Unit 6, lesson 79

Vector Projections and Work

How to split a vector into a part along a given direction and a part perpendicular to it, and how the part of a force along the motion gives the work it does, with its units and sign.

  • Theory part 157
  • Practice part 158
  • Vectors

Pull a sled with a rope that slants upward and only part of your effort moves the sled forward. The rest pulls straight up and does no work as the sled slides along the snow. Splitting a vector into the part along a direction and the part straight across it tells you how much of a force really helps, and the helpful part is exactly what goes into the work the force does.

Key terms

Scalar component

Definition Let v\mathbf{v} be a nonzero vector. The scalar component of u\mathbf{u} along v\mathbf{v} is

comp⁡vu=u⋅v∥v∥.\operatorname{comp}_{\mathbf{v}} \mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\lVert \mathbf{v} \rVert}.

If u≠0\mathbf{u} \ne \mathbf{0} and θ\theta is the angle between u\mathbf{u} and v\mathbf{v}, this equals ∥u∥cos⁡θ\lVert \mathbf{u} \rVert \cos \theta. It is positive when 0≤θ<π/20 \le \theta < \pi/2, zero when the vectors are orthogonal, and negative when π/2<θ≤π\pi/2 < \theta \le \pi. It depends only on the direction of v\mathbf{v}, not on its length. Some books call it the scalar projection.

In plain words Shine a light straight across the line through v\mathbf{v}. The scalar component is the length of the shadow of u\mathbf{u} on that line, counted positive if the shadow points the way v\mathbf{v} points and negative if it points back.

Example Take u=⟨7,1⟩\mathbf{u} = \langle 7, 1 \rangle and v=⟨1,1⟩\mathbf{v} = \langle 1, 1 \rangle. Then u⋅v=8\mathbf{u} \cdot \mathbf{v} = 8 and ∥v∥=2\lVert \mathbf{v} \rVert = \sqrt{2}, so

comp⁡vu=82=42≈5.66.\operatorname{comp}_{\mathbf{v}} \mathbf{u} = \frac{8}{\sqrt{2}} = 4\sqrt{2} \approx 5.66.

For ⟨−3,1⟩\langle -3, 1 \rangle along the same v\mathbf{v} the dot product is −2-2, and the scalar component is −2/2=−2-2/\sqrt{2} = -\sqrt{2}, a shadow pointing backward.

Vector projection onto a nonzero vector

Definition Let v\mathbf{v} be a nonzero vector. The vector projection of u\mathbf{u} onto v\mathbf{v} is

proj⁡vu=u⋅v∥v∥2 v=(comp⁡vu)v∥v∥.\begin{aligned} \operatorname{proj}_{\mathbf{v}} \mathbf{u} &= \frac{\mathbf{u} \cdot \mathbf{v}}{\lVert \mathbf{v} \rVert^2}\,\mathbf{v} \\[0.8em] &= \left( \operatorname{comp}_{\mathbf{v}} \mathbf{u} \right) \frac{\mathbf{v}}{\lVert \mathbf{v} \rVert}. \end{aligned}

It is a scalar multiple of v\mathbf{v} with length ∣comp⁡vu∣\lvert \operatorname{comp}_{\mathbf{v}} \mathbf{u} \rvert. It points along v\mathbf{v} when the scalar component is positive and against v\mathbf{v} when it is negative. Projection onto the zero vector is not defined, because the formula would divide by 00.

In plain words The scalar component says how long the shadow is. The vector projection is the shadow itself, an arrow lying on the line of v\mathbf{v}.

Example With u=⟨7,1⟩\mathbf{u} = \langle 7, 1 \rangle and v=⟨1,1⟩\mathbf{v} = \langle 1, 1 \rangle as before, ∥v∥2=2\lVert \mathbf{v} \rVert^2 = 2 and

proj⁡vu=82⟨1,1⟩=⟨4,4⟩.\operatorname{proj}_{\mathbf{v}} \mathbf{u} = \frac{8}{2} \langle 1, 1 \rangle = \langle 4, 4 \rangle.

Its length is 32=42\sqrt{32} = 4\sqrt{2}, the scalar component found above.

Orthogonal decomposition

Definition Let v\mathbf{v} be a nonzero vector. Every vector u\mathbf{u} can be written in exactly one way as

u=u∥+u⊥,\mathbf{u} = \mathbf{u}_{\parallel} + \mathbf{u}_{\perp},

where u∥\mathbf{u}_{\parallel} is a scalar multiple of v\mathbf{v} and u⊥\mathbf{u}_{\perp} is orthogonal to v\mathbf{v}. The two parts are

u∥=proj⁡vuandu⊥=u−proj⁡vu.\begin{aligned} \mathbf{u}_{\parallel} &= \operatorname{proj}_{\mathbf{v}} \mathbf{u} \quad \text{and} \\[0.4em] \mathbf{u}_{\perp} &= \mathbf{u} - \operatorname{proj}_{\mathbf{v}} \mathbf{u}. \end{aligned}

The second part is orthogonal to v\mathbf{v} because

(u−u⋅v∥v∥2 v)⋅v=u⋅v−u⋅v∥v∥2 ∥v∥2=0.\begin{aligned} &\left( \mathbf{u} - \frac{\mathbf{u} \cdot \mathbf{v}}{\lVert \mathbf{v} \rVert^2}\,\mathbf{v} \right) \cdot \mathbf{v} \\[0.8em] &\quad = \mathbf{u} \cdot \mathbf{v} - \frac{\mathbf{u} \cdot \mathbf{v}}{\lVert \mathbf{v} \rVert^2}\,\lVert \mathbf{v} \rVert^2 = 0. \end{aligned}

The split is unique, since dotting u=cv+w\mathbf{u} = c\mathbf{v} + \mathbf{w} with v\mathbf{v}, where w⋅v=0\mathbf{w} \cdot \mathbf{v} = 0, gives u⋅v=c∥v∥2\mathbf{u} \cdot \mathbf{v} = c \lVert \mathbf{v} \rVert^2 and forces the value of cc. The parts are orthogonal, so the Pythagorean theorem gives

∥u∥2=∥u∥∥2+∥u⊥∥2.\lVert \mathbf{u} \rVert^2 = \lVert \mathbf{u}_{\parallel} \rVert^2 + \lVert \mathbf{u}_{\perp} \rVert^2.

In plain words Any vector splits into a part along a chosen direction and a part straight across it. The two parts are the legs of a right triangle whose hypotenuse is the original vector.

Example A 1010 kilogram box sits on a ramp that rises to the right at 30∘30^\circ. Gravity pulls it straight down with a force of about 9898 newtons, so F=⟨0,−98⟩\mathbf{F} = \langle 0, -98 \rangle. The unit vector pointing up the ramp is

v=⟨cos⁡30∘,sin⁡30∘⟩=⟨32,12⟩.\begin{aligned} \mathbf{v} &= \langle \cos 30^\circ, \sin 30^\circ \rangle \\[0.4em] &= \left\langle \frac{\sqrt{3}}{2}, \frac{1}{2} \right\rangle. \end{aligned}

Since F⋅v=−49\mathbf{F} \cdot \mathbf{v} = -49 and ∥v∥=1\lVert \mathbf{v} \rVert = 1,

F∥=−49 v≈⟨−42.44,  −24.5⟩,F⊥=F−F∥≈⟨42.44,  −73.5⟩.\begin{aligned} \mathbf{F}_{\parallel} &= -49\,\mathbf{v} \approx \langle -42.44, \; -24.5 \rangle, \\[0.4em] \mathbf{F}_{\perp} &= \mathbf{F} - \mathbf{F}_{\parallel} \approx \langle 42.44, \; -73.5 \rangle. \end{aligned}

The first part, 4949 newtons, pulls the box down the slope. The second, 493≈84.949\sqrt{3} \approx 84.9 newtons, presses it into the ramp. As a check, the exact perpendicular part is F⊥=⟨493/2, −147/2⟩\mathbf{F}_{\perp} = \langle 49\sqrt{3}/2, \, -147/2 \rangle, and

F⊥⋅v=4932⋅32−1472⋅12=1474−1474=0.\begin{aligned} \mathbf{F}_{\perp} \cdot \mathbf{v} &= \frac{49\sqrt{3}}{2} \cdot \frac{\sqrt{3}}{2} - \frac{147}{2} \cdot \frac{1}{2} \\[0.8em] &= \frac{147}{4} - \frac{147}{4} = 0. \end{aligned}

Work done by a constant force

Definition Suppose a constant force F\mathbf{F} acts on an object while the object moves in a straight line from PP to QQ, with displacement vector D=PQ→\mathbf{D} = \overrightarrow{PQ}. The work done by F\mathbf{F} is

W=F⋅D=∥F∥ ∥D∥cos⁡θ,W = \mathbf{F} \cdot \mathbf{D} = \lVert \mathbf{F} \rVert \, \lVert \mathbf{D} \rVert \cos \theta,

where θ\theta is the angle between F\mathbf{F} and D\mathbf{D} when both are nonzero. Equivalently, for D≠0\mathbf{D} \ne \mathbf{0}, the work is the scalar component of F\mathbf{F} along D\mathbf{D} times the distance ∥D∥\lVert \mathbf{D} \rVert. When the force points along the motion this is force times distance. Work is a scalar. A force that changes along the way needs the integrals of calculus.

In plain words Only the part of the force that points along the motion does work. The part straight across the motion neither helps nor hinders.

Example A rope pulls a sled 2020 meters across level snow with a constant force of 5050 newtons at 30∘30^\circ above the horizontal. With F=⟨253, 25⟩\mathbf{F} = \langle 25\sqrt{3}, \, 25 \rangle and D=⟨20,0⟩\mathbf{D} = \langle 20, 0 \rangle,

W=253⋅20+25⋅0=5003≈866 joules.\begin{aligned} W &= 25\sqrt{3} \cdot 20 + 25 \cdot 0 \\[0.4em] &= 500\sqrt{3} \approx 866 \text{ joules.} \end{aligned}

The angle form gives the same number, 50⋅20⋅cos⁡30∘=500350 \cdot 20 \cdot \cos 30^\circ = 500\sqrt{3}. The upward 2525 newtons contribute nothing.

Units of work

Definition Work is measured in units of force times units of distance. With force in newtons and distance in meters the unit is the joule,

1 joule=1 newton⋅1 meter.1 \text{ joule} = 1 \text{ newton} \cdot 1 \text{ meter}.

With force in pounds and distance in feet the unit is the foot-pound, and 11 foot-pound is about 1.3561.356 joules. Before taking the dot product, express the force and the displacement in one consistent system of units. A scalar component or a vector projection carries the units of the vector being projected.

In plain words Newtons times meters give joules, and pounds times feet give foot-pounds. Convert first and multiply second, because a product of mixed units is easy to misread.

Example A force F=⟨12,5⟩\mathbf{F} = \langle 12, 5 \rangle newtons moves a cart through the displacement D=⟨150,0⟩\mathbf{D} = \langle 150, 0 \rangle centimeters. Converting first gives D=⟨1.5,0⟩\mathbf{D} = \langle 1.5, 0 \rangle meters and

W=12⋅1.5+5⋅0=18 joules.W = 12 \cdot 1.5 + 5 \cdot 0 = 18 \text{ joules.}

Using centimeters directly gives 18001800, which counts newton-centimeters and is a hundred times too large as a number of joules. In customary units, lifting a 1010 pound bag 33 feet straight up at a steady speed takes 10⋅3=3010 \cdot 3 = 30 foot-pounds of work.

Sign of work

Definition For a nonzero force F\mathbf{F} and a nonzero displacement D\mathbf{D} with angle θ\theta between them, the work W=∥F∥ ∥D∥cos⁡θW = \lVert \mathbf{F} \rVert \, \lVert \mathbf{D} \rVert \cos \theta has the sign of cos⁡θ\cos \theta. It is positive when 0∘≤θ<90∘0^\circ \le \theta < 90^\circ, zero when θ=90∘\theta = 90^\circ, and negative when 90∘<θ≤180∘90^\circ < \theta \le 180^\circ. Positive work means the force passes energy to the moving object, and negative work means it takes energy from the object. The work is also zero when the force or the displacement is the zero vector.

In plain words A force that pushes with the motion does positive work, a force that pushes against it does negative work, and a force that pushes straight sideways does none.

Example Four forces act on the sled from before while it moves through D=⟨20,0⟩\mathbf{D} = \langle 20, 0 \rangle meters. The sled weighs 9898 newtons, friction from the snow pulls back with 1515 newtons, and the snow pushes up with 7373 newtons because the rope carries the other 2525.

ForceVector in newtonsWork in joules
Rope⟨253,25⟩\langle 25\sqrt{3}, 25 \rangle5003≈866500\sqrt{3} \approx 866
Friction⟨−15,0⟩\langle -15, 0 \rangle−300-300
Gravity⟨0,−98⟩\langle 0, -98 \rangle00
Snow pushing up⟨0,73⟩\langle 0, 73 \rangle00

The rope does positive work, friction does negative work, and gravity and the snow do none, since they act straight across the motion. The total work is about 866−300=566866 - 300 = 566 joules.

Check yourself

  1. For u=⟨5,5⟩\mathbf{u} = \langle 5, 5 \rangle and v=⟨3,1⟩\mathbf{v} = \langle 3, 1 \rangle, find the scalar component of u\mathbf{u} along v\mathbf{v}, the vector projection of u\mathbf{u} onto v\mathbf{v}, and the orthogonal decomposition of u\mathbf{u}.
Show the answer

Here u⋅v=15+5=20\mathbf{u} \cdot \mathbf{v} = 15 + 5 = 20 and ∥v∥2=10\lVert \mathbf{v} \rVert^2 = 10. The scalar component is 20/10=21020/\sqrt{10} = 2\sqrt{10}, and the projection is 2⟨3,1⟩=⟨6,2⟩2\langle 3, 1 \rangle = \langle 6, 2 \rangle. So

⟨5,5⟩=⟨6,2⟩+⟨−1,3⟩,\langle 5, 5 \rangle = \langle 6, 2 \rangle + \langle -1, 3 \rangle,

and the second part is orthogonal to v\mathbf{v} because ⟨−1,3⟩⋅⟨3,1⟩=−3+3=0\langle -1, 3 \rangle \cdot \langle 3, 1 \rangle = -3 + 3 = 0.

  1. A lawn mower is pushed 88 meters across flat ground by a constant force of 4040 newtons along the handle, which points 60∘60^\circ below the horizontal. How much work does the force do?
Show the answer

The mower moves horizontally and the force points 60∘60^\circ below the horizontal, so the angle between the force and the motion is 60∘60^\circ and

W=40⋅8⋅cos⁡60∘=160 joules.W = 40 \cdot 8 \cdot \cos 60^\circ = 160 \text{ joules.}
  1. A constant force F=⟨2,−6⟩\mathbf{F} = \langle 2, -6 \rangle newtons acts on an object that moves in a straight line from P(1,1)P(1, 1) to Q(4,3)Q(4, 3), with coordinates in meters. Find the work done by F\mathbf{F} and say what its sign means.
Show the answer

The displacement is D=⟨3,2⟩\mathbf{D} = \langle 3, 2 \rangle, so W=2⋅3+(−6)⋅2=−6W = 2 \cdot 3 + (-6) \cdot 2 = -6 joules. The work is negative, so the angle between the force and the motion is obtuse (about 105∘105^\circ) and the force works against the motion.

How this lesson connects