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A dot product measures how much two Euclidean vectors point along each other. This article uses real Euclidean vectors in ${\mathbb{R}}^n$, with the usual Cartesian basis: multiply matching coordinates and add,The result is a scalar, while $\Vert \mathbf{a}\Vert$ is the Euclidean norm. This real convention has no complex conjugation; for vectors over the complex numbers, use the convention specified in inner product.

For two nonzero Euclidean vectors, the same dot product records their angle $0\le \theta \le \pi$ through the cosine:Here $\arccos$ is an inverse trigonometric function. The Cauchy-Schwarz inequality guarantees that the normalized ratio lies in $[-1,1]$. The zero vector has dot product zero with every vector, but its angle is undefined: the ratio would divide by zero. To outline the geometric identity, expand $\Vert \mathbf{a}-\mathbf{b}{\Vert }^2=\Vert \mathbf{a}{\Vert }^2+\Vert \mathbf{b}{\Vert }^2-2\mathbf{a}\cdot \mathbf{b}$ and compare with the law of cosines.^[ref]

Take $\mathbf{a}=(3,4)$ and $\mathbf{c}=(2,0)$. Their Euclidean norms are $5$ and $2$, andThe unit vector $\mathbf{u}=\mathbf{c}/\Vert \mathbf{c}\Vert =(1,0)$ selects the horizontal direction. The scalar projection $\mathbf{a}\cdot \mathbf{u}=3$ says how far $\mathbf{a}$ reaches along that direction; its vector projection is $3\mathbf{u}=(3,0)$. Thus $\mathbf{a}=(3,0)+(0,4)$ splits into an aligned part and an orthogonal projection residual.

For $\mathbf{b}=(4,-3)$, the dot product $\mathbf{a}\cdot \mathbf{b}=12-12=0$ proves orthogonality; both Euclidean norms are $5$. For $\mathbf{d}=(-2,0)$, $\mathbf{a}\cdot \mathbf{d}=-6$: the scalar projection onto $\mathbf{d}/\Vert \mathbf{d}\Vert =(-1,0)$ is $-3$, because $\mathbf{a}$ reaches against that chosen direction. Its angle with $\mathbf{d}$ is $126.8699^{\circ }$.

Keeping $\mathbf{a}=(3,4)$ fixed and choosing $\mathbf{b}$ of Euclidean norm $2$ gives $\mathbf{a}\cdot \mathbf{b}=10\cos \theta$. At angles $0^{\circ },60^{\circ },90^{\circ },120^{\circ },180^{\circ }$, the dot products are $10,5,0,-5,-10$. Positive values mean partial alignment, zero means orthogonality, and negative values mean a component against the other direction.

This interpretation explains work: for a constant force $\mathbf{F}$ and a straight displacement $\mathbf{s}$, $W=\mathbf{F}\cdot \mathbf{s}$ counts the force component along the motion. A sideways force contributes zero work, and an opposing force contributes negative work.^[ref]

Choose a unit vector ${\mathbf{e}}_d$ pointing downhill and a perpendicular unit vector ${\mathbf{e}}_n$ pointing outward from the inclined plane. The gravitational weight points vertically downward. Its scalar projections onto these directions giveThe angle between the downward vertical and the inward normal is $\theta$: the inward component uses cosine, and the downhill component uses sine. These are two components of one force. They replace its vertical arrow in a component calculation; adding them to that same arrow would count gravitational weight twice.

Continued contact with a stationary straight inclined plane gives zero perpendicular acceleration. The perpendicular component of Newton's second law therefore givesThis is why the normal force generally differs from $mg$: it balances the inward component, rather than the whole gravitational weight. An extra lifting force, a curved surface, or motion of the support would require a new perpendicular balance.^[ref]

The same geometry explains the ramp's usefulness. A frictionless block pulled uphill at constant velocity needs a parallel force $P=mg\sin \theta$. Travelling a distance $L$ raises it through height $h=L\sin \theta$, so the pulling work is $PL=mgh$, its gain in gravitational potential energy. The smaller force costs a larger displacement; the work-energy theorem keeps the accounting consistent.

The direction must first be represented by a unit vector: $\hat{\mathbf{q}}=(1,1,1)/\sqrt{3}$. The scalar projection of the vector area is thereforeIf a vector projection is intended, multiply this scalar by $\hat{\mathbf{q}}$:

The scalar $\mathbf{r}\cdot \hat{\mathbf{u}}$ is the signed scalar projection along the unit vector $\hat{\mathbf{u}}$. Multiplying by that unit vector makes $P_{\hat{\mathbf{u}}}(\mathbf{r})$ the orthogonal projection of $\mathbf{r}$ onto its span.

For the other expression, the vector triple product identity givesThis is the orthogonal projection onto a plane perpendicular to $\hat{\mathbf{u}}$ through the origin. Its dot product with $\hat{\mathbf{u}}$ is zero, while the difference from $\mathbf{r}$ is parallel to $\hat{\mathbf{u}}$.

Using the complex conjugate, $b\bar{z}+\bar{b}z=2\mathrm{Re}(\bar{b}z)$. The locus is therefore the straight lineIn the complex plane, the quantity $\mathrm{Re}(\bar{b}z)/|b|$ is the signed scalar projection of $z$ in the direction $b$. Thus the straight line is perpendicular to $b$ and passes through $-cb/(2|b{|}^2)$. This description uses no separation of $z$ into real coordinates.

For nonzero $x$, let $\varphi$ be its angle to the unit vector $q_1$. The scalar projection on the oriented $q_1$ axis is $\Vert x\Vert \cos \varphi$. Since $\Vert q_1\Vert =1$, the dot product identity gives $q_1^{T}x=\Vert x\Vert \cos \varphi$. Multiplying that signed scalar by $q_1$ places the foot at $p=(q_1^{T}x)q_1$.

The residual is perpendicular to the axis, because $q_1^T(x-p)=q_1^{T}x-(q_1^{T}x)(q_1^T{q}_1)=0$. An obtuse angle gives a negative scalar and a foot on the negative side of the line; $x=0$ has projection zero without needing an angle.

The first panel is a constructed geometric illustration. The second illustrates the directional stretch used later in Question 6.

The scalar projection of a real Euclidean vector $\mathbf{a}$ onto a nonzero Euclidean vector $\mathbf{c}$ is its signed component along the unit vector $\mathbf{c}/\Vert \mathbf{c}\Vert$:For $\mathbf{a}=(3,4)$, the scalar projection onto $(2,0)$ is $3$, while that onto $(-2,0)$ is $-3$. Reversing the measuring direction reverses the sign; the scalar projection is not necessarily a nonnegative length. For nonzero $\mathbf{a}$, it also equals $\Vert \mathbf{a}\Vert \cos \theta$, using the angle with $\mathbf{c}$ and its cosine. For the zero vector it is zero, even though that angle is undefined.

The vector projection multiplies the scalar projection by the chosen unit vector, giving the orthogonal projection onto the linear span of a nonzero real Euclidean vector $\mathbf{c}$:For $\mathbf{a}=(3,4)$ and $\mathbf{c}=(2,0)$, this gives $\mathbf{p}=(6/4)(2,0)=(3,0)$ and $\mathbf{r}=(0,4)$; the dot product $\mathbf{r}\cdot \mathbf{c}=0$ checks the orthogonal projection residual. Replacing $\mathbf{c}$ by $(-2,0)$ gives the same $\mathbf{p}=(-6/4)(-2,0)=(3,0)$: reversing direction changes the scalar projection, but preserves the vector projection onto the line.

For a short proof of the nearest-point property, any vector $\mathbf{z}$ on this line satisfies $\Vert \mathbf{a}-\mathbf{z}{\Vert }^2=\Vert \mathbf{r}{\Vert }^2+\Vert \mathbf{p}-\mathbf{z}{\Vert }^2$ by the Pythagorean identity. The second term is uniquely minimized at $\mathbf{z}=\mathbf{p}$. This connects a directional measurement to the orthogonal projection used in approximation.^[ref]