The Dot Product

Multiply matching components, then add

The dot product of two vectors is written a · b and read “a dot b”. To work it out, multiply the matching components, the first with the first and the second with the second, and add the two products: $\left(a_1,a_2\right)\cdot \left(b_1,b_2\right)=a_1{b}_1+a_2{b}_2$(a₁, a₂) · (b₁, b₂) = a₁b₁ + a₂b₂.

For (3, 1) and (2, 4), the first components give 3 × 2 = 6 and the second components give 1 × 4 = 4, so (3, 1) · (2, 4) = 6 + 4 = 10.

The answer is a single number, not a vector. A number like this is a scalar, so the dot product is also called the scalar product.

What the number measures

Drop a perpendicular from the head of b onto the line of a. The piece of a’s line between the tail and the foot of that perpendicular is the part of b lying along a, called the projection of b on a. The dot product is the length of a times that projection.

In a right triangle with hypotenuse |b| and the angle $\theta$θ between a and b, the projection is the side next to $\theta$θ, which is $|b|\cos \theta$|b| cos θ. So the same number can be written $a\cdot b=|a||b|\cos \theta$a · b = |a| |b| cos θ.

In the lesson’s second figure, a = (4, 3) and b = (1, 4). The components give a · b = 4 × 1 + 3 × 4 = 4 + 12 = 16. The length of a is $\sqrt{4^2+3^2}=5$√(4² + 3²) = 5, so the projection of b on a is 16 ÷ 5 = 3.2.

a = (3, 1) and b = (2, 4). The gold projection of b along a is $10÷\sqrt{10}=\sqrt{10}$10 ÷ √10 = √10, which is 3.16 to 2 decimal places, and the length of a is also $\sqrt{10}$√10, so $a\cdot b=\sqrt{10}\times \sqrt{10}=10$a · b = √10 × √10 = 10, the number the components gave.

Why the two give the same number

Draw a and b from one point, and join their heads. The third side is the vector b − a, and the angle between a and b is $\theta$θ. The cosine rule on this triangle gives $|b-a{|}^2=|a{|}^2+|b{|}^2-2|a||b|\cos \theta$|b − a|² = |a|² + |b|² − 2|a| |b| cos θ.

Now work out the left side from the components. $|b-a{|}^2={\left(b_1-a_1\right)}^2+{\left(b_2-a_2\right)}^2$|b − a|² = (b₁ − a₁)² + (b₂ − a₂)², which multiplies out to $a_1^2+a_2^2+b_1^2+b_2^2-2\left(a_1{b}_1+a_2{b}_2\right)$a₁² + a₂² + b₁² + b₂² − 2(a₁b₁ + a₂b₂), and that is $|a{|}^2+|b{|}^2-2\left(a_1{b}_1+a_2{b}_2\right)$|a|² + |b|² − 2(a₁b₁ + a₂b₂).

The two right-hand sides are equal, and the squared lengths cancel, so $a_1{b}_1+a_2{b}_2=|a||b|\cos \theta$a₁b₁ + a₂b₂ = |a| |b| cos θ. The rule with components and the length times the projection are the same number.

Check it with a = (3, 1) and b = (2, 4). Then $|a{|}^2=10$|a|² = 10, $|b{|}^2=20$|b|² = 20 and b − a = (−1, 3), so $|b-a{|}^2=1+9=10$|b − a|² = 1 + 9 = 10. The cosine rule gives $10=10+20-2|a||b|\cos \theta$10 = 10 + 20 − 2|a| |b| cos θ, so $|a||b|\cos \theta =10$|a| |b| cos θ = 10, the same as the components gave.

a = (3, 1), b = (2, 4), and the third side b − a = (−1, 3), from the head of a to the head of b.

Biggest when the two line up

For two vectors of fixed lengths, only $\cos \theta$cos θ can change, and $\cos \theta$cos θ is largest, 1, when $\theta =0^{\circ }$θ = 0°. So the dot product is biggest when the two vectors point the same way, and then it is |a| |b|.

The sign tells you about the angle. For an acute angle $\cos \theta$cos θ is positive, so the dot product is positive. For an obtuse angle $\cos \theta$cos θ is negative, the projection points backward along a, and the dot product is negative: (2, 1) · (−3, 1) = −6 + 1 = −5, so those two vectors meet at an obtuse angle.

a has length 4 and b has length 3. At $\theta =60^{\circ }$θ = 60°, the projection of b on a’s line is 3 cos 60° = 1.5, so a · b = 4 × 1.5 = 6. Swing b round: at 0° the dot product is 12, its largest; at 90° the projection shrinks to nothing; past 90° it points against a and the dot product is negative, down to −12 at 180°.

Zero at a right angle

When the two vectors are perpendicular, the perpendicular from the head of b lands at the tail of a, so the projection has length 0 and a · b = 0. In the other form, cos 90° = 0. Two vectors that are not zero are perpendicular exactly when their dot product is 0.

For a = (4, 1) and b = (−1, 4): a · b = 4 × (−1) + 1 × 4 = −4 + 4 = 0, so they meet at a right angle.

That pair shows a quick way to find a perpendicular vector: swap the two components and change the sign of one. Turning (p, q) a quarter turn gives (−q, p), and (p, q) · (−q, p) = −pq + qp = 0 for every p and q.

a = (4, 1) and b = (−1, 4) meet at a right angle, and a · b = −4 + 4 = 0.

A vector with itself, and two rules

A vector dotted with itself gives its length squared, because the angle is 0° and cos 0° = 1: $\left(3,4\right)\cdot \left(3,4\right)=9+16=25=5^2$(3, 4) · (3, 4) = 9 + 16 = 25 = 5².

The order does not matter, a · b = b · a, because 3 × 2 + 1 × 4 is the same sum as 2 × 3 + 4 × 1. And the dot product distributes over addition: a · (b + c) = a · b + a · c. With a = (3, 1), b = (2, 4) and c = (1, −1), b + c = (3, 3) and a · (b + c) = 9 + 3 = 12; separately, a · b = 10 and a · c = 3 − 1 = 2, which also add to 12.

With three components, each matching pair is multiplied in the same way and all three products are added.

The usual mistakes

Leaving the answer as a pair. (3, 1) · (2, 4) is not (6, 4): the two products are added, and the dot product is the single number 10.

Pairing across with up. 3 × 4 + 1 × 2 = 14 multiplies components in different places; only matching places are multiplied.

Swapping the components without changing a sign. (1, 4) is not perpendicular to (4, 1): their dot product is 4 + 4 = 8. The quarter turn is (−1, 4).

Reversing the vector to make it perpendicular. (−4, −1) points the opposite way to (4, 1), and their dot product is −16 − 1 = −17, as negative as two vectors of those lengths can give, not zero.

Work done by a force

In mechanics, the work done by a force F that moves an object through a displacement d is the dot product F · d. Only the part of the force along the motion does work, and that is exactly what the dot product counts. A force pointing partly against the motion gives a negative dot product: it does negative work. The application below works out both for a trolley pulled up a ramp.