Precalculus

Vector and scalar projection

A force pulls at a slant, but the object can only move along one line. How much of the pull actually points along that line, and how much work does it do?

You need

The dot product and the angle between vectorsPrecalculus

Where you are 3 in The dot product

Trigonometry7

Polar coordinates and vectors10

The dot product3

The dot product and the angle between vectors Orthogonal vectors

Vector and scalar projection

Why this exists

A rope pulls a raft at a slant to the riverbank, but the raft can only travel along the river. A person pushes a loaded dolly partly sideways, yet the dolly rolls in a straight line. In both cases only part of the force goes along the motion, and the question is how big that part is.

The dot product already holds the answer. It equals \|\mathbf{u}\|\,\|\mathbf{v}\|\cos\theta, and \|\mathbf{u}\|\cos\theta is the length of the side next to \theta in a right triangle with hypotenuse \|\mathbf{u}\|. Divide the dot product by one length instead of two and that side appears: the part of \mathbf{u} that runs along \mathbf{v}. This lesson names that number, turns it into a vector, uses it to split any vector into two perpendicular pieces, and uses it to measure the work a force does.

Definition

Let \mathbf{u} be any vector and \mathbf{v} a nonzero vector, with \theta the angle between them when \mathbf{u} is nonzero.

The scalar projection of \mathbf{u} onto \mathbf{v}, also called the component of \mathbf{u} in the direction of \mathbf{v}, is the number

\operatorname{comp}_{\mathbf{v}}\mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|}.

It is the signed length of the shadow of \mathbf{u} on the line of \mathbf{v}: positive when the shadow points the same way as \mathbf{v}, negative when it points the opposite way, zero when \mathbf{u} is perpendicular to \mathbf{v}.

The vector projection of \mathbf{u} onto \mathbf{v} is that shadow as an arrow:

\operatorname{proj}_{\mathbf{v}}\mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2}\,\mathbf{v}.

The work done by a constant force \mathbf{F} on an object that moves along the straight displacement \mathbf{d} is the dot product

W = \mathbf{F} \cdot \mathbf{d}.

With force in newtons and distance in meters, work is in newton-meters, called joules (\text{J}). With force in pounds and distance in feet, it is in foot-pounds (\text{ft-lb}).

Intuition

Figure 1The projection is the shadow \mathbf{u} casts on the line of \mathbf{v} when light falls at right angles to that line. Its signed length is \operatorname{comp}_{\mathbf{v}}\mathbf{u}, negative when the shadow falls behind the tail.

Drop a line from the head of \mathbf{u} to the line of \mathbf{v}, meeting it at a right angle. The arrow from the tail to that foot is \operatorname{proj}_{\mathbf{v}}\mathbf{u}. When \theta is acute the foot lands on the side \mathbf{v} points to. When \theta is obtuse it lands behind the tail, so the shadow points against \mathbf{v} and its signed length is negative. That sign is the sign of \mathbf{u} \cdot \mathbf{v}, since \|\mathbf{v}\| is positive.

Only the direction of \mathbf{v} matters. Doubling \mathbf{v} does not move the line it lies on, so it cannot change the shadow; the formulas divide by \|\mathbf{v}\| and \|\mathbf{v}\|^2 exactly to cancel the length of \mathbf{v}.

Work follows the same picture. A force at a slant to the path does two things: the shadow along the path pushes the object forward, and the perpendicular part only presses sideways. Work counts the first part, times the distance moved.

Results

Scalar projection. For \mathbf{v} \ne \mathbf{0} and \mathbf{u} \ne \mathbf{0} at angle \theta,

\operatorname{comp}_{\mathbf{v}}\mathbf{u} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|} = \|\mathbf{u}\|\cos\theta.

Vector projection. It is the scalar projection times the unit vector in the direction of \mathbf{v}:

\begin{aligned} \operatorname{proj}_{\mathbf{v}}\mathbf{u} &= \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2}\,\mathbf{v} \\ &= \left(\operatorname{comp}_{\mathbf{v}}\mathbf{u}\right)\frac{\mathbf{v}}{\|\mathbf{v}\|}, \end{aligned}

so its length is \left|\operatorname{comp}_{\mathbf{v}}\mathbf{u}\right|.

Splitting a vector. Every vector \mathbf{u} is the sum of a part parallel to \mathbf{v} and a part orthogonal to \mathbf{v},

\begin{aligned} \mathbf{u} &= \operatorname{proj}_{\mathbf{v}}\mathbf{u} + \mathbf{w}, \\ \mathbf{w} &= \mathbf{u} - \operatorname{proj}_{\mathbf{v}}\mathbf{u}, \quad \mathbf{w} \cdot \mathbf{v} = 0, \end{aligned}

and this split is the only one of its kind. Because the two parts are perpendicular, \|\mathbf{u}\|^2 = \|\operatorname{proj}_{\mathbf{v}}\mathbf{u}\|^2 + \|\mathbf{w}\|^2.

Work. For a constant force \mathbf{F} and a straight displacement \mathbf{d} at angle \theta,

\begin{aligned} W &= \mathbf{F} \cdot \mathbf{d} = \|\mathbf{F}\|\,\|\mathbf{d}\|\cos\theta \\ &= \left(\operatorname{comp}_{\mathbf{d}}\mathbf{F}\right)\|\mathbf{d}\|. \end{aligned}

Why it is true

The scalar projection. For nonzero \mathbf{u}, the cosine formula \mathbf{u} \cdot \mathbf{v} = \|\mathbf{u}\|\,\|\mathbf{v}\|\cos\theta divided by \|\mathbf{v}\| gives \displaystyle \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|} = \|\mathbf{u}\|\cos\theta. When \theta is acute, the dashed drop in Figure 1 makes a right triangle with hypotenuse \|\mathbf{u}\| and the shadow as the side next to \theta, so the shadow has length \|\mathbf{u}\|\cos\theta. When \theta is obtuse, the triangle sits behind the tail with angle 180^\circ - \theta at the tail, so the shadow has length \|\mathbf{u}\|\cos(180^\circ - \theta) = -\|\mathbf{u}\|\cos\theta and points against \mathbf{v}; the signed length is \|\mathbf{u}\|\cos\theta again.

The vector projection. The unit vector \displaystyle \frac{\mathbf{v}}{\|\mathbf{v}\|} has length 1 and the direction of \mathbf{v}. Scaling it by the signed length of the shadow gives an arrow of that length, pointing along \mathbf{v} when the number is positive and against it when negative, which is the shadow. Multiplying out, \displaystyle \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|} \cdot \frac{\mathbf{v}}{\|\mathbf{v}\|} = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2}\,\mathbf{v}.

The split. Write \displaystyle k = \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} and \mathbf{w} = \mathbf{u} - k\mathbf{v}. By the distributive law and \mathbf{v} \cdot \mathbf{v} = \|\mathbf{v}\|^2,

\begin{aligned} \mathbf{w} \cdot \mathbf{v} &= \mathbf{u} \cdot \mathbf{v} - k\,\|\mathbf{v}\|^2 \\ &= \mathbf{u} \cdot \mathbf{v} - \mathbf{u} \cdot \mathbf{v} = 0. \end{aligned}

So \mathbf{w} is orthogonal to \mathbf{v}, and \mathbf{u} = k\mathbf{v} + \mathbf{w} by construction. For uniqueness, suppose \mathbf{u} = c\mathbf{v} + \mathbf{z} with \mathbf{z} \cdot \mathbf{v} = 0. Dotting both sides with \mathbf{v} gives \mathbf{u} \cdot \mathbf{v} = c\,\|\mathbf{v}\|^2, so c = k and then \mathbf{z} = \mathbf{w}. The length equation follows by expanding (k\mathbf{v} + \mathbf{w}) \cdot (k\mathbf{v} + \mathbf{w}): the cross terms are 2k\,\mathbf{v} \cdot \mathbf{w} = 0.

Work. In physics, the work a constant force does on an object moved along a straight line is the part of the force in the direction of motion times the distance moved. That part is \displaystyle \operatorname{comp}_{\mathbf{d}}\mathbf{F} = \frac{\mathbf{F} \cdot \mathbf{d}}{\|\mathbf{d}\|}, and multiplying by \|\mathbf{d}\| leaves \mathbf{F} \cdot \mathbf{d}.

Worked examples

Example 1

Both projections from components

Find \operatorname{comp}_{\mathbf{v}}\mathbf{u} and \operatorname{proj}_{\mathbf{v}}\mathbf{u} for \mathbf{u} = \langle 9, 2 \rangle and \mathbf{v} = \langle 8, 15 \rangle.

Solution

Both formulas need \mathbf{u} \cdot \mathbf{v} and the length of \mathbf{v}:

\begin{aligned} \mathbf{u} \cdot \mathbf{v} &= (9)(8) + (2)(15) = 72 + 30 = 102, \\ \|\mathbf{v}\| &= \sqrt{64 + 225} = \sqrt{289} = 17. \end{aligned}

So \displaystyle \operatorname{comp}_{\mathbf{v}}\mathbf{u} = \frac{102}{17} = 6. The vector projection divides by \|\mathbf{v}\|^2 = 289:

\begin{aligned} \operatorname{proj}_{\mathbf{v}}\mathbf{u} &= \frac{102}{289}\langle 8, 15 \rangle = \frac{6}{17}\langle 8, 15 \rangle \\ &= \left\langle \frac{48}{17}, \frac{90}{17} \right\rangle \approx \langle 2.82, 5.29 \rangle. \end{aligned}

Check: this arrow is \displaystyle \frac{6}{17} of \mathbf{v}, whose length is 17, so its length is 6, the scalar projection. It is shorter than \displaystyle \|\mathbf{u}\| = \sqrt{85} \approx 9.22, as a shadow must be.

Example 2

A negative scalar projection

Find \operatorname{comp}_{\mathbf{v}}\mathbf{u} and \operatorname{proj}_{\mathbf{v}}\mathbf{u} for \mathbf{u} = \langle -7, 4 \rangle and \mathbf{v} = \langle 1, -2 \rangle.

Solution

Here \mathbf{u} \cdot \mathbf{v} = -7 - 8 = -15 and \displaystyle \|\mathbf{v}\| = \sqrt{5}, so

\operatorname{comp}_{\mathbf{v}}\mathbf{u} = \frac{-15}{\sqrt{5}} = -3\sqrt{5} \approx -6.71.

The minus sign says the angle is obtuse and the shadow points against \mathbf{v}. With \|\mathbf{v}\|^2 = 5,

\begin{aligned} \operatorname{proj}_{\mathbf{v}}\mathbf{u} &= \frac{-15}{5}\langle 1, -2 \rangle \\ &= -3\langle 1, -2 \rangle = \langle -3, 6 \rangle. \end{aligned}

It is a negative multiple of \mathbf{v}, so it does point the opposite way, and its length is \displaystyle 3\sqrt{5}, the size of the scalar projection.

Example 3

Splitting a vector into perpendicular parts

Write \mathbf{u} = \langle 1, 7 \rangle as a vector parallel to \mathbf{v} = \langle 10, 5 \rangle plus a vector orthogonal to \mathbf{v}, and check the second part.

Solution

Figure 2Example 3. \mathbf{u} = \langle 1, 7 \rangle splits into \operatorname{proj}_{\mathbf{v}}\mathbf{u} = \langle 3.6, 1.8 \rangle along \mathbf{v} and \mathbf{w} = \langle -2.6, 5.2 \rangle at right angles to it.

The parallel part is the projection. Since \mathbf{u} \cdot \mathbf{v} = 10 + 35 = 45 and \|\mathbf{v}\|^2 = 100 + 25 = 125,

\begin{aligned} \operatorname{proj}_{\mathbf{v}}\mathbf{u} &= \frac{45}{125}\langle 10, 5 \rangle = \frac{9}{25}\langle 10, 5 \rangle \\ &= \langle 3.6, 1.8 \rangle. \end{aligned}

The rest of \mathbf{u} is what is left after removing it:

\begin{aligned} \mathbf{w} &= \mathbf{u} - \operatorname{proj}_{\mathbf{v}}\mathbf{u} = \langle 1 - 3.6, 7 - 1.8 \rangle \\ &= \langle -2.6, 5.2 \rangle. \end{aligned}

Check: \mathbf{w} \cdot \mathbf{v} = (-2.6)(10) + (5.2)(5) = -26 + 26 = 0, so \mathbf{w} is orthogonal to \mathbf{v}. The lengths agree with the right triangle in the figure: 3.6^2 + 1.8^2 = 16.2 and 2.6^2 + 5.2^2 = 33.8, which add to 50 = 1^2 + 7^2 = \|\mathbf{u}\|^2.

Example 4

From a length and an angle

A vector \mathbf{u} has length 8 and makes an angle of 60^\circ with \mathbf{v} = \langle -15, 20 \rangle. The components of \mathbf{u} are not given. Find \operatorname{comp}_{\mathbf{v}}\mathbf{u} and \operatorname{proj}_{\mathbf{v}}\mathbf{u}.

Solution

Without components, use the cosine form: \operatorname{comp}_{\mathbf{v}}\mathbf{u} = \|\mathbf{u}\|\cos\theta = 8\cos 60^\circ = 4. The vector projection is that number times the unit vector along \mathbf{v}, and \displaystyle \|\mathbf{v}\| = \sqrt{225 + 400} = 25:

\begin{aligned} \operatorname{proj}_{\mathbf{v}}\mathbf{u} &= 4 \cdot \frac{\langle -15, 20 \rangle}{25} \\ &= \left\langle -\frac{12}{5}, \frac{16}{5} \right\rangle = \langle -2.4, 3.2 \rangle. \end{aligned}

Its length is \displaystyle \sqrt{5.76 + 10.24} = 4, as it should be. The answer is a definite vector even though \mathbf{u} itself could be either of two arrows at 60^\circ to \mathbf{v}: both cast the same shadow.

Example 5

Work from components

Fiifi pushes a loaded dolly across a level floor. On a floor plan marked in meters the dolly moves in a straight line from (-3, 1) to (18, 21), while he pushes with the constant force \mathbf{F} = \langle 30, -4 \rangle newtons. How much work does his push do, and how much of the force acts along the path?

Solution

The displacement runs from start to finish: \mathbf{d} = \langle 18 - (-3), 21 - 1 \rangle = \langle 21, 20 \rangle. Work is the dot product:

\begin{aligned} W = \mathbf{F} \cdot \mathbf{d} &= (30)(21) + (-4)(20) \\ &= 630 - 80 = 550 \text{ J}. \end{aligned}

The path has length \displaystyle \|\mathbf{d}\| = \sqrt{441 + 400} = 29 meters, so the part of the force along it is \displaystyle \operatorname{comp}_{\mathbf{d}}\mathbf{F} = \frac{550}{29} \approx 18.97 newtons. His whole push is \displaystyle \|\mathbf{F}\| = \sqrt{916} \approx 30.27 newtons; the rest presses sideways and does no work.

Example 6

A rope at an angle

Mansa walks along a straight riverbank towing a raft with a rope. The rope pulls with a constant 18 pounds at 35^\circ to the raft’s path, and the raft moves 40 feet straight along the river. Find the work done by the rope.

Solution

Figure 3Example 6. Only the part of the rope’s pull along the path, 18\cos 35^\circ pounds, does work. The arrows are not drawn to one scale.

Only the lengths and the angle are known, so use the cosine form of the dot product:

\begin{aligned} W &= \|\mathbf{F}\|\,\|\mathbf{d}\|\cos\theta = (18)(40)\cos 35^\circ \\ &= 720\cos 35^\circ \approx 589.8 \text{ ft-lb}. \end{aligned}

Read the other way, the rope’s useful pull is \operatorname{comp}_{\mathbf{d}}\mathbf{F} = 18\cos 35^\circ \approx 14.74 pounds, and 14.74 \times 40 \approx 589.8. The other part, 18\sin 35^\circ \approx 10.32 pounds, pulls the raft toward the bank and does no work along the river. A steeper rope angle would shrink \cos\theta and the work with it.

Boundaries

\mathbf{v} must be nonzero. Both formulas divide by \|\mathbf{v}\|, and the zero vector has no direction to cast a shadow on. The vector \mathbf{u} may be zero; its projections are then 0 and \mathbf{0}.
The order matters. \operatorname{proj}_{\mathbf{v}}\mathbf{u} lies along \mathbf{v} and \operatorname{proj}_{\mathbf{u}}\mathbf{v} lies along \mathbf{u}, so for nonzero vectors they are different unless \mathbf{u} = \mathbf{v}, or \mathbf{u} and \mathbf{v} are orthogonal and both projections are \mathbf{0}.
The scalar projection is signed, not a length. It can be negative; the length of the vector projection is its absolute value.
W = \mathbf{F} \cdot \mathbf{d} needs a constant force and a straight path. A force that changes along the way, or a curved path, needs the path cut into short pieces and their works added, which is Work as an integral.
Work here is done by one force. Other forces on the same object (gravity, friction) do their own work, each computed with its own dot product.

Where this goes next

The orthogonal part \mathbf{w} gives the shortest route from a point to a line, which is Distance from a point to a line in vector form. Repeating the split, removing from each new vector its projections onto the earlier ones, is the The Gram-Schmidt process, which builds perpendicular directions for any space, and projecting onto a whole subspace leads to Orthogonality and least squares, the idea behind fitting a line to data. Work with a force that changes along the path becomes Work as an integral. The parent idea, The dot product, gathers these together.

Exercises

Scalar and vector projections

Find \operatorname{comp}_{\mathbf{v}}\mathbf{u} and \operatorname{proj}_{\mathbf{v}}\mathbf{u} for \mathbf{u} = \langle -5, 8 \rangle and \mathbf{v} = \langle 0, -3 \rangle.
Find \operatorname{comp}_{\mathbf{v}}\mathbf{u} and \operatorname{proj}_{\mathbf{v}}\mathbf{u} for \mathbf{u} = \langle -1, 8 \rangle and \mathbf{v} = \langle -12, 9 \rangle.
Find \operatorname{comp}_{\mathbf{v}}\mathbf{u} and \operatorname{proj}_{\mathbf{v}}\mathbf{u} for \mathbf{u} = \langle 5, -9 \rangle and \mathbf{v} = \langle -2, 2 \rangle.
Find the scalar and vector projections of 6\mathbf{i} + \mathbf{j} onto \mathbf{i} - 3\mathbf{j}.
A vector of length 10 makes an angle of 135^\circ with \mathbf{v}. Find its scalar projection onto \mathbf{v}.
A vector \mathbf{u} of length 9 makes an angle of 40^\circ with \mathbf{v} = \langle -20, 21 \rangle. Find \operatorname{comp}_{\mathbf{v}}\mathbf{u} and \operatorname{proj}_{\mathbf{v}}\mathbf{u} to two decimal places.

Splitting a vector

For Exercises 7 to 10, write \mathbf{u} as \operatorname{proj}_{\mathbf{v}}\mathbf{u} + \mathbf{w} with \mathbf{w} orthogonal to \mathbf{v}, and check that \mathbf{w} \cdot \mathbf{v} = 0.

\mathbf{u} = \langle 3, 9 \rangle, \mathbf{v} = \langle -6, 3 \rangle.
\mathbf{u} = \langle -5, 7 \rangle, \mathbf{v} = \langle 6, 5 \rangle.
\mathbf{u} = \langle 7, 7 \rangle, \mathbf{v} = \langle 9, -9 \rangle.
\mathbf{u} = \langle 15, -10 \rangle, \mathbf{v} = \langle -9, 6 \rangle.

Work and other uses

A constant force \mathbf{F} = \langle 20, -6 \rangle newtons moves an object in a straight line from (-2, 3) to (13, 7), with distances in meters. Find the work done.
A constant force of 50 newtons acts at 30^\circ to the direction of motion while an object moves 8 meters. Find the exact work and its value to one decimal place.
Kojo pulls a rickshaw 150 feet along a straight level street, pulling with a constant 70 pounds at 15^\circ above the street. Find the work he does, to the nearest foot-pound.
A constant force of 30 pounds acts at 120^\circ to the direction of motion while an object moves 10 feet. Find the work done and say what its sign means.
Mansa must do 900 joules of work dragging a load 15 meters, pulling at 20^\circ to the direction of motion. What constant force, to two decimal places, does she need?
A box slides in a straight line from (0, 3) to (8, 0), distances in meters. Its weight is the constant force \mathbf{F} = \langle 0, -49 \rangle newtons. Find the work its weight does.
Find the distance from the point P(8, 1) to the line through the origin with direction \mathbf{v} = \langle -3, 9 \rangle, using the part of \overrightarrow{OP} orthogonal to \mathbf{v}.

Reasoning

Prove that for k \ne 0, \operatorname{proj}_{k\mathbf{v}}\mathbf{u} = \operatorname{proj}_{\mathbf{v}}\mathbf{u}, and that \operatorname{comp}_{k\mathbf{v}}\mathbf{u} equals \operatorname{comp}_{\mathbf{v}}\mathbf{u} when k > 0 and its negative when k < 0. Explain the difference in words.
For \mathbf{u} = \langle 4, 8 \rangle and \mathbf{v} = \langle 10, 3 \rangle, find \operatorname{proj}_{\mathbf{v}}\mathbf{u} and \operatorname{proj}_{\mathbf{u}}\mathbf{v}, and explain why they could not have been equal.
Explain why \left|\operatorname{comp}_{\mathbf{v}}\mathbf{u}\right| \le \|\mathbf{u}\|, and say when the two sides are equal.
Fiifi carries a box across a level room at a constant height, holding it up with a vertical force. Explain why, by W = \mathbf{F} \cdot \mathbf{d}, his holding force does no work on the box.
Kojo says \operatorname{proj}_{\mathbf{v}}\mathbf{u} is always a positive multiple of \mathbf{v}. Say when he is right, and give an example where he is wrong.

Answers

\operatorname{comp}_{\mathbf{v}}\mathbf{u} = -8 and \operatorname{proj}_{\mathbf{v}}\mathbf{u} = \langle 0, 8 \rangle. \mathbf{u} \cdot \mathbf{v} = 0 - 24 = -24 and \|\mathbf{v}\| = 3, so the scalar projection is -8; the vector is \displaystyle \frac{-24}{9}\langle 0, -3 \rangle. It is the vertical part of \mathbf{u}, which points up while \mathbf{v} points down.
\displaystyle \operatorname{comp}_{\mathbf{v}}\mathbf{u} = \frac{28}{5} and \displaystyle \operatorname{proj}_{\mathbf{v}}\mathbf{u} = \left\langle -\frac{112}{25}, \frac{84}{25} \right\rangle. \mathbf{u} \cdot \mathbf{v} = 12 + 72 = 84 and \|\mathbf{v}\| = 15; the vector is \displaystyle \frac{84}{225}\langle -12, 9 \rangle = \frac{28}{75}\langle -12, 9 \rangle.
\displaystyle \operatorname{comp}_{\mathbf{v}}\mathbf{u} = -7\sqrt{2} \approx -9.90 and \operatorname{proj}_{\mathbf{v}}\mathbf{u} = \langle 7, -7 \rangle. \mathbf{u} \cdot \mathbf{v} = -10 - 18 = -28 and \displaystyle \|\mathbf{v}\| = 2\sqrt{2}; the vector is \displaystyle \frac{-28}{8}\langle -2, 2 \rangle.
Scalar \displaystyle \frac{3\sqrt{10}}{10} \approx 0.95; vector \displaystyle \frac{3}{10}\mathbf{i} - \frac{9}{10}\mathbf{j}. The dot product is 6 - 3 = 3 and the length of \mathbf{i} - 3\mathbf{j} is \displaystyle \sqrt{10}, so the vector is \displaystyle \frac{3}{10}(\mathbf{i} - 3\mathbf{j}).
\displaystyle -5\sqrt{2} \approx -7.07. \displaystyle 10\cos 135^\circ = 10\left(-\frac{\sqrt{2}}{2}\right).
\operatorname{comp}_{\mathbf{v}}\mathbf{u} \approx 6.89 and \operatorname{proj}_{\mathbf{v}}\mathbf{u} \approx \langle -4.75, 4.99 \rangle. 9\cos 40^\circ \approx 6.894, \displaystyle \|\mathbf{v}\| = \sqrt{400 + 441} = 29, and the vector is \displaystyle \frac{6.894}{29}\langle -20, 21 \rangle.
\displaystyle \operatorname{proj}_{\mathbf{v}}\mathbf{u} = \left\langle -\frac{6}{5}, \frac{3}{5} \right\rangle and \displaystyle \mathbf{w} = \left\langle \frac{21}{5}, \frac{42}{5} \right\rangle. \mathbf{u} \cdot \mathbf{v} = -18 + 27 = 9 and \|\mathbf{v}\|^2 = 45, so the projection is \displaystyle \frac{1}{5}\langle -6, 3 \rangle. Check: \displaystyle \mathbf{w} \cdot \mathbf{v} = -\frac{126}{5} + \frac{126}{5} = 0.
\displaystyle \operatorname{proj}_{\mathbf{v}}\mathbf{u} = \left\langle \frac{30}{61}, \frac{25}{61} \right\rangle and \displaystyle \mathbf{w} = \left\langle -\frac{335}{61}, \frac{402}{61} \right\rangle. \mathbf{u} \cdot \mathbf{v} = -30 + 35 = 5 and \|\mathbf{v}\|^2 = 61. Check: \displaystyle \mathbf{w} \cdot \mathbf{v} = \frac{-2010 + 2010}{61} = 0.
\operatorname{proj}_{\mathbf{v}}\mathbf{u} = \langle 0, 0 \rangle and \mathbf{w} = \langle 7, 7 \rangle. \mathbf{u} \cdot \mathbf{v} = 63 - 63 = 0: \mathbf{u} is already orthogonal to \mathbf{v}, so it casts no shadow and is all perpendicular part.
\operatorname{proj}_{\mathbf{v}}\mathbf{u} = \langle 15, -10 \rangle and \mathbf{w} = \langle 0, 0 \rangle. \mathbf{u} \cdot \mathbf{v} = -135 - 60 = -195 and \|\mathbf{v}\|^2 = 117, so the projection is \displaystyle -\frac{5}{3}\langle -9, 6 \rangle = \mathbf{u}. Here \mathbf{u} is parallel to \mathbf{v}, so it is all parallel part; \mathbf{w} = \mathbf{0} is orthogonal to everything.
276 J. \mathbf{d} = \langle 15, 4 \rangle and W = 300 - 24.
\displaystyle 200\sqrt{3} \approx 346.4 J. \displaystyle (50)(8)\cos 30^\circ = 400 \cdot \frac{\sqrt{3}}{2}.
About 10{,}142 ft-lb. (70)(150)\cos 15^\circ = 10{,}500\cos 15^\circ \approx 10{,}142.2.
-150 ft-lb. \displaystyle (30)(10)\cos 120^\circ = 300\left(-\frac{1}{2}\right). The force acts partly against the motion, so it takes energy away rather than adding it, as a brake does.
About 63.85 N. 900 = \|\mathbf{F}\|(15)\cos 20^\circ, so \displaystyle \|\mathbf{F}\| = \frac{60}{\cos 20^\circ}.
147 J. \mathbf{d} = \langle 8, -3 \rangle and W = (0)(8) + (-49)(-3). Only the 3-meter drop counts, because the weight has no horizontal part.
\displaystyle \frac{5\sqrt{10}}{2} \approx 7.91. \overrightarrow{OP} = \langle 8, 1 \rangle, its dot product with \mathbf{v} is -24 + 9 = -15 and \|\mathbf{v}\|^2 = 90, so the projection is \displaystyle -\frac{1}{6}\langle -3, 9 \rangle = \left\langle \frac{1}{2}, -\frac{3}{2} \right\rangle and the orthogonal part is \displaystyle \left\langle \frac{15}{2}, \frac{5}{2} \right\rangle, of length \displaystyle \sqrt{\frac{250}{4}}.
\displaystyle \operatorname{proj}_{k\mathbf{v}}\mathbf{u} = \frac{\mathbf{u} \cdot k\mathbf{v}}{k^2\|\mathbf{v}\|^2}\,k\mathbf{v} = \frac{k^2\,\mathbf{u} \cdot \mathbf{v}}{k^2\|\mathbf{v}\|^2}\,\mathbf{v} = \operatorname{proj}_{\mathbf{v}}\mathbf{u}. And \displaystyle \operatorname{comp}_{k\mathbf{v}}\mathbf{u} = \frac{k\,\mathbf{u} \cdot \mathbf{v}}{|k|\,\|\mathbf{v}\|}, which is \operatorname{comp}_{\mathbf{v}}\mathbf{u} times \displaystyle \frac{k}{|k|} = \pm 1. In words: k\mathbf{v} lies on the same line, so the shadow is the same arrow, but its sign is measured against the direction of k\mathbf{v}, which reverses when k < 0.
\displaystyle \operatorname{proj}_{\mathbf{v}}\mathbf{u} = \frac{64}{109}\langle 10, 3 \rangle = \left\langle \frac{640}{109}, \frac{192}{109} \right\rangle and \displaystyle \operatorname{proj}_{\mathbf{u}}\mathbf{v} = \frac{64}{80}\langle 4, 8 \rangle = \left\langle \frac{16}{5}, \frac{32}{5} \right\rangle, from \mathbf{u} \cdot \mathbf{v} = 40 + 24 = 64. The first is a multiple of \mathbf{v} and the second a multiple of \mathbf{u}; these two vectors are not parallel, so nonzero multiples of them can never be equal.
\left|\operatorname{comp}_{\mathbf{v}}\mathbf{u}\right| = \|\mathbf{u}\|\,|\cos\theta| and |\cos\theta| \le 1. Equality needs \theta = 0^\circ or 180^\circ, that is \mathbf{u} parallel to \mathbf{v} (or \mathbf{u} = \mathbf{0}). In the split, \|\mathbf{u}\|^2 = \|\operatorname{proj}_{\mathbf{v}}\mathbf{u}\|^2 + \|\mathbf{w}\|^2 says the same: the shadow is a leg and \mathbf{u} the hypotenuse.
The holding force points straight up and the displacement is horizontal, so they are perpendicular and \mathbf{F} \cdot \mathbf{d} = \|\mathbf{F}\|\,\|\mathbf{d}\|\cos 90^\circ = 0. The force has no part along the motion.
He is right exactly when \mathbf{u} \cdot \mathbf{v} > 0, since the multiplier is \displaystyle \frac{\mathbf{u} \cdot \mathbf{v}}{\|\mathbf{v}\|^2} and the denominator is positive. Example 2 is a counterexample: the multiplier is -3. When \mathbf{u} \cdot \mathbf{v} = 0 the projection is \mathbf{0}, which is 0\mathbf{v}, not a positive multiple either.