Vectors, Dot Products, and Cross Products (2024)

When studying the flight of an airplane, a rocket firing, or countless other applications, we can use vectors to mathematically describe what's going on. A vector has two components: magnitude and direction. This is similar to the use of polar coordinates, where we describe a point by its distance from the origin (like the magnitude of a vector) and the angle formed by the line from the origin to the point (like the direction of a vector).

Since all that distinguishes a vector from other vectors is its magnitude and direction, we can move it around at will. Therefore, for the sake of convenience, we often place the tail of the vector at the origin and describe the vector by the location of the head. This way of describing a vector looks just like describing a point, so to be clear that we're talking about a vector (that when the tail is placed at the origin has its head at this location), we use angled brackets, like so:v=1,3,5\vec{v} = \langle 1,3,5 \ranglev=1,3,5This vector is shown below.

Vectors, Dot Products, and Cross Products (1)

Magnitude

The magnitude of this vector is given byv=x2+y2+z2=12+32+52=35\begin{aligned}|\vec{v}| &= \sqrt{x^2 + y^2 + z^2}\&= \sqrt{1^2 + 3^2 + 5^2}\&= \sqrt{35}\end{aligned}v=x2+y2+z2=12+32+52=35

Standard Unit Vectors

A unit vector is a vector with magnitude 111. The standard unit vectors point in the direction of the xxx axis, yyy axis, and zzz axis. Note that the hat on these vectors denotes that they are unit vectors; whenever we write a vector with a hat like this, we'll mean a unit vector.

i^=1,0,0j^=0,1,0k^=0,0,1\begin{aligned}\hat{i} &= \langle 1,0,0 \rangle\\hat{j} &= \langle 0,1,0 \rangle\\hat{k} &= \langle 0,0,1 \rangle\end{aligned}i^j^k^=1,0,0=0,1,0=0,0,1

In R2\mathbb{R}^2R2, we only need two standard unit vectors:i^=1,0andj^=0,1.\hat{i} = \langle 1,0 \rangle \text{ and } \hat{j} = \langle 0,1 \rangle.i^=1,0andj^=0,1.

In Rn\mathbb{R}^nRn, we need nnn unit vectors, usually writtene1^,e2^,e3^,,en^\hat{e_1}, \hat{e_2}, \hat{e_3}, \ldots, \hat{e_n}e1^,e2^,e3^,,en^where en^\hat{e_n}en^ has a 111 in the nth position and 000's elsewhere.

Vectors in rectangular coordinates can be expressed in terms of these standard unit vectors.

Write v=2,6,17\vec{v} = \langle −2,6,17 \ranglev=2,6,17 in terms of the standard unit vectors.

Solution

v=2,6,17=2i^+6j^+17k^\vec{v} = \langle −2,6,17 \rangle = \ans{−2\hat{i}+6\hat{j}+17\hat{k}}v=2,6,17=2i^+6j^+17k^

If v=5,2,3=5i^+2j^3k^\vec{v} = \langle 5,2,−3 \rangle = 5\hat{i} + 2\hat{j} − 3\hat{k}v=5,2,3=5i^+2j^3k^, find 2v2\vec{v}2v.

Solution

2v=10,4,6=10i^+4j^6k^2\vec{v} = \langle 10,4,−6 \rangle = \ans{10\hat{i}+4\hat{j}−6\hat{k}}2v=10,4,6=10i^+4j^6k^

Adding / Subtracting Vectors

Visually, to add two vectors, place the tail of one at the head of the other and find the resulting vector. Numerically, simply add/subtract the components.

If u=3i^2j^+5k^u⃗ =3\hat{i}−2\hat{j}+5\hat{k}u=3i^2j^+5k^ and v=4i^+6k^v⃗ =4\hat{i}+6\hat{k}v=4i^+6k^, find 3u+v3u⃗ +v⃗3u+v.

Solution

3u+v=9i^6j^+15k^+4i^+6k^=13i^6j^+21k^\begin{aligned}3u⃗ +v⃗ &= 9\hat{i}−6\hat{j}+15\hat{k}+4\hat{i}+6\hat{k}\&= \ans{13\hat{i}−6\hat{j}+21\hat{k}}\end{aligned}3u+v=9i^6j^+15k^+4i^+6k^=13i^6j^+21k^

Unit Vectors

If uu⃗u is any vector, u^\hat{u}u^ is a unit vector in the direction of uu⃗u. To find this unit vector, divide (u⃗) by its magnitude:

u^=uu\ans{\hat{u} = \dfrac{u⃗}{|u⃗|}}u^=uu

If u=3,2,2u⃗ =⟨3,2,−2⟩u=3,2,2, find u^\hat{u}u^.

Solution

The magnitude of uu⃗u is

u=32+22+(2)2=17,|u⃗|= \sqrt{3^2+2^2+(−2)^2} = \sqrt{17},u=32+22+(2)2=17,so the unit vector isu^=1173,2,2=31717,21717,21717\hat{u}=\dfrac{1}{\sqrt{17}} ⟨3,2,−2⟩ = \ans{\left⟨ \dfrac{3\sqrt{17}}{17}, \dfrac{2\sqrt{17}}{17}, −\dfrac{2\sqrt{17}}{17} \right⟩}u^=1713,2,2=17317,17217,17217

Find a vector in the same direction as v=8i^j^+4k^v⃗ =8\hat{i}−\hat{j}+4\hat{k}v=8i^j^+4k^ that is 444 units long.

Solution

To find this, first find a unit vector in the same direction as vv⃗v and then multiply it by 444.

v=82+(1)2+42=81=9,|v⃗| = \sqrt{8^2+(−1)^2+4^2} = \sqrt{81} = 9,v=82+(1)2+42=81=9,so the unit vector isv^=198,1,4.\hat{v} = \dfrac{1}{9} ⟨8,−1,4⟩.v^=918,1,4.Therefore,4v^=498,1,44\hat{v} = \ans{\dfrac{4}{9} ⟨8,-1,4⟩}4v^=948,1,4

Application: Statics

Find T1T_1T1 and T2T_2T2, the tension in the two cables below.

Vectors, Dot Products, and Cross Products (2)

Start by representing the forces with vectors:Vectors, Dot Products, and Cross Products (3)

If this system is in static equilibrium (nothing is moving), it means that the three forces must cancel each other, meaning that if we add the three vectors, the result must be 000:T1+T2400j^=0T1+T2=400j^T1ii^+T1jj^+T2ii^+T2jj^=400j^    T1i+T2i=0T1j+T2j=400    T1cos60+T2cos30=0T1sin60+T2sin30=400    12T1+32T2=032T1+12T2=400    T1=346.41lbandT2=200lb\begin{aligned}&\vec{T_1} + \vec{T_2} - 400\hat{j} = 0 \longrightarrow \vec{T_1}+\vec{T_2} = 400\hat{j}\&T_{1i}\hat{i}+T_{1j}\hat{j}+T_{2i}\hat{i}+T_{2j}\hat{j} = 400\hat{j}\\\implies &T_{1i}+T_{2i}=0\&T_{1j}+T_{2j}=400\\\implies &-|T_1| \cos 60^{\circ} + |T_2|\cos 30^{\circ} = 0\&|T_1|\sin 60^{\circ}+|T_2|\sin 30^{\circ} = 400\\\implies &-\dfrac{1}{2}|T_1|+\dfrac{\sqrt{3}}{2}|T_2| = 0\&\dfrac{\sqrt{3}}{2}|T_1| + \dfrac{1}{2}|T_2| = 400\\\implies &\ans{T_1 = 346.41 \textrm{ lb and } T_2 = 200 \textrm{ lb}}\end{aligned}T1+T2400j^=0T1+T2=400j^T1ii^+T1jj^+T2ii^+T2jj^=400j^T1i+T2i=0T1j+T2j=400T1cos60+T2cos30=0T1sin60+T2sin30=40021T1+23T2=023T1+21T2=400T1=346.41lbandT2=200lb

Dot Product (Scalar Product)

The dot product, one of two ways to multiply a vector, is defined as follows: if u=u1,u2,u3u⃗ = ⟨u_1,u_2,u_3⟩u=u1,u2,u3 and v=v1,v2,v3v⃗ = ⟨v_1,v_2,v_3⟩v=v1,v2,v3, thenuv=u1v1+u2v2+u3v3=uvcosθ\ans{\begin{aligned}\vec{u} \cdot \vec{v} &= u_1v_1 + u_2v_2 + u_3v_3\&= |\vec{u}| |\vec{v}| \cos \theta\end{aligned}}uv=u1v1+u2v2+u3v3=uvcosθ

where θ\thetaθ is the angle between the two vectors. Note that the result is a scalar, which is why this is called the scalar product.

Also note that uu=u2\vec{u} \cdot \vec{u} = |\vec{u}|^2uu=u2

Let u=2,3,4u⃗ = ⟨2,−3,4⟩u=2,3,4, v=4,3,5v⃗ = ⟨4,−3,5⟩v=4,3,5, and w=1,1,3w⃗ =⟨1,1,−3⟩w=1,1,3. Find uvu⃗ ⋅ v⃗uv , 2vw2v⃗ ⋅ w⃗2vw, and the angle between uu⃗u and vv⃗v.

Solution

  1. uv\vec{u} \cdot \vec{v}uv:uv=(2)(4)+(3)(3)+(4)(5)=37\vec{u} \cdot \vec{v} = (2)(4)+(-3)(-3)+(4)(5) = \ans{37}uv=(2)(4)+(3)(3)+(4)(5)=37

  2. 2vw2v⃗ ⋅ w⃗2vw:2vw=(8)(1)+(6)(1)+(10)(3)=282\vec{v} \cdot \vec{w} = (8)(1)+(-6)(1)+(10)(-3) = \ans{-28}2vw=(8)(1)+(6)(1)+(10)(3)=28[note that (2v)w=2(vw)(2\vec{v})\cdot \vec{w} = 2(\vec{v} \cdot \vec{w})(2v)w=2(vw)]

  3. The angle between u\vec{u}u and v\vec{v}v:uv=37=uvcosθ=(4+9+16)(16+9+25)cosθcosθ=372950=13.7\begin{aligned}\vec{u} \cdot \vec{v} = 37 &= |\vec{u}| |\vec{v}| \cos \theta\&= (\sqrt{4+9+16}) (\sqrt{16+9+25}) \cos \theta\\longrightarrow \cos \theta &= \dfrac{37}{\sqrt{29}\ \sqrt{50}} = \ans{13.7^{\circ}}\end{aligned}uv=37cosθ=uvcosθ=(4+9+16)(16+9+25)cosθ=295037=13.7

Orthogonal (or Perpendicular) Vectors

Two vectors are orthogonal if the angle between them is θ=π2θ=\dfrac{\pi}{2}θ=2π. In terms of the dot product, this means that two vectors uu⃗u and vv⃗v are orthogonal iff (if and only if) uv=0u⃗ ⋅ v⃗ = 0uv=0.

Let u=3i^=3,0,0u⃗ = 3\hat{i} = ⟨3,0,0⟩u=3i^=3,0,0 and v=2j^=0,2,0v⃗ = −2\hat{j} = ⟨0,−2,0⟩v=2j^=0,2,0. Show that uu⃗u and vv⃗v are orthogonal.

Solution

uv=0,sothevectorsareorthogonal.\vec{u} \cdot \vec{v} = 0 \text{, so the vectors are orthogonal.}uv=0,sothevectorsareorthogonal.

Projection of uu⃗u onto vv⃗v (or vice versa)

The projection of one vector onto another can be thought of as the shadow that one vector casts onto the other, as shown below.

Vectors, Dot Products, and Cross Products (4)

To find the projection of uu⃗u onto vv⃗v, find the magnitude of the projection and multiply it byv^=vv.\hat{v} = \dfrac{v⃗}{|v⃗|}.v^=vv.The magnitude of the projection isprojvu=ucosθ,|proj_v\ u⃗| = |u⃗| \cos \theta,projvu=ucosθ,and since uv=uvcosθ,u⃗ ⋅ v⃗ = |u⃗||v⃗| \cos \theta,uv=uvcosθ,projvu=uvv.|proj_v\ u⃗| = \dfrac{u⃗ ⋅ v⃗}{|v⃗|}.projvu=vuv.Therefore,projvu=uvvvv=(uvvv)v.\ans{proj_v\ \vec{u} = \dfrac{\vec{u} \cdot \vec{v}}{|\vec{v}|} \ \dfrac{\vec{v}}{|\vec{v}|} = \left(\dfrac{\vec{u} \cdot \vec{v}}{\vec{v} \cdot \vec{v}}\right) \vec{v}.}projvu=vuvvv=(vvuv)v.

Find the projection of u=3,3,4\vec{u} = \langle 3,3,4 \rangleu=3,3,4 onto v=3,2,0\vec{v} = \langle 3,2,0 \ranglev=3,2,0.

Solution

projvu=(uvvv)v=(9+6+09+4+0)3,2,0=15133,2,0=4513,3013,0\begin{aligned}proj_v\ \vec{u} &= \left(\dfrac{\vec{u} \cdot \vec{v}}{\vec{v} \cdot \vec{v}}\right) \vec{v}\&= \left(\dfrac{9+6+0}{9+4+0}\right) \langle 3,2,0 \rangle = \dfrac{15}{13} \langle 3,2,0 \rangle\&= \ans{\left\langle \dfrac{45}{13},\dfrac{30}{13},0 \right\rangle}\end{aligned}projvu=(vvuv)v=(9+4+09+6+0)3,2,0=13153,2,0=1345,1330,0

For the next example, note that the work done by a force F\vec{F}F acting to move an object along a vector d\vec{d}d is given byW=Fd.W = \vec{F} \cdot \vec{d}.W=Fd.

Work Calculation

Find the work done in the figure below (think of pulling a cart with a rope).

Vectors, Dot Products, and Cross Products (5)

Solution

W=Fd=(15lb)(4ft)cos40=45.96ftlb\begin{aligned}W = \vec{F} \cdot \vec{d} &= (15 \ lb)(4\ ft) \cos 40^{\circ}\&= \ans{45.96 \ ft-lb}\end{aligned}W=Fd=(15lb)(4ft)cos40=45.96ftlb

Cross Product (Vector Product)

Application: Torque (Moments)

Vectors, Dot Products, and Cross Products (6)

Let's say you want to tighten a nut as shown above. To tighten it as much as possible, you have two options: you can push harder on the wrench, or you can get a longer wrench. Think about pushing a door open; pushing closer to the hinges is harder, but if you push at the end away from the hinges, it's much easier. When you apply a force to turn something like the nut or the door, you are creating a moment or torque around the point of rotation. This torque depends on the force you apply and how far from the rotation point you apply that force; the magnitude of the torque is the product of these two.

M=rF|\vec{M}| = |\vec{r}| |\vec{F}|M=rFwhere M\vec{M}M is the moment, rr⃗r is the moment arm, and F\vec{F}F is the force that is applied perpendicular to the moment arm.

What if the force is applied at an angle, though? Without going into all the details, it turns out that only the perpendicular part of the force contributes to the moment, and the moment can be calculated using the cross product of rr⃗r and F\vec{F}F:M=r×F\vec{M} = r⃗ × \vec{F}M=r×F

Finding the Cross Product

The cross product u×vu⃗ × v⃗u×v yields a third vector that is perpendicular to both uu⃗u and vv⃗v. The direction is specified by the right hand rule (RHR): if you start by pointing the fingers of your right hand along the first vector, then curl them in the direction of the second vector, your thumb will point in the direction of the cross product.

Vectors, Dot Products, and Cross Products (7)

Vectors, Dot Products, and Cross Products (8)

Vectors, Dot Products, and Cross Products (9)

The magnitude of the cross product isu×v=uvsinθ|u⃗ × v⃗| = |u⃗| |v⃗| \sin \thetau×v=uvsinθwhere θθθ is the angle between uu⃗u and vv⃗v. For instance,i^×j^=k^.\hat{i} × \hat{j} = \hat{k}.i^×j^=k^.

We rarely use that formula to find the cross product; instead, we find the determinant of a specific matrix. We'll leave most of the discussion of determinants for Linear Algebra; for now, we'll just cover what we need.

If u=u1,u2,u3\vec{u} = ⟨u_1,u_2,u_3⟩u=u1,u2,u3 and v=v1,v2,v3\vec{v} = ⟨v_1,v_2,v_3⟩v=v1,v2,v3, then u×vu⃗ × v⃗u×v is given by the following determinant:

u×v=i^j^k^u1u2u3v1v2v3\vec{u} \times \vec{v} = \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\u_1 & u_2 & u_3\v_1 & v_2 & v_3\end{vmatrix}u×v=i^u1v1j^u2v2k^u3v3

To evaluate this determinant, cross out the row and column that intersect at i^\hat{i}i^, then the ones that intersect at j^\hat{j}j^, and finally the ones that intersect at k^\hat{k}k^. Find the determinants of these smaller matrices (called minors) by multiplying along the diagonals and subtracting, as shown:

Vectors, Dot Products, and Cross Products (10)

This probably all looks very confusing right now, but you end up with this:

Vectors, Dot Products, and Cross Products (11)

Be sure to note the minus sign on the j^\hat{j}j^ minor. The first and third terms are positive, but this second term is negative. Also note thatu×v=v×u\vec{u} \times \vec{v}\ \cancel{=}\ \vec{v} \times \vec{u}u×v=v×u

More specifically, if you switch the order in the cross product, the result will have the same magnitude, but it will point in the opposite direction.

If u=2i^3j^+k^u⃗ = 2\hat{i}−3\hat{j}+\hat{k}u=2i^3j^+k^ and v=3i^+2j^+2k^v⃗ = 3\hat{i}+2\hat{j}+2\hat{k}v=3i^+2j^+2k^, find u×vu⃗ × v⃗u×v.

Solution

u×v=i^j^k^231322=(62)i^(43)j^+(4+9)k^=8i^j^+13k^\begin{aligned}\vec{u} \times \vec{v} &= \begin{vmatrix}\hat{i} & \hat{j} & \hat{k}\2 & -3 & 1\3 & 2 & 2\end{vmatrix}\&= (-6-2)\hat{i} - (4-3)\hat{j} + (4+9)\hat{k}\&= -8\hat{i} - \hat{j} + 13\hat{k}\end{aligned}u×v=i^23j^32k^12=(62)i^(43)j^+(4+9)k^=8i^j^+13k^

If u=2i^3j^+k^u⃗ = 2\hat{i}−3\hat{j}+\hat{k}u=2i^3j^+k^ and v=3i^+2j^+2k^v⃗ = 3\hat{i}+2\hat{j}+2\hat{k}v=3i^+2j^+2k^, find a vector that is orthogonal to both uu⃗u and vv⃗v.

Solution

Remember that the cross product of two vectors is orthogonal to both, and we already found u×vu⃗ × v⃗u×v. Therefore,w=8,1,13w⃗ = ⟨−8,−1,13⟩w=8,1,13or any multiple of ww⃗w will be orthogonal to both uu⃗u and vv⃗v.

Vectors, Dot Products, and Cross Products (2024)
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