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The dot product and projections

Geometry · level 8, lesson 5 of 6

A question to start

Here are two arrows: \((3, 4)\) and \((5, -1)\). What is the angle between them?

You could draw them and use a protractor. Or you could do this: multiply the first components, \(3 \times 5 = 15\), multiply the second components, \(4 \times (-1) = -4\), and add, getting 11. From that single number 11, and the two lengths, the angle drops out: about 64.4°. No drawing, no protractor, and it works just as well for arrows in three dimensions, or in a hundred. How can multiplying and adding components possibly know about angles? The answer is the cosine law, in disguise.

The idea

Two definitions of one number

The dot product of two vectors is a number (a scalar), not a vector. Its algebraic definition is

\[\vec u \cdot \vec v = u_1 v_1 + u_2 v_2 \qquad \text{(in 3D, add } u_3 v_3\text{)}\]

Its geometric meaning is

\[\vec u \cdot \vec v = |\vec u|\,|\vec v|\cos\theta\]

where \(\theta\) is the angle between the two arrows when they start from the same point (\(0 \le \theta \le \pi\)). The surprising part is that these are always equal.

Why the two agree

Draw \(\vec u\) and \(\vec v\) from the same point. The third side of the triangle they make is \(\vec u - \vec v\) (the arrow from the tip of \(\vec v\) to the tip of \(\vec u\), from the last lesson). The cosine law for that triangle says

\[|\vec u - \vec v|^2 = |\vec u|^2 + |\vec v|^2 - 2|\vec u|\,|\vec v|\cos\theta\]

Now work out the left side with components instead:

\[|\vec u - \vec v|^2 = (u_1 - v_1)^2 + (u_2 - v_2)^2 = (u_1^2 + u_2^2) + (v_1^2 + v_2^2) - 2(u_1v_1 + u_2v_2)\]

which is \(|\vec u|^2 + |\vec v|^2 - 2(u_1v_1 + u_2v_2)\). Both lines describe the same length. Comparing them, the \(|\vec u|^2 + |\vec v|^2\) parts match, so the rest must match too:

\[u_1v_1 + u_2v_2 = |\vec u|\,|\vec v|\cos\theta \qquad ∎\]

The same argument works in three dimensions, with one more term. So the simple sum of products secretly carries the cosine of the angle.

What the dot product tells you

Rearranging gives the angle between any two non-zero vectors:

\[\cos\theta = \frac{\vec u \cdot \vec v}{|\vec u|\,|\vec v|}\]

The sign alone tells you a lot, because the lengths are positive:

\(\vec u \cdot \vec v\) \(\cos\theta\) The angle is
positive positive acute: the arrows point roughly the same way
zero zero exactly 90°: the vectors are perpendicular (also called orthogonal)
negative negative obtuse: the arrows point roughly opposite ways

The test "perpendicular exactly when the dot product is 0" is one of the most used facts in all of applied mathematics.

Rules of the dot product

From the algebraic definition, these follow at once:

  • \(\vec u \cdot \vec v = \vec v \cdot \vec u\) (order does not matter);
  • \(\vec u \cdot (\vec v + \vec w) = \vec u \cdot \vec v + \vec u \cdot \vec w\) (it spreads over addition, like ordinary multiplication);
  • \((k\vec u) \cdot \vec v = k(\vec u \cdot \vec v)\);
  • \(\vec v \cdot \vec v = v_1^2 + v_2^2 = |\vec v|^2\) (a vector dotted with itself is its length squared).

Since \(\cos\theta\) is between \(-1\) and \(1\), also \(|\vec u \cdot \vec v| \le |\vec u|\,|\vec v|\). This is the Cauchy-Schwarz inequality, which is one of the most important inequalities in mathematics.

Projections: the shadow of one vector on another

Shine a light straight down onto the line of \(\vec v\). The shadow that \(\vec u\) casts on that line is its projection onto \(\vec v\).

Ouv θprojection: |u| cos θ

The right triangle shows that the shadow's length is \(|\vec u|\cos\theta\). Using the dot product to replace the cosine:

\[\text{scalar projection} = |\vec u|\cos\theta = \frac{\vec u \cdot \vec v}{|\vec v|}\]

(negative if the angle is obtuse, meaning the shadow falls backwards). To get the shadow as a vector, multiply this length by the unit vector \(\dfrac{\vec v}{|\vec v|}\):

\[\operatorname{proj}_{\vec v}\vec u = \frac{\vec u \cdot \vec v}{|\vec v|^2}\,\vec v\]

What is left over, \(\vec u - \operatorname{proj}_{\vec v}\vec u\), is perpendicular to \(\vec v\). So every vector splits into a part along \(\vec v\) and a part across it. Engineers do this constantly: the part of gravity pulling a cart down a slope, the part of the wind pushing a sail forward.

Work

In physics, when a constant force \(\vec F\) moves something through a displacement \(\vec d\), the work done is

\[W = \vec F \cdot \vec d = |\vec F|\,|\vec d|\cos\theta\]

Only the part of the force along the motion does work. Pushing straight down on a box sliding across the floor does no work at all, because the angle is 90°.

Worked examples

1. The opening question. Find the angle between \((3, 4)\) and \((5, -1)\).

  1. Dot product: \(3 \times 5 + 4 \times (-1) = 11\).
  2. Lengths: \(\sqrt{9 + 16} = 5\) and \(\sqrt{25 + 1} = \sqrt{26}\).
  3. \(\cos\theta = \dfrac{11}{5\sqrt{26}} \approx 0.4315\), so \(\theta \approx 64.4°\).

2. In three dimensions. Find the angle between \((1, 2, 2)\) and \((2, -1, 2)\).

  1. Dot product: \(2 - 2 + 4 = 4\).
  2. Lengths: \(\sqrt{1 + 4 + 4} = 3\) and \(\sqrt{4 + 1 + 4} = 3\).
  3. \(\cos\theta = \tfrac{4}{9}\), so \(\theta \approx 63.6°\).

3. A projection. Project \(\vec u = (5, 1)\) onto \(\vec v = (3, 4)\), and split \(\vec u\) into parts along and across \(\vec v\).

  1. \(\vec u \cdot \vec v = 15 + 4 = 19\) and \(|\vec v|^2 = 25\).
  2. Scalar projection: \(\tfrac{19}{5} = 3.8\).
  3. Vector projection: \(\tfrac{19}{25}(3, 4) = (2.28,\ 3.04)\).
  4. Part across: \((5, 1) - (2.28, 3.04) = (2.72,\ -2.04)\).
  5. Check it is perpendicular to \(\vec v\): \(2.72 \times 3 + (-2.04) \times 4 = 8.16 - 8.16 = 0\). Good.

4. Work on a sled. A rope pulls a sled 50 m across flat snow with a force of 80 newtons, at 30° above the horizontal. Work done: \(80 \times 50 \times \cos 30° \approx 3464\) joules. The upward part of the pull lifts a little weight off the runners but does no work moving the sled forward.

5. Uphill on Mars. On Mars, gravity pulls with about 3.71 newtons on each kilogram. A 1000 kg rover drives up a straight ramp from \((0, 0)\) to \((40, 3)\), in metres (40 across, 3 up).

  1. Gravity: \(\vec F = (0,\ -3710)\) N. Displacement: \(\vec d = (40, 3)\).
  2. Work done by gravity: \(\vec F \cdot \vec d = 0 \times 40 + (-3710) \times 3 = -11\,130\) J. Negative, because gravity pulls against the climb.
  3. So the motors must supply at least 11 130 J for the climb (plus whatever friction takes). Notice the 40 m across did not matter: only the part of the motion along the force counts. On Earth the same climb needs about \(9.81 \times 1000 \times 3 \approx 29\,400\) J.

6. Thales' theorem in one line. Let a circle have centre \(O\) at the origin and radius \(r\), with diameter from \(A\) to \(B\), so \(\vec b = -\vec a\), and let \(C\) be any point on the circle, so \(|\vec c| = r = |\vec a|\). Then

\[\overrightarrow{AC} \cdot \overrightarrow{BC} = (\vec c - \vec a)\cdot(\vec c + \vec a) = \vec c \cdot \vec c - \vec a \cdot \vec a = r^2 - r^2 = 0\]

So \(AC\) is perpendicular to \(BC\): the angle in a semicircle is a right angle. ∎ Compare the proof with isosceles triangles.

Try it

  1. Find \((2, -3) \cdot (4, 5)\). Is the angle between these vectors acute, right or obtuse?
  2. Find the angle between \((2, 1)\) and \((1, 3)\) exactly.
  3. For which \(k\) is \((k, 3)\) perpendicular to \((2, -4)\)?
  4. Find the projection of \((6, 2)\) onto \((1, 1)\), and the part of \((6, 2)\) perpendicular to \((1, 1)\).
  5. Find the distance from the point \((5, 1)\) to the line through the origin in the direction \((3, 4)\).
  6. A triangle has corners \(A(0, 0)\), \(B(4, 0)\) and \(C(1, 3)\). Find the angle at \(A\).
  7. Show that \(|\vec u + \vec v|^2 = |\vec u|^2 + |\vec v|^2 + 2\,\vec u \cdot \vec v\). What does it say when \(\vec u\) and \(\vec v\) are perpendicular?
  8. Prove with the dot product that the diagonals of a rhombus (a parallelogram with all sides equal) are perpendicular.
Answers
  1. \(8 - 15 = -7\). Negative, so the angle is obtuse.
  2. Dot product \(2 + 3 = 5\); lengths \(\sqrt5\) and \(\sqrt{10}\). \(\cos\theta = \dfrac{5}{\sqrt{50}} = \dfrac{5}{5\sqrt2} = \dfrac{1}{\sqrt2}\), so \(\theta = 45°\), that is \(\tfrac{\pi}{4}\).
  3. \(2k - 12 = 0\), so \(k = 6\).
  4. \((6, 2) \cdot (1, 1) = 8\) and \(|(1, 1)|^2 = 2\), so the projection is \(4(1, 1) = (4, 4)\). The perpendicular part is \((6, 2) - (4, 4) = (2, -2)\). Check: \((2, -2) \cdot (1, 1) = 0\).
  5. It is the length of the part of \((5, 1)\) across the line. From worked example 3 that part is \((2.72, -2.04)\), with length \(\sqrt{7.3984 + 4.1616} = \sqrt{11.56} = 3.4\).
  6. \(\overrightarrow{AB} = (4, 0)\) and \(\overrightarrow{AC} = (1, 3)\). Dot product 4; lengths 4 and \(\sqrt{10}\). \(\cos A = \dfrac{4}{4\sqrt{10}} = \dfrac{1}{\sqrt{10}}\), so \(A \approx 71.6°\).
  7. \(|\vec u + \vec v|^2 = (\vec u + \vec v)\cdot(\vec u + \vec v) = \vec u\cdot\vec u + 2\,\vec u\cdot\vec v + \vec v\cdot\vec v\), using the rules above. When \(\vec u \perp \vec v\), the middle term is 0 and \(|\vec u + \vec v|^2 = |\vec u|^2 + |\vec v|^2\): that is Pythagoras' theorem, for the right triangle with legs \(\vec u\) and \(\vec v\).
  8. Let the rhombus have sides \(\vec a\) and \(\vec c\) from one corner, with \(|\vec a| = |\vec c|\). Its diagonals are \(\vec a + \vec c\) and \(\vec c - \vec a\). Then \((\vec a + \vec c)\cdot(\vec c - \vec a) = |\vec c|^2 - |\vec a|^2 = 0\), so the diagonals are perpendicular. ∎

Common mistake

Expecting a vector. \(\vec u \cdot \vec v\) is a single number. Writing \((3, 4)\cdot(5, -1) = (15, -4)\) is wrong: add the products to get 11.

Projecting onto the wrong vector. \(\operatorname{proj}_{\vec v}\vec u\) (the shadow of \(\vec u\) on \(\vec v\)) divides by \(|\vec v|^2\), the length of the vector you project onto. Swapping the roles gives a different answer.

Where it leads