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Dot Product

Dot product in 2D and 3D

For vectors

a⃗=(a1,a2,a3),b⃗=(b1,b2,b3),\vec{a} = (a_1, a_2, a_3), \quad \vec{b} = (b_1, b_2, b_3),

the dot product in 3D is

a⃗⋅b⃗=a1b1+a2b2+a3b3.\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3.

In 2D, if

a⃗=(a1,a2),b⃗=(b1,b2),\vec{a} = (a_1, a_2), \quad \vec{b} = (b_1, b_2),

then

a⃗⋅b⃗=a1b1+a2b2.\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2.

Example in 3D:

(2,−1,3)⋅(4,5,−2)=2⋅4+(−1)⋅5+3⋅(−2)=−3.(2,-1,3)\cdot(4,5,-2) = 2\cdot 4 + (-1)\cdot 5 + 3\cdot(-2) = -3.

Compute

(3,−4)⋅(2,5).(3,-4)\cdot(2,5).

Answer:

3⋅2+(−4)⋅5=6−20=−14.3\cdot 2 + (-4)\cdot 5 = 6 - 20 = -14.

Compute

(1,2,−1)⋅(4,−3,5).(1,2,-1)\cdot(4,-3,5).

Answer:

1⋅4+2⋅(−3)+(−1)⋅5=4−6−5=−7.1\cdot 4 + 2\cdot(-3) + (-1)\cdot 5 = 4 - 6 - 5 = -7.

If

(x,2,1)⋅(3,−1,4)=9,(x,2,1)\cdot(3,-1,4)=9,

find xx.

Answer:

3x+2(−1)+1⋅4=9⇒3x+2=9⇒x=73.3x + 2(-1) + 1\cdot4 = 9 \Rightarrow 3x + 2 = 9 \Rightarrow x = \frac{7}{3}.

Arithmetic properties of dot product

For vectors a⃗,b⃗,c⃗\vec{a},\vec{b},\vec{c} and scalar kk, the dot product satisfies:

  1. Commutative law
a⃗⋅b⃗=b⃗⋅a⃗\vec{a}\cdot\vec{b} = \vec{b}\cdot\vec{a}
  1. Distributive over addition
a⃗⋅(b⃗+c⃗)=a⃗⋅b⃗+a⃗⋅c⃗\vec{a}\cdot(\vec{b}+\vec{c}) = \vec{a}\cdot\vec{b} + \vec{a}\cdot\vec{c}
  1. Distributive in the first slot
(a⃗+b⃗)⋅c⃗=a⃗⋅c⃗+b⃗⋅c⃗(\vec{a}+\vec{b})\cdot\vec{c} = \vec{a}\cdot\vec{c} + \vec{b}\cdot\vec{c}
  1. Scalar multiplication compatibility
(ka⃗)⋅b⃗=k(a⃗⋅b⃗)=a⃗⋅(kb⃗)(k\vec{a})\cdot\vec{b} = k(\vec{a}\cdot\vec{b}) = \vec{a}\cdot(k\vec{b})
  1. Zero vector property
a⃗⋅0⃗=0,0⃗⋅a⃗=0\vec{a}\cdot\vec{0}=0, \quad \vec{0}\cdot\vec{a}=0
  1. Self-dot gives squared magnitude
a⃗⋅a⃗=∥a⃗∥2≥0\vec{a}\cdot\vec{a} = \|\vec{a}\|^2 \ge 0

These can be summarized as: dot product is symmetric and bilinear.

Given

a⃗⋅b⃗=7,a⃗⋅c⃗=−2,\vec{a}\cdot\vec{b}=7,\quad \vec{a}\cdot\vec{c}=-2,

find

a⃗⋅(b⃗+c⃗).\vec{a}\cdot(\vec{b}+\vec{c}).

Answer:

a⃗⋅(b⃗+c⃗)=7+(−2)=5.\vec{a}\cdot(\vec{b}+\vec{c}) = 7 + (-2) = 5.

If u⃗⋅v⃗=−3\vec{u}\cdot\vec{v}=-3, compute

(4u⃗)⋅(−2v⃗).(4\vec{u})\cdot(-2\vec{v}).

Answer:

(4u⃗)⋅(−2v⃗)=−8(u⃗⋅v⃗)=−8(−3)=24.(4\vec{u})\cdot(-2\vec{v}) = -8(\vec{u}\cdot\vec{v}) = -8(-3)=24.

If

w⃗⋅w⃗=49,\vec{w}\cdot\vec{w}=49,

what is ∥w⃗∥\|\vec{w}\|?

Answer:

∥w⃗∥=7.\|\vec{w}\|=7.

The angle theorem for dot product

Theorem. For nonzero vectors a⃗,b⃗\vec{a},\vec{b} with angle θ\theta between them,

a⃗⋅b⃗=∥a⃗∥ ∥b⃗∥cos⁡θ.\vec{a}\cdot\vec{b} = \|\vec{a}\|\,\|\vec{b}\|\cos\theta.

Proof (using the law of cosines)

Consider the triangle formed by a⃗\vec{a}, b⃗\vec{b}, and a⃗−b⃗\vec{a}-\vec{b}. By law of cosines,

∥a⃗−b⃗∥2=∥a⃗∥2+∥b⃗∥2−2∥a⃗∥∥b⃗∥cos⁡θ.\|\vec{a}-\vec{b}\|^2 = \|\vec{a}\|^2 + \|\vec{b}\|^2 - 2\|\vec{a}\|\|\vec{b}\|\cos\theta.

On the other hand,

∥a⃗−b⃗∥2=(a⃗−b⃗)⋅(a⃗−b⃗)=a⃗⋅a⃗−2a⃗⋅b⃗+b⃗⋅b⃗=∥a⃗∥2+∥b⃗∥2−2a⃗⋅b⃗.\|\vec{a}-\vec{b}\|^2 = (\vec{a}-\vec{b})\cdot(\vec{a}-\vec{b}) = \vec{a}\cdot\vec{a} - 2\vec{a}\cdot\vec{b} + \vec{b}\cdot\vec{b} = \|\vec{a}\|^2 + \|\vec{b}\|^2 - 2\vec{a}\cdot\vec{b}.

Compare the two expressions for ∥a⃗−b⃗∥2\|\vec{a}-\vec{b}\|^2:

∥a⃗∥2+∥b⃗∥2−2a⃗⋅b⃗=∥a⃗∥2+∥b⃗∥2−2∥a⃗∥∥b⃗∥cos⁡θ.\|\vec{a}\|^2 + \|\vec{b}\|^2 - 2\vec{a}\cdot\vec{b} = \|\vec{a}\|^2 + \|\vec{b}\|^2 - 2\|\vec{a}\|\|\vec{b}\|\cos\theta.

Therefore,

a⃗⋅b⃗=∥a⃗∥∥b⃗∥cos⁡θ.\vec{a}\cdot\vec{b}=\|\vec{a}\|\|\vec{b}\|\cos\theta.

In this graph, O=(0,0)O=(0,0), A=a⃗A=\vec{a}, and B=b⃗B=\vec{b}. The side ABAB is exactly a⃗−b⃗\vec{a}-\vec{b}, so the triangle makes the law-of-cosines identity visible.

Triangle proof view: why |a-b|^2 = |a|^2 + |b|^2 - 2|a||b|cos(theta)

a⃗ = (4.00, 1.00)b⃗ = (2.00, 3.00)a⃗ · b⃗ = 11.000θ = 42.3°

Drag the coloured dots at the arrow tips to move each vector.

-8-7-6-5-4-3-2-112345678-6-5-4-3-2-1123456a⃗−b⃗θa⃗b⃗O

If

a⃗=(1,2),b⃗=(2,1),\vec{a}=(1,2),\quad \vec{b}=(2,1),

find the angle between them.

Answer:

a⃗⋅b⃗=4,∥a⃗∥=∥b⃗∥=5.\vec{a}\cdot\vec{b}=4, \quad \|\vec{a}\|=\|\vec{b}\|=\sqrt{5}.

So

cos⁡θ=45,heta=cos⁡−1(45)≈36.87∘.\cos\theta = \frac{4}{5}, \quad heta=\cos^{-1}\left(\frac{4}{5}\right)\approx 36.87^\circ.

Determine whether

u⃗=(3,−1,2),v⃗=(1,5,1)\vec{u}=(3,-1,2),\quad \vec{v}=(1,5,1)

are perpendicular.

Answer:

u⃗⋅v⃗=3⋅1+(−1)⋅5+2⋅1=0.\vec{u}\cdot\vec{v} = 3\cdot1 + (-1)\cdot5 + 2\cdot1 = 0.

Hence cos⁡θ=0\cos\theta=0, so θ=90∘\theta=90^\circ and they are perpendicular.

Find kk so that

(k,1) is perpendicular to (2,−6).(k,1)\ \text{is perpendicular to}\ (2,-6).

Answer:

Perpendicular means dot product =0=0:

2k+1⋅(−6)=0⇒2k−6=0⇒k=3.2k + 1\cdot(-6)=0 \Rightarrow 2k-6=0 \Rightarrow k=3.

Geometric meaning of dot product

The formula

a⃗⋅b⃗=∥a⃗∥∥b⃗∥cos⁡θ\vec{a}\cdot\vec{b}=\|\vec{a}\|\|\vec{b}\|\cos\theta

gives several geometric interpretations:

  1. Angle information:
cos⁡θ=a⃗⋅b⃗∥a⃗∥∥b⃗∥\cos\theta = \frac{\vec{a}\cdot\vec{b}}{\|\vec{a}\|\|\vec{b}\|}

  1. Sign tells relative direction:

    • a⃗⋅b⃗>0\vec{a}\cdot\vec{b}>0 means acute angle (0∘<θ<90∘0^\circ<\theta<90^\circ)
    • a⃗⋅b⃗=0\vec{a}\cdot\vec{b}=0 means right angle (θ=90∘\theta=90^\circ)
    • a⃗⋅b⃗<0\vec{a}\cdot\vec{b}<0 means obtuse angle (90∘<θ<180∘90^\circ<\theta<180^\circ)

  2. Orthogonality test:

a⃗⊥b⃗  ⟺  a⃗⋅b⃗=0\vec{a}\perp\vec{b} \iff \vec{a}\cdot\vec{b}=0
  1. Directional component / projection relation:
a⃗⋅b⃗=∥b⃗∥ (scalar projection of a⃗ on b⃗).\vec{a}\cdot\vec{b} = \|\vec{b}\|\,(\text{scalar projection of }\vec{a}\text{ on }\vec{b}).

Use this to explore acute/obtuse/perpendicular cases by changing a⃗\vec{a} and b⃗\vec{b}. The annotation shows cos⁡θ\cos\theta and both sides of the identity.

Geometric meaning: angle sign and dot-product identity

a⃗ = (3.00, 2.00)b⃗ = (4.00, -1.00)a⃗ · b⃗ = 10.000θ = 47.7°

Drag the coloured dots at the arrow tips to move each vector.

-8-7-6-5-4-3-2-112345678-6-5-4-3-2-1123456a⃗−b⃗θa⃗b⃗O

Without finding the exact angle, classify the angle between

p⃗=(2,1,−1),q⃗=(1,3,0).\vec{p}=(2,1,-1),\quad \vec{q}=(1,3,0).

Answer:

p⃗⋅q⃗=2⋅1+1⋅3+(−1)⋅0=5>0.\vec{p}\cdot\vec{q}=2\cdot1+1\cdot3+(-1)\cdot0=5>0.

So the angle is acute.

Are

p⃗=(1,2,3),q⃗=(2,−1,0)\vec{p}=(1,2,3),\quad \vec{q}=(2,-1,0)

orthogonal?

Answer:

p⃗⋅q⃗=1⋅2+2⋅(−1)+3⋅0=0.\vec{p}\cdot\vec{q}=1\cdot2+2\cdot(-1)+3\cdot0=0.

Yes, they are orthogonal.

For

u⃗=(1,0,1),v⃗=(2,2,1),\vec{u}=(1,0,1),\quad \vec{v}=(2,2,1),

find cos⁡θ\cos\theta.

Answer:

u⃗⋅v⃗=3,∥u⃗∥=2,∥v⃗∥=3.\vec{u}\cdot\vec{v}=3, \quad \|\vec{u}\|=\sqrt{2}, \quad \|\vec{v}\|=3.

Thus

cos⁡θ=332=12.\cos\theta=\frac{3}{3\sqrt{2}}=\frac{1}{\sqrt{2}}.

Scalar projection vs vector projection

Let a⃗\vec{a} be projected onto a nonzero vector b⃗\vec{b}.

Scalar projection (component)

The scalar projection of a⃗\vec{a} onto b⃗\vec{b} is

comp⁡b⃗a⃗=a⃗⋅b⃗∥b⃗∥.\operatorname{comp}_{\vec{b}}\vec{a} = \frac{\vec{a}\cdot\vec{b}}{\|\vec{b}\|}.

This is a number. It can be positive, negative, or zero.

Vector projection

The vector projection of a⃗\vec{a} onto b⃗\vec{b} is

proj⁡b⃗a⃗=(a⃗⋅b⃗∥b⃗∥2)b⃗.\operatorname{proj}_{\vec{b}}\vec{a} = \left(\frac{\vec{a}\cdot\vec{b}}{\|\vec{b}\|^2}\right)\vec{b}.

This is a vector in the direction of b⃗\vec{b} (or opposite if the coefficient is negative).

Relationship:

proj⁡b⃗a⃗=(comp⁡b⃗a⃗)b⃗∥b⃗∥.\operatorname{proj}_{\vec{b}}\vec{a} = \left(\operatorname{comp}_{\vec{b}}\vec{a}\right)\frac{\vec{b}}{\|\vec{b}\|}.

In the graph below, the purple vector is proj⁡b⃗a⃗\operatorname{proj}_{\vec{b}}\vec{a} and the red dashed segment is the rejection a⃗−proj⁡b⃗a⃗\vec{a}-\operatorname{proj}_{\vec{b}}\vec{a}.

Projection view: scalar component vs projection vector

a⃗ = (5.00, 2.00)b⃗ = (3.00, 0.00)a⃗ · b⃗ = 15.000θ = 21.8°comp_b(a⃗) = 5.000proj_b(a⃗) = (5.00, 0.00)

Drag the coloured dots at the arrow tips to move each vector.

-8-7-6-5-4-3-2-112345678-6-5-4-3-2-1123456projθa⃗b⃗O

Let

a⃗=(3,4),b⃗=(1,0).\vec{a}=(3,4),\quad \vec{b}=(1,0).

Find comp⁡b⃗a⃗\operatorname{comp}_{\vec{b}}\vec{a}.

Answer:

comp⁡b⃗a⃗=a⃗⋅b⃗∥b⃗∥=31=3.\operatorname{comp}_{\vec{b}}\vec{a} =\frac{\vec{a}\cdot\vec{b}}{\|\vec{b}\|} =\frac{3}{1}=3.

Let

a⃗=(3,4),b⃗=(1,0).\vec{a}=(3,4),\quad \vec{b}=(1,0).

Find proj⁡b⃗a⃗\operatorname{proj}_{\vec{b}}\vec{a}.

Answer:

proj⁡b⃗a⃗=(a⃗⋅b⃗∥b⃗∥2)b⃗=(31)(1,0)=(3,0).\operatorname{proj}_{\vec{b}}\vec{a} =\left(\frac{\vec{a}\cdot\vec{b}}{\|\vec{b}\|^2}\right)\vec{b} =\left(\frac{3}{1}\right)(1,0)=(3,0).

Let

a⃗=(2,1),b⃗=(−4,0).\vec{a}=(2,1),\quad \vec{b}=(-4,0).

Find both comp⁡b⃗a⃗\operatorname{comp}_{\vec{b}}\vec{a} and proj⁡b⃗a⃗\operatorname{proj}_{\vec{b}}\vec{a}.

Answer:

a⃗⋅b⃗=−8,∥b⃗∥=4,∥b⃗∥2=16.\vec{a}\cdot\vec{b}= -8, \quad \|\vec{b}\|=4, \quad \|\vec{b}\|^2=16.

So

comp⁡b⃗a⃗=−84=−2,\operatorname{comp}_{\vec{b}}\vec{a}=\frac{-8}{4}=-2,

and

proj⁡b⃗a⃗=(−816)(−4,0)=(−12)(−4,0)=(2,0).\operatorname{proj}_{\vec{b}}\vec{a} =\left(\frac{-8}{16}\right)(-4,0) =\left(-\frac{1}{2}\right)(-4,0) =(2,0).

The scalar projection is a signed number (−2-2), while the vector projection is an actual vector ((2,0)(2,0)).

Mixed practice on dot product

Given

u⃗=(1,−2,2),v⃗=(2,1,0),\vec{u}=(1,-2,2),\quad \vec{v}=(2,1,0),

compute u⃗⋅v⃗\vec{u}\cdot\vec{v} and decide whether the angle is acute, right, or obtuse.

Compute the dot product first, then use its sign.

u⃗⋅v⃗=1⋅2+(−2)⋅1+2⋅0=0.\vec{u}\cdot\vec{v}=1\cdot2+(-2)\cdot1+2\cdot0=0.

So the vectors are orthogonal and the angle is right.

Find the angle between

a⃗=(1,1,0),b⃗=(1,0,1).\vec{a}=(1,1,0),\quad \vec{b}=(1,0,1).

Use cos⁡θ=a⃗⋅b⃗∥a⃗∥∥b⃗∥\cos\theta=\dfrac{\vec{a}\cdot\vec{b}}{\|\vec{a}\|\|\vec{b}\|}.

a⃗⋅b⃗=1,∥a⃗∥=2,∥b⃗∥=2.\vec{a}\cdot\vec{b}=1, \quad \|\vec{a}\|=\sqrt{2}, \quad \|\vec{b}\|=\sqrt{2}.

Therefore

cos⁡θ=12,θ=60∘.\cos\theta=\frac{1}{2},\quad \theta=60^\circ.

For

a⃗=(4,2,0),b⃗=(2,0,0),\vec{a}=(4,2,0),\quad \vec{b}=(2,0,0),

find both comp⁡b⃗a⃗\operatorname{comp}_{\vec{b}}\vec{a} and proj⁡b⃗a⃗\operatorname{proj}_{\vec{b}}\vec{a}.

Compute a⃗⋅b⃗\vec{a}\cdot\vec{b}, then divide by ∥b⃗∥\|\vec{b}\| or ∥b⃗∥2\|\vec{b}\|^2 as needed.

a⃗⋅b⃗=8,∥b⃗∥=2,∥b⃗∥2=4.\vec{a}\cdot\vec{b}=8, \quad \|\vec{b}\|=2, \quad \|\vec{b}\|^2=4.

So

comp⁡b⃗a⃗=82=4,\operatorname{comp}_{\vec{b}}\vec{a}=\frac{8}{2}=4,

and

proj⁡b⃗a⃗=(84)(2,0,0)=2(2,0,0)=(4,0,0).\operatorname{proj}_{\vec{b}}\vec{a} =\left(\frac{8}{4}\right)(2,0,0) =2(2,0,0)=(4,0,0).

Find all kk such that

(k,2,−1)⊥(3,k,6).(k,2,-1)\perp(3,k,6).

Perpendicular vectors have dot product zero.

(k,2,−1)⋅(3,k,6)=3k+2k−6=0(k,2,-1)\cdot(3,k,6)=3k+2k-6=0

so

5k=6⇒k=65.5k=6 \Rightarrow k=\frac{6}{5}.