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The Dot Product

Calculus 2 · Chapter 10: Vectors & the Geometry of SpaceSection 10.2 · lesson 52 of 549 min · free

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In this lesson

Two vectors in, one number out.

  • Compute the dot product two equivalent ways
  • Read the angle - and orthogonality - from its sign
  • Separate scalar from vector projection
  • Recover work as a dot product

Transcript

Everything said in the video, section by section (1,270 words). Read along, or search it.

MAT-265 - Calculus 2 · Section 10.2

The Dot Product

Section ten point one showed you how to add, subtract and scale vectors. But none of those operations multiplies two vectors together. The dot product does. And it does something unusual. It takes two vectors in, and returns a single number out. A scalar. Which sounds like it should lose information. But that one number turns out to encode the angle between them.

Two Definitions

Componentwise, and Geometric

There are two definitions, and they're equal. The componentwise definition: multiply corresponding components, and add. u dot v equals u one v one, plus u two v two, plus u three v three. And the geometric definition: the magnitude of u, times the magnitude of v, times cosine theta. Where theta is the angle between them. The first is what you compute with. The second is what it means. Two consequences are worth noting immediately. Take v equal to u. Then the angle is zero, cosine is one, and you get u dot u equals the magnitude of u, squared. So the dot product of a vector with itself is its squared length. And the operation is commutative — u dot v equals v dot u — because multiplication of components is.

The Angle

The Sign Reports the Geometry

Solving the geometric form for the cosine gives the angle between two nonzero vectors. Cosine theta equals u dot v, over the product of the magnitudes. And theta is the arccosine of that. Which returns a value between zero and pi. The range of arccosine. But here's the practical part, and it's often all you need. Since both magnitudes are positive, the sign of the dot product is exactly the sign of cosine theta. So: u dot v positive means theta is less than ninety degrees. The vectors point in broadly the same direction. u dot v equals zero means theta is exactly ninety degrees. Perpendicular. And u dot v negative means theta exceeds ninety degrees. They lean apart. Look at the three panels. Same v in each. And u leaning three different ways. The middle case is important enough to have its own name. Two vectors are called orthogonal when their dot product is zero. Orthogonality is the vector generalization of perpendicularity. And the test is remarkably cheap. Multiply components, add, and check whether the result is zero. No square roots. No inverse trigonometry. One convention deserves note. The zero vector is considered orthogonal to every vector — since its dot product with anything is zero. That's a definitional choice rather than a geometric fact. Because the zero vector has no direction, and so forms no angle at all.

Worked Example

Finding an Angle

Let's find an angle. Between u equals two comma three, and v equals five comma minus one. Step one. Compute the dot product. Two times five, plus three times minus one. Ten minus three. Seven. Step two. Compute the magnitudes. The magnitude of u is root of four plus nine. Root thirteen. About three point six zero six. The magnitude of v is root of twenty-five plus one. Root twenty-six. About five point zero nine nine. Step three. Divide, and take the arccosine. Seven over root thirteen times root twenty-six. That's about zero point three eight zero seven. And the arccosine of that is about sixty-seven point six degrees. Now notice — the sign told you the answer before the arithmetic did. The dot product was positive. So theta had to be less than ninety degrees. The arccosine only supplied the precise value.

Projections

Splitting One Vector Along Another

Now a genuinely useful application. Splitting one vector along another. Let v be a nonzero vector. The scalar projection of u onto v — also called the component of u along v — is u dot v, over the magnitude of v. And the vector projection of u onto v is u dot v, over the magnitude of v squared, times v. The scalar projection is a number. And it may be negative — which indicates that u leans away from v. The vector projection is a vector. Always parallel to v. Now, the distinction in the denominators is worth pausing on. Because it's where students lose marks. The scalar projection divides by the magnitude of v once. Converting the dot product into a length. The vector projection divides by the magnitude twice. Once to obtain that length. And once more to reduce v to unit length, before scaling it. Forgetting that second division is the most frequent error in this section. Look at the picture. The gold arrow is the vector projection — the piece of u that lies along v. Its length is the scalar projection. The dashed segment is what's left over. Perpendicular to v. And together, those two pieces reconstruct u exactly.

Worked Example

Both Projections, and the Leftover

Let's compute both. With u equals six comma four, and v equals eight comma zero. Step one. Dot product and magnitude. u dot v is six times eight, plus four times zero. Forty-eight. The magnitude of v is eight. And its square is sixty-four. Step two. The scalar projection. Divide once. Forty-eight over eight. Six. Step three. The vector projection. Divide twice — that is, by sixty-four. Forty-eight over sixty-four is three quarters. Times the vector eight comma zero. That gives six comma zero. Step four. Extract the perpendicular piece, by subtracting the projection from u. Six comma four, minus six comma zero. Equals zero comma four. Step five. Verify orthogonality. Zero comma four, dotted with eight comma zero. Zero times eight, plus four times zero. Zero. Confirmed. So the decomposition is exact. Six comma four equals six comma zero, plus zero comma four. Splitting u into a piece along v, and a piece perpendicular to it. Exactly as the figure depicts.

Key Takeaways

Work, and What to Carry Forward

Four related quantities have now been introduced, all built from the same dot product. The dot product itself. A scalar. It measures the overall alignment of u with v. The angle. Also a scalar. The opening between the two directions. The scalar projection. A signed length — how far u reaches along v. And the vector projection. The only one that returns a vector. The portion of u lying along v. Plus the work application, which is also a scalar. Only the vector projection returns a vector. Everything else returns a plain number. Now, about work. Chapter four point six treated work as the product of a force and the distance over which it acts. That approach served well when force and motion pointed the same way. The vector formulation removes that restriction entirely. Work equals F dot d. Which is the magnitude of F, times the magnitude of d, times cosine theta. And in doing so, it explains why the earlier definition worked. When the force and the displacement point the same way, theta is zero, cosine theta is one, and the formula reduces to force times distance. The old rule was the special case. And notice a consequence. A force perpendicular to the motion does no work at all — because cosine ninety is zero. Five takeaways. The dot product takes two vectors and returns a scalar. Componentwise it's a sum of products. Geometrically it's the product of magnitudes times cosine theta. The sign reports the angle. Positive means acute, zero means orthogonal, negative means obtuse. Orthogonality is tested by checking whether the dot product is zero. No square roots required. The scalar projection divides by the magnitude once. The vector projection divides twice — and forgetting that is the classic error. And work is a dot product. Which explains why force times distance only worked when the two pointed the same way. Next up is section ten point three — the Cross Product. Which multiplies two vectors into a third vector.

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FAQ

Questions students ask us

About The Dot Product

What is the dot product and how do you calculate it?
The dot product combines two vectors into a single scalar by multiplying corresponding components and adding: \mathbf{u}\cdot\mathbf{v} = u_1v_1 + u_2v_2 + u_3v_3. It also equals |\mathbf{u}||\mathbf{v}|\cos\theta, where \theta is the angle between the vectors. Use the componentwise form to compute and the geometric form to interpret. Setting \mathbf{v}=\mathbf{u} gives \mathbf{u}\cdot\mathbf{u} = |\mathbf{u}|^2, the squared length.
How do you find the angle between two vectors using the dot product?
For nonzero vectors, use \cos\theta = \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{u}||\mathbf{v}|}, then take the arccosine, which gives an angle between 0 and \pi. For example, with \mathbf{u}=\langle 2,3\rangle and \mathbf{v}=\langle 5,-1\rangle, the dot product is 10-3=7. The magnitudes are \sqrt{13} and \sqrt{26}. Then \cos\theta \approx 0.3807, so \theta \approx 67.6^\circ.
What does the sign of the dot product tell you about two vectors?
For nonzero vectors, the sign of the dot product matches the sign of \cos\theta, so it reveals the type of angle. If \mathbf{u}\cdot\mathbf{v}>0, then \theta<90^\circ. If \mathbf{u}\cdot\mathbf{v}=0, then \theta=90^\circ and the vectors are orthogonal. If \mathbf{u}\cdot\mathbf{v}<0, then \theta>90^\circ. By convention, the zero vector counts as orthogonal to every vector, even though it forms no angle.
What is the difference between scalar projection and vector projection?
For a nonzero \mathbf{v}, the scalar projection \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{v}|} is a signed number, while the vector projection \frac{\mathbf{u}\cdot\mathbf{v}}{|\mathbf{v}|^2}\mathbf{v} is a vector parallel to \mathbf{v}. Forgetting the second division by |\mathbf{v}| is the classic error. For \mathbf{u}=\langle 6,4\rangle and \mathbf{v}=\langle 8,0\rangle, the dot product is 48. The scalar projection is 48/8=6, and the vector projection is \tfrac{48}{64}\langle 8,0\rangle=\langle 6,0\rangle.

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