Precalculus · N.VM.B.4C · N.VM.B.5

Scaling a vector

A trip of so much across and so much up can be taken twice, taken halfway, or taken backward. What does multiplying a vector by a number do to its components, its length and its direction?

You need

Component form of a vectorPrecalculus

Where you are 3 in Vector addition and scalar multiplication

Trigonometry7

Polar coordinates and vectors10

Vector addition and scalar multiplication3

Adding and subtracting vectors Forces and velocities as vectors

Scaling a vector

Why this exists

Merey, Nursultan and Sanzhar are building a pen plotter, a drawing machine that drags a pen across a sheet of paper. Its controller moves the pen by a vector each second, measured in centimeters: so much across, so much up, in component form. Running the motor for three seconds moves the pen three times as far the same way. Running it at half speed moves it half as far. Running it in reverse sends the pen back the way it came.

Each of these is a vector multiplied by a number. Numbers and vectors are different kinds of objects, so the product has to be defined, and then its effect on the two things a vector carries, its magnitude and direction, has to be worked out. The answer is simple: the length is multiplied by the size of the number, and the direction is kept or reversed by its sign.

Definition

In vector work an ordinary real number is called a scalar, because multiplying by it scales.

For a scalar k and a vector \mathbf{v} = \langle a, b \rangle, the scalar multiple k\mathbf{v} is the vector whose components are k times those of \mathbf{v}:

k\mathbf{v} = k\langle a, b \rangle = \langle ka, kb \rangle

In \mathbf{i}, \mathbf{j} form the same rule reads k(a\mathbf{i} + b\mathbf{j}) = ka\,\mathbf{i} + kb\,\mathbf{j}.

Three multiples get names. The opposite of \mathbf{v} is -\mathbf{v} = (-1)\mathbf{v} = \langle -a, -b \rangle. The zero vector is \mathbf{0} = \langle 0, 0 \rangle, and 0\mathbf{v} = \mathbf{0} for every \mathbf{v}. Dividing a vector by a nonzero number k means multiplying it by \displaystyle \frac{1}{k}.

Two nonzero vectors are parallel, written \mathbf{v} \parallel \mathbf{w}, when one is a scalar multiple of the other: \mathbf{w} = k\mathbf{v} for some number k. They have the same direction when k > 0 and opposite directions when k < 0.

Intuition

Figure 1 draws \mathbf{v} = \langle 2, 1 \rangle and four of its multiples. 2\mathbf{v} = \langle 4, 2 \rangle is two copies of the trip laid end to end along one line: twice as far across, twice as far up, so twice as long and pointed the same way. \displaystyle \frac{1}{2}\mathbf{v} = \langle 1, \frac{1}{2} \rangle is half the trip. -\mathbf{v} = \langle -2, -1 \rangle undoes the trip, the same length pointed the other way, and \displaystyle -\frac{3}{2}\mathbf{v} reverses and stretches at once.

Figure 1Every scalar multiple of \mathbf{v} = \langle 2, 1 \rangle lies along the same slant. A positive multiplier keeps the way \mathbf{v} points, a negative one reverses it, and the size of the multiplier sets the length.

All five arrows share the slope \displaystyle \frac{1}{2}. Multiplying both components by the same k leaves their ratio \displaystyle \frac{kb}{ka} = \frac{b}{a} unchanged, so a scalar multiple never turns a vector off its line. It can only slide the head along that line, past the tail if k is negative. Drawn from the origin with k > 0, k\mathbf{v} is the image of \mathbf{v} under the dilation with center O and scale factor k; a negative k adds a half turn about O to the dilation by |k|.

Results

Length. For every scalar k and vector \mathbf{v},

\|k\mathbf{v}\| = |k|\,\|\mathbf{v}\|

Direction. Let \mathbf{v} \ne \mathbf{0} have direction angle \theta.

If k > 0, then k\mathbf{v} has the same direction angle \theta.
If k < 0, then k\mathbf{v} has direction angle \theta + 180^\circ, less 360^\circ if that reaches or passes 360^\circ.
If k = 0, then k\mathbf{v} = \mathbf{0}, which has no direction.

Test for parallel by components. Nonzero vectors \langle a, b \rangle and \langle c, d \rangle are parallel exactly when

ad - bc = 0

Properties. For all scalars k and m and every vector \mathbf{v}:

Property	Statement
Associative property for scalars	(km)\mathbf{v} = k(m\mathbf{v})
Multiplying by one	1\mathbf{v} = \mathbf{v}
Multiplying by negative one	(-1)\mathbf{v} = -\mathbf{v}
Multiplying by zero	0\mathbf{v} = \mathbf{0} and k\mathbf{0} = \mathbf{0}

Why it is true

Length. With \mathbf{v} = \langle a, b \rangle,

\begin{aligned} \|k\mathbf{v}\| &= \sqrt{(ka)^2 + (kb)^2} = \sqrt{k^2(a^2 + b^2)} \\ &= \sqrt{k^2}\,\sqrt{a^2 + b^2} = |k|\,\|\mathbf{v}\| \end{aligned}

The square root of a product of two nonnegative numbers is the product of their square roots, and \displaystyle \sqrt{k^2} = |k| because a square root is never negative: \displaystyle \sqrt{(-3)^2} = 3, not -3.

Direction. Let k \ne 0. By the length result \|k\mathbf{v}\| = |k|\,\|\mathbf{v}\|, so the cosine and sine of the direction angle of k\mathbf{v} are

\begin{aligned} \frac{ka}{|k|\,\|\mathbf{v}\|} &= \frac{k}{|k|}\cos\theta \\ \frac{kb}{|k|\,\|\mathbf{v}\|} &= \frac{k}{|k|}\sin\theta \end{aligned}

The factor \displaystyle \frac{k}{|k|} is 1 when k > 0, so both values match those of \theta and the direction angle is \theta. It is -1 when k < 0, so both change sign, and \cos(\theta + 180^\circ) = -\cos\theta, \sin(\theta + 180^\circ) = -\sin\theta: the direction angle is \theta + 180^\circ.

Parallel test. If \langle c, d \rangle = k\langle a, b \rangle, then c = ka and d = kb, so ad - bc = a(kb) - b(ka) = 0. Conversely, suppose ad - bc = 0 with \langle a, b \rangle \ne \mathbf{0}. If a \ne 0, put \displaystyle k = \frac{c}{a}: then ka = c, and \displaystyle kb = \frac{bc}{a} = \frac{ad}{a} = d. If a = 0, then b \ne 0 and the condition reads -bc = 0, so c = 0; put \displaystyle k = \frac{d}{b}. Either way \langle c, d \rangle = k\langle a, b \rangle.

Properties. Each is the matching fact about numbers, applied to both components at once. For example

\begin{aligned} (km)\langle a, b \rangle &= \langle (km)a, (km)b \rangle = \langle k(ma), k(mb) \rangle \\ &= k\langle ma, mb \rangle = k(m\langle a, b \rangle) \end{aligned}

using the associative property of multiplying numbers in the middle step.

Worked examples

Example 1

Multiples and their lengths

Let \mathbf{v} = \langle -8, 6 \rangle. (a) Find 3\mathbf{v}, \displaystyle -\frac{1}{2}\mathbf{v} and 0\mathbf{v}. (b) Find the magnitude of each, both from components and from |k|\,\|\mathbf{v}\|. (c) For \mathbf{w} = 5\mathbf{i} - 2\mathbf{j}, find -3\mathbf{w}.

Solution

(a) Multiply each component by the scalar:

\begin{aligned} 3\mathbf{v} &= \langle -24, 18 \rangle \\ -\tfrac{1}{2}\mathbf{v} &= \langle 4, -3 \rangle \\ 0\mathbf{v} &= \langle 0, 0 \rangle = \mathbf{0} \end{aligned}

(b) First \displaystyle \|\mathbf{v}\| = \sqrt{64 + 36} = 10. From components, \displaystyle \|3\mathbf{v}\| = \sqrt{576 + 324} = \sqrt{900} = 30, and the rule gives 3 \cdot 10 = 30. Next \displaystyle \left\|-\frac{1}{2}\mathbf{v}\right\| = \sqrt{16 + 9} = 5, and the rule gives \displaystyle \left|-\frac{1}{2}\right| \cdot 10 = 5: the length is positive although the scalar is negative. Last, \|\mathbf{0}\| = 0 = 0 \cdot 10.

Check. \mathbf{v} points into quadrant II, as does 3\mathbf{v}; \displaystyle -\frac{1}{2}\mathbf{v} has both signs flipped and points into quadrant IV.

Example 2

Direction angles

A vector \mathbf{v} has magnitude 10 and direction angle 310^\circ. Find the magnitude and direction angle of -2.5\mathbf{v} and of 0.4\mathbf{v}.

Solution

No components are needed. For -2.5\mathbf{v} the length is |-2.5| \cdot 10 = 25, and the negative scalar reverses the direction: 310^\circ + 180^\circ = 490^\circ, which passes 360^\circ, so the direction angle is 490^\circ - 360^\circ = 130^\circ.

For 0.4\mathbf{v} the length is 0.4 \cdot 10 = 4, and a positive scalar keeps the direction angle, 310^\circ.

Check. \mathbf{v} points into quadrant IV, so its opposite must point into quadrant II, and 130^\circ is in quadrant II.

Example 3

Parallel or not

Let \mathbf{u} = \langle 6, -10 \rangle, \mathbf{w} = \langle -9, 15 \rangle and \mathbf{z} = \langle 9, -14 \rangle. Is \mathbf{w} parallel to \mathbf{u}? Is \mathbf{z}?

Solution

For \mathbf{u} and \mathbf{w}, ad - bc = 6(15) - (-10)(-9) = 90 - 90 = 0, so they are parallel. To find the multiplier, divide matching components: \displaystyle \frac{-9}{6} = -\frac{3}{2} and \displaystyle \frac{15}{-10} = -\frac{3}{2}. So \displaystyle \mathbf{w} = -\frac{3}{2}\mathbf{u}, and the negative multiplier means they point in opposite directions.

For \mathbf{u} and \mathbf{z}, 6(-14) - (-10)(9) = -84 + 90 = 6 \ne 0, so they are not parallel. The ratios show why: \displaystyle \frac{9}{6} = \frac{3}{2} but \displaystyle \frac{-14}{-10} = \frac{7}{5}, and no single k does both components. The two vectors look close, but \mathbf{z} is turned slightly off the line of \mathbf{u}.

Example 4

An unknown component

Find every t for which \langle t, 2 \rangle and \langle 8, t \rangle are parallel. For each, say whether they point the same way or opposite ways.

Solution

The parallel test gives an equation in t:

\begin{aligned} t \cdot t - 2 \cdot 8 &= 0 \\ t^2 &= 16 \\ t = 4 \ \text{or} \ t &= -4 \end{aligned}

With t = 4 the vectors are \langle 4, 2 \rangle and \langle 8, 4 \rangle = 2\langle 4, 2 \rangle: a positive multiple, the same direction. With t = -4 they are \langle -4, 2 \rangle and \langle 8, -4 \rangle = -2\langle -4, 2 \rangle: opposite directions. Both vectors are nonzero for every t, so the test applies.

Example 5

The plotter runs forward, then backs up

Sanzhar’s plotter moves its pen by \mathbf{d} = \langle 0.8, 0.6 \rangle centimeters each second. The pen starts at (2, 1) and runs forward for 7.5 seconds. Then Merey reverses the motor at half speed for 3 seconds, so each second the pen moves by \displaystyle -\frac{1}{2}\mathbf{d}. Where does the pen stop, and how far is it from its start?

Solution

The forward run is 7.5\mathbf{d} = \langle 6, 4.5 \rangle, so the pen reaches (2 + 6, 1 + 4.5) = (8, 5.5). Since \displaystyle \|\mathbf{d}\| = \sqrt{0.64 + 0.36} = 1, that run is 7.5 centimeters long.

The backward run is three seconds of \displaystyle -\frac{1}{2}\mathbf{d}, that is, \displaystyle 3\left(-\frac{1}{2} \mathbf{d}\right). By the associative property for scalars this is \displaystyle \left(3 \cdot \left(-\frac{1}{2} \right)\right)\mathbf{d} = -1.5\mathbf{d}, which is \langle -1.2, -0.9 \rangle. The pen stops at (8 - 1.2,\ 5.5 - 0.9) = (6.8, 4.6).

From (2, 1) to (6.8, 4.6) the vector is \langle 4.8, 3.6 \rangle = 6\mathbf{d}, so the pen ends |6| \cdot 1 = 6 centimeters from its start, although it traveled 7.5 + 1.5 = 9 centimeters of line.

Example 6

The segment joining two midpoints

Triangle ABC has A(-2, 5), B(-4, -1) and C(6, 1). Let M and N be the midpoints of \overline{AB} and \overline{AC}. Show that \overline{MN} is parallel to \overline{BC} and half as long.

Solution

By the midpoint formula, \displaystyle M = \left(\frac{-2 - 4}{2}, \frac{5 - 1}{2}\right) = (-3, 2) and \displaystyle N = \left(\frac{-2 + 6}{2}, \frac{5 + 1}{2}\right) = (2, 3). Terminal minus initial,

\begin{aligned} \overrightarrow{MN} &= \langle 5, 1 \rangle \\ \overrightarrow{BC} &= \langle 10, 2 \rangle = 2\,\overrightarrow{MN} \end{aligned}

Figure 2Example 6. \overrightarrow{BC} = 2\,\overrightarrow{MN}, so the segment joining the two midpoints is parallel to the third side and half as long.

Because \overrightarrow{BC} is a positive multiple of \overrightarrow{MN}, the two vectors are parallel and point the same way, so the segments lie on parallel lines (they do not share a line, since B is not on line MN: that line rises 1 for every 5 across and passes through (-3, 2), so at x = -4 it is at y = 1.8, not -1). The length result gives \|\overrightarrow{BC}\| = 2\|\overrightarrow{MN}\|, so \displaystyle MN = \frac{1}{2}BC. In numbers, \displaystyle MN = \sqrt{26} and \displaystyle BC = \sqrt{104} = 2\sqrt{26}. This is the midsegment theorem for this triangle; Exercise 24 proves it for every triangle.

Boundaries

The zero vector has no direction. 0\mathbf{v} and k\mathbf{0} are \mathbf{0}, which has length 0 and no direction angle, so “same or opposite direction” says nothing about it. The parallel test is stated for nonzero vectors: \mathbf{0} = 0\mathbf{v} is a multiple of every vector, and some books call it parallel to every vector by convention, but it lies along no line.

Length uses |k|, not k. A length is never negative, so \|k\mathbf{v}\| = k\|\mathbf{v}\| holds only for k \ge 0. The sign of k shows up in the direction instead.

Parallel is about direction, not position. Parallel vectors may be drawn on one line or on two different lines; deciding which (as in Example 6) needs a point, not just the components.

Only scalar times vector so far. This lesson multiplies a vector by a number. A vector times a vector is a different operation; the dot product is one such product, and its result is a number.

Where this goes next

Adding vectors places arrows tip to tail, and together with scalar multiples it gives every combination k\mathbf{u} + m\mathbf{v}, the distributive laws that link the two operations, and subtraction as adding -\mathbf{v}. Then a\mathbf{i} + b\mathbf{j} becomes a true sum of the two scalar multiples a\mathbf{i} and b\mathbf{j}, and the unit vector \displaystyle \frac{\mathbf{v}}{\|\mathbf{v}\|} of Unit vectors and i, j notation is the scalar multiple \displaystyle \frac{1}{\|\mathbf{v}\|}\mathbf{v}, which has length \displaystyle \frac{1}{\|\mathbf{v}\|} \cdot \|\mathbf{v}\| = 1 by the length result. In forces and velocities, a speed times a unit heading is a velocity, and a velocity times a time is a displacement, exactly as in Example 5.

Exercises

Computing scalar multiples

Let \mathbf{u} = \langle 7, -3 \rangle.

Find 4\mathbf{u}.
Find -2\mathbf{u}.
Find \displaystyle \frac{1}{3}\langle -9, 12 \rangle.
Find -0.5(6\mathbf{i} - 10\mathbf{j}).
Find 0\langle 5, -8 \rangle.
Find \displaystyle \frac{2}{3}(-6\mathbf{i} + \mathbf{j}).

Length and direction

\|\mathbf{v}\| = 6. Find \|-5\mathbf{v}\|.
\|\mathbf{v}\| = 12. Find \displaystyle \left\|-\frac{3}{4}\mathbf{v}\right\|.
For \mathbf{v} = \langle -7, 24 \rangle, find \|-3\mathbf{v}\| two ways: from the components of -3\mathbf{v}, and from |k|\,\|\mathbf{v}\|.
\mathbf{v} has magnitude 4 and direction angle 75^\circ. Find the magnitude and direction angle of -2\mathbf{v}.
\mathbf{v} has magnitude 9 and direction angle 200^\circ. Find the magnitude and direction angle of \displaystyle -\frac{1}{3}\mathbf{v}.
\mathbf{v} has magnitude 5 and direction angle 140^\circ. Find the magnitude and direction angle of 1.6\mathbf{v}.

Parallel or not

Decide whether the two vectors are parallel. If they are, find k with second = k \cdot first and say whether they point the same way or opposite ways.

\langle -6, 9 \rangle and \langle 10, -15 \rangle
\langle 3, 7 \rangle and \langle 9, 20 \rangle
2\mathbf{i} - 5\mathbf{j} and 0.8\mathbf{i} - 2\mathbf{j}
\langle 0, 3 \rangle and \langle 0, -8 \rangle

Unknown parts

Find a so that \langle a, -6 \rangle is parallel to \langle 2, 3 \rangle.
Find every t for which \langle t, 9 \rangle and \langle 4, t \rangle are parallel, and say for each whether they point the same way or opposite ways.
Find every k for which k\langle 5, -12 \rangle has magnitude 26, and the vectors that result.
Find the vector of magnitude 51 pointing opposite to \langle 8, -15 \rangle.

Using scalar multiples

Nursultan’s plotter moves its pen by \langle 1.2, -0.5 \rangle centimeters each second. The pen starts at (-3, 10) and runs for 10 seconds. Where does it stop, and how far did it move?
Show that A(1, 2), B(4, 8) and C(-1, -2) lie on one line, using \overrightarrow{AB} and \overrightarrow{AC}.

Reasoning

Merey writes \|-3\mathbf{v}\| = -3\|\mathbf{v}\|. Explain what is wrong and correct it.
Prove the midsegment result for every triangle: if M and N are the midpoints of \overline{AB} and \overline{AC}, then \overrightarrow{BC} = 2\,\overrightarrow{MN}. Use A(x_1, y_1), B(x_2, y_2), C(x_3, y_3).
Show that k\mathbf{v} = \mathbf{0} only when k = 0 or \mathbf{v} = \mathbf{0}.

Answers

\langle 28, -12 \rangle. \langle 4 \cdot 7,\ 4 \cdot (-3) \rangle.
\langle -14, 6 \rangle. Both signs flip and both sizes double.
\langle -3, 4 \rangle. Dividing each component by 3.
-3\mathbf{i} + 5\mathbf{j}. -0.5 \cdot 6 = -3 and -0.5 \cdot (-10) = 5.
\langle 0, 0 \rangle = \mathbf{0}. Zero times any component is 0.
\displaystyle -4\mathbf{i} + \frac{2}{3}\mathbf{j}. \displaystyle \frac{2}{3} \cdot (-6) = -4 and \displaystyle \frac{2}{3} \cdot 1 = \frac{2}{3}.
30. |-5| \cdot 6.
9. \displaystyle \frac{3}{4} \cdot 12.
75. From components, -3\mathbf{v} = \langle 21, -72 \rangle and \displaystyle \sqrt{441 + 5184} = \sqrt{5625} = 75. From the rule, \displaystyle \|\mathbf{v}\| = \sqrt{49 + 576} = 25 and |-3| \cdot 25 = 75.
Magnitude 8, direction angle 255^\circ. 2 \cdot 4 = 8, and a negative scalar adds 180^\circ.
Magnitude 3, direction angle 20^\circ. \displaystyle \frac{1}{3} \cdot 9 = 3, and 200^\circ + 180^\circ = 380^\circ, which is 20^\circ after subtracting 360^\circ.
Magnitude 8, direction angle 140^\circ. 1.6 \cdot 5 = 8, and a positive scalar keeps the direction.
Parallel, \displaystyle k = -\frac{5}{3}, opposite. (-6)(-15) - 9 \cdot 10 = 90 - 90 = 0, and \displaystyle \frac{10}{-6} = \frac{-15}{9} = -\frac{5}{3}.
Not parallel. 3 \cdot 20 - 7 \cdot 9 = 60 - 63 = -3 \ne 0; the ratios 3 and \displaystyle \frac{20}{7} differ.
Parallel, k = 0.4, same direction. 2(-2) - (-5)(0.8) = -4 + 4 = 0, and \displaystyle \frac{0.8}{2} = \frac{-2}{-5} = 0.4.
Parallel, \displaystyle k = -\frac{8}{3}, opposite. 0 \cdot (-8) - 3 \cdot 0 = 0; both lie along the y-axis, and \displaystyle \frac{-8}{3} scales the second components.
a = -4. 3a - (-6)(2) = 0 gives 3a = -12; check: \langle -4, -6 \rangle = -2\langle 2, 3 \rangle.
t = 6, same direction, or t = -6, opposite directions. t^2 - 36 = 0. With t = 6, \displaystyle \langle 4, 6 \rangle = \frac{2}{3}\langle 6, 9 \rangle; with t = -6, \displaystyle \langle 4, -6 \rangle = -\frac{2}{3}\langle -6, 9 \rangle.
k = 2 or k = -2, giving \langle 10, -24 \rangle and \langle -10, 24 \rangle. \|\langle 5, -12 \rangle\| = 13, so |k| \cdot 13 = 26 and |k| = 2.
\langle -24, 45 \rangle. \|\langle 8, -15 \rangle\| = 17; opposite and 3 times as long means k = -3.
(9, 5), 13 centimeters. 10\langle 1.2, -0.5 \rangle = \langle 12, -5 \rangle, added to (-3, 10); \displaystyle \sqrt{144 + 25} = 13, which is also 10 \cdot \|\langle 1.2, -0.5 \rangle\| = 10 \cdot 1.3.
\overrightarrow{AB} = \langle 3, 6 \rangle and \displaystyle \overrightarrow{AC} = \langle -2, -4 \rangle = -\frac{2}{3}\overrightarrow{AB}. The two vectors are parallel, so lines AB and AC have the same slope 2, and they share the point A, so they are the same line. The negative k puts C and B on opposite sides of A.
A length cannot be negative. The rule uses the absolute value of the scalar: \|-3\mathbf{v}\| = |-3|\,\|\mathbf{v}\| = 3\|\mathbf{v}\|. The minus sign reverses the direction of \mathbf{v}; it does not change the length.
\displaystyle M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) and \displaystyle N = \left(\frac{x_1 + x_3}{2}, \frac{y_1 + y_3}{2}\right), so \overrightarrow{MN} has components \displaystyle \frac{x_3 - x_2}{2} and \displaystyle \frac{y_3 - y_2}{2}, since the x_1 and y_1 terms cancel. And \overrightarrow{BC} = \langle x_3 - x_2,\ y_3 - y_2 \rangle, which is exactly 2\,\overrightarrow{MN}.
If k\langle a, b \rangle = \langle ka, kb \rangle = \langle 0, 0 \rangle, then ka = 0 and kb = 0. If k \ne 0, dividing by k gives a = 0 and b = 0, so \mathbf{v} = \mathbf{0}. Otherwise k = 0. Equivalently, \|k\mathbf{v}\| = |k|\,\|\mathbf{v}\| = 0 forces one factor to be 0.