection
The numerator measures how strongly \(\mathbf{b}\) points along the chosen direction \(\mathbf{u}\); the denominator adjusts for the length of that direction. The denominator is the squared length, not the length itself. A reliable calculation has three parts: compute both dot products, multiply \(\mathbf{u}\) by their ratio, and check that the residual is perpendicular to \(\mathbf{u}\). Membership in the line is checked by confirming that the proposed projection is a scalar multiple of \(\mathbf{u}\).
Worked Example: Projecting in \(\mathbb{R}^2\)
Find the projection of \(\mathbf{b}=\begin{bmatrix}3\\4\end{bmatrix}\) onto the line spanned by \(\mathbf{u}=\begin{bmatrix}1\\2\end{bmatrix}\), and verify the result.
First calculate the dot products:
The formula gives
This is in \(L\), since it is \(11/5\) times \(\mathbf{u}\). The residual is
Check: its dot product with the direction vector is \(\frac45(1)+(-\frac25)(2)=\frac45-\frac45=0\). Thus the projection lies on the line and the residual is perpendicular to it, as required.
Worked Example: A Projection in \(\mathbb{R}^3\)
Project \(\mathbf{b}=\begin{bmatrix}2\\-1\\3\end{bmatrix}\) onto the line spanned by \(\mathbf{u}=\begin{bmatrix}2\\1\\0\end{bmatrix}\).
Compute the numerator and denominator separately, retaining the negative sign in the second coordinate of \(\mathbf{b}\):
Therefore,
Subtract coordinate by coordinate to obtain the residual:
Check: the proposed projection is \(3/5\) times \(\mathbf{u}\), so it lies on \(L\). The residual's dot product with \(\mathbf{u}\) is \(\frac45(2)+(-\frac85)(1)+3(0)=\frac85-\frac85+0=0\). Both defining conditions hold.
Zero and Orthogonal Inputs
The formula covers two useful special cases without needing a separate method. If the vector being projected is zero, its dot product with any direction is zero, so the projection is zero. If \(\mathbf{b}\) is already perpendicular to the line, then \(\mathbf{b}\cdot\mathbf{u}=0\) and again the projection is zero. In the second case, \(\mathbf{b}\) need not itself be zero; it has no component along the line.
Worked Example: Projecting the Zero Vector
Let \(\mathbf{b}=\begin{bmatrix}0\\0\\0\end{bmatrix}\) and \(\mathbf{u}=\begin{bmatrix}3\\-2\\1\end{bmatrix}\). Calculate the projection onto \(L=\operatorname{span}\{\mathbf{u}\}\).
The direction is nonzero, and its squared length is \(3^2+(-2)^2+1^2=9+4+1=14\). The numerator is
Hence
Check: the result is \(0\mathbf{u}\), so it lies on \(L\). The residual is \(\mathbf{b}-\operatorname{proj}_{L}\mathbf{b}=\mathbf{0}\), and \(\mathbf{0}\cdot\mathbf{u}=0\). The formula is defined because the denominator is \(14\), not zero.
Worked Example: A Nonzero Vector Perpendicular to the Line
Project \(\mathbf{b}=\begin{bmatrix}2\\-1\end{bmatrix}\) onto the line generated by \(\mathbf{u}=\begin{bmatrix}1\\2\end{bmatrix}\).
The vector being projected is nonzero, but its dot product with the direction is \(2(1)+(-1)(2)=2-2=0\). Also, \(\mathbf{u}\cdot\mathbf{u}=1+4=5\). Thus
Check: the zero vector belongs to every subspace, including \(L\). The residual is \(\begin{bmatrix}2\\-1\end{bmatrix}\), and its dot product with \(\mathbf{u}\) is \(2(1)+(-1)(2)=0\). Therefore the residual is perpendicular to the entire line.
The Direction Vector Is Not Unique
A line has many possible direction vectors. If \(c\in\mathbb{R}\) and \(c\ne0\), then \(c\mathbf{u}\) generates the same line as \(\mathbf{u}\): every multiple of \(c\mathbf{u}\) is a multiple of \(\mathbf{u}\), and every multiple of \(\mathbf{u}\) is a multiple of \(c\mathbf{u}\). The projection formula must therefore give the same point when the chosen generator is rescaled.
Proof. Since \(c\ne0\), the vectors \(\mathbf{u}\) and \(c\mathbf{u}\) span the same line. Applying the formula with \(c\mathbf{u}\) gives \(\frac{\mathbf{b}\cdot(c\mathbf{u})}{(c\mathbf{u})\cdot(c\mathbf{u})}(c\mathbf{u})\). By scalar multiplication in the dot product, the numerator is \(c(\mathbf{b}\cdot\mathbf{u})\), and the denominator is \(c^2(\mathbf{u}\cdot\mathbf{u})\). Since \(c\ne0\), the expression simplifies to \(\frac{c(\mathbf{b}\cdot\mathbf{u})}{c^2(\mathbf{u}\cdot\mathbf{u})}(c\mathbf{u})=\frac{\mathbf{b}\cdot\mathbf{u}}{\mathbf{u}\cdot\mathbf{u}}\mathbf{u}\). This is the original projection, proving the claim. \(\square\)
Rescaling can make the arithmetic simpler, but the denominator must be rescaled too. For instance, using \(-2\mathbf{u}\) reverses the direction and doubles its length; the coefficient changes accordingly, while the final projected vector stays fixed. The zero scalar is excluded because \(0\mathbf{u}=\mathbf{0}\) no longer generates the same line.
Common Mistake: Using the Wrong Denominator
A frequent error is to divide by \(\mathbf{b}\cdot\mathbf{b}\) instead of \(\mathbf{u}\cdot\mathbf{u}\). The scalar being found is the coefficient \(t\) in the candidate \(t\mathbf{u}\). The perpendicularity equation contains \(t(\mathbf{u}\cdot\mathbf{u})\), so it is the direction vector's squared length that must appear in the denominator.
For the first worked example, the incorrect calculation would be \(t=(\mathbf{b}\cdot\mathbf{u})/(\mathbf{b}\cdot\mathbf{b})=11/(3^2+4^2)=11/25\), producing the wrong candidate \(\frac{11}{25}\begin{bmatrix}1\\2\end{bmatrix}=\begin{bmatrix}11/25\\22/25\end{bmatrix}\). Its residual is \(\begin{bmatrix}64/25\\78/25\end{bmatrix}\), whose dot product with \(\mathbf{u}\) is \(64/25+2(78/25)=220/25=44/5\ne0\). The candidate lies on the line, but its residual is not perpendicular to it, so it cannot be the projection. The corrected denominator is \(\mathbf{u}\cdot\mathbf{u}=5\), giving the previously checked projection \(\begin{bmatrix}11/5\\22/5\end{bmatrix}\).
Another error is to omit the squared length altogether. Unless \(\mathbf{u}\) is a unit vector, the ratio \(\mathbf{b}\cdot\mathbf{u}\) alone is not generally the coefficient of \(\mathbf{u}\). The displayed formula works for any nonzero direction vector, whether or not it has length one.
Using the Formula Reliably
For a vector \(\mathbf{b}\) and a nonzero line generator \(\mathbf{u}\), calculate \(\mathbf{b}\cdot\mathbf{u}\) and \(\mathbf{u}\cdot\mathbf{u}\), divide, and multiply the result by \(\mathbf{u}\). Then verify two facts: the answer is a multiple of \(\mathbf{u}\), and the residual has dot product zero with \(\mathbf{u}\). The second check confirms perpendicularity to every vector on the line because every vector on it is a multiple of \(\mathbf{u}\). In particular, if \(\mathbf{b}=\mathbf{0}\), the numerator is zero and the projection is \(\mathbf{0}\); no special exception to the formula is needed.
Check Your Understanding
Use the line-projection formula and check each result by testing line membership and residual orthogonality.
- Project \(\begin{bmatrix}1\\3\end{bmatrix}\) onto the line spanned by \(\begin{bmatrix}2\\1\end{bmatrix}\).
- Project \(\begin{bmatrix}0\\0\\0\end{bmatrix}\) onto the line spanned by \(\begin{bmatrix}1\\-1\\2\end{bmatrix}\). Which dot product in the formula makes the result zero?
- Find the projection of \(\begin{bmatrix}4\\2\end{bmatrix}\) onto the line spanned by \(\begin{bmatrix}1\\-2\end{bmatrix}\). Explain why the answer is zero even though the vector being projected is nonzero.
- For a nonzero scalar \(c\), explain why \(\operatorname{span}\{\mathbf{u}\}=\operatorname{span}\{c\mathbf{u}\}\).
- A proposed projection is a multiple of \(\mathbf{u}\), but its residual has nonzero dot product with \(\mathbf{u}\). Which defining condition does it fail?
