HSN.VM.B.5Common CoreMathNumber and QuantityGrades 9-12
HSN.VM.B.5: Multiplying a Vector by a Scalar
In plain English: HSN.VM.B.5 is an advanced (+) Common Core number and quantity standard, usually taught in Precalculus, that asks students to multiply a vector by a scalar. Students scale arrows and reverse them for negative scalars, compute c⟨vx, vy⟩ = ⟨cvx, cvy⟩, find the magnitude with ||cv|| = |c| ||v||, and state that cv points along v for a positive scalar and against v for a negative one.
(+) Multiply a vector by a scalar.
a.Represent scalar multiplication graphically by scaling vectors and possibly reversing their direction; perform scalar multiplication component-wise, e.g., as c(vx, vy) = (cvx, cvy).
b.Compute the magnitude of a scalar multiple cv using ||cv|| = |c|v. Compute the direction of cv knowing that when |c|v ≠ 0, the direction of cv is either along v (for c > 0) or against v (for c < 0).
Common Core State Standards for Mathematics · Domain: Vector and Matrix Quantities (VM) · Cluster: Perform operations on vectors. Also written as HSN-VM.B.5 or N-VM.5 · Official standard
Students learn what multiplying a vector by a real number (a scalar) does: it stretches or shrinks the arrow, and a negative scalar also reverses it. They connect the picture to the rule c⟨vx, vy⟩ = ⟨cvx, cvy⟩, then compute magnitudes with ||cv|| = |c| ||v|| and directions from the sign of c.
The lesson ends with velocity problems in magnitude and direction form, where a scalar changes speed or reverses motion. Notation used on this page: a bold letter such as v names a vector, ||v|| is its magnitude, ⟨3, 4⟩ is component form, and directions are angles measured counterclockwise from the positive x-axis. The official text writes the magnitude rule as ||cv|| = |c|v, where the plain v stands for the magnitude of v.
Learning Objectives
By the end of this lesson, students will be able to:
Draw a scalar multiple cv by scaling v, and reverse the arrow when c is negative
Multiply a vector by a scalar component-wise, c⟨vx, vy⟩ = ⟨cvx, cvy⟩, and match the result to the drawing
Compute the magnitude of cv with ||cv|| = |c| ||v|| and explain why the absolute value is needed
State the direction of cv: along v when c > 0, against v (add 180°) when c < 0, and none when cv is the zero vector
Prior Knowledge Required
Students should already be comfortable with:
Vectors as directed segments and vector notation, including ||v|| for magnitude HSN.VM.A.1
Component form of a vector from its initial and terminal points HSN.VM.A.2
Absolute value and the distance formula 8.G.B.8
Similar triangles and dilations with a scale factor HSG.SRT.A.1
Put one arrow on the board: from (0, 0) to (3, 1). Ask:
Warm-Up Prompt
"A robot moves along this arrow each second. Where is it after 2 seconds? Draw one arrow for the 2-second trip. Now it runs in reverse at the same speed for 2 seconds from the start. Draw that arrow. How are the three arrows alike, and how are they different?"
Students should draw ⟨6, 2⟩ and ⟨-6, -2⟩ and notice that all three arrows lie on one line through the origin. Name the two new arrows 2v and -2v. Ask what the arrow for "half a second" would be, to set up scalars that are fractions.
Direct Instruction20 minutes
Part 1: The picture (standard a). Use Diagram 1. Multiplying v by c gives a vector parallel to v whose length is |c| times the length of v. When c > 0 it points the same way, when c < 0 it points the opposite way, and 0v is the zero vector. Drawing each multiple from a different point shows that a vector is defined by its length and direction, not its position.
Part 2: Components (standard a). Multiply each component by c: c⟨vx, vy⟩ = ⟨cvx, cvy⟩. The right triangle under cv is the right triangle under v dilated by |c|, which is why the components scale and the direction stays on the same line.
Component-wise, positive scalar
Find 4v for v = ⟨-3, 2⟩ and describe it.
Equation: 4⟨-3, 2⟩ = ⟨-12, 8⟩: same direction as v, 4 times as long
Negative scalar, graphically
Draw v = ⟨2, 1⟩ and -1.5v (Diagram 1).
Equation: -1.5v = ⟨-3, -1.5⟩: parallel to v, reversed, 1.5 times as long
Part 3: Magnitude and direction (standard b). Derive the magnitude rule from components: ||cv|| = √((cvx)² + (cvy)²) = √(c²) · √(vx² + vy²) = |c| ||v||. The absolute value appears because √(c²) = |c|; a magnitude can never be negative. For the direction: if c > 0, cv has the same direction angle as v; if c < 0, add or subtract 180°; if c = 0 or v is the zero vector, there is no direction to find. Diagram 2 graphs the magnitude against c.
Magnitude: multiply ||v|| by |c|.
Direction when c > 0: keep the direction angle of v.
Direction when c < 0: add 180° to the direction angle of v (subtract 360° if the result is 360° or more).
Check with components: the components of cv are c times the components of v.
v has magnitude 12 and direction 35°. Find the magnitude and direction of 3v and of -2v.
Equation: 3v: 36 at 35°. -2v: 24 at 35° + 180° = 215°
Velocity in context
A drone flies with velocity ⟨4, 3⟩ m/s (speed 5 m/s). The pilot changes the velocity to 2.5 times the original, then later to -0.6 times the original. Describe both new velocities.
Equation: 2.5⟨4, 3⟩ = ⟨10, 7.5⟩, 12.5 m/s in the same direction (about 36.9°). -0.6⟨4, 3⟩ = ⟨-2.4, -1.8⟩, 3 m/s in the opposite direction (about 216.9°)
Guided Practice15-20 minutes
Pairs complete the table, drawing each multiple on grid paper before computing. After each row, one pair explains its sketch.
Guided practice problems and answers
Problem
Answer
v = ⟨-4, 6⟩: find 0.5v and -3v and sketch them with v
⟨-2, 3⟩ and ⟨12, -18⟩
Find ||v|| and ||-3v|| for v = ⟨-4, 6⟩, first with components, then with |c| ||v||
√52 = 2√13 ≈ 7.21 and 6√13 ≈ 21.63
w has magnitude 8 and direction 120°. Find the magnitude and direction of 1.5w and of -0.25w
12 at 120°, and 2 at 300°
Listen for these errors: multiplying only one component, writing a negative magnitude such as -24, keeping the same direction angle for a negative scalar, and adding 180° when the scalar is a positive fraction.
Independent Practice15 minutes
Students work alone:
For v = ⟨5, -12⟩, find 2v and ||2v|| (⟨10, -24⟩, magnitude 26)
Find -⅓⟨9, -6⟩ and sketch both vectors (⟨-3, 2⟩)
Find ||-4⟨1, 1⟩|| (4√2 ≈ 5.66)
A vector has magnitude 6 and direction 300°. Find the magnitude and direction of -2 times it (12 at 120°)
Find the scalar c with c⟨2, -3⟩ = ⟨-8, 12⟩ and say whether the result points along or against ⟨2, -3⟩ (c = -4, against)
Closure5 minutes
Exit ticket: (1) For v = ⟨-1, 3⟩, find -5v and ||-5v||. (Answers: ⟨5, -15⟩ and 5√10 ≈ 15.81.) (2) A vector u has magnitude 4 and direction 10°. Give the magnitude and direction of -3u. (12 at 190°.) (3) In one sentence, explain why ||cv|| uses |c| and not c.
Differentiation Strategies
For Struggling Students
Start with whole-number scalars on grid paper and count squares: 3v means "repeat the steps of v three times"
Use two colors: one for positive multiples and one for negative multiples, so the reversal is visible
Give a three-column organizer: sign of c, size of |c|, and what happens to the arrow
For Advanced Students
Find a unit vector in the direction of v = ⟨-5, 12⟩ by choosing c = 1/||v||, and explain why it works
Prove that c(u + w) = cu + cw component-wise and draw a picture that shows it with similar triangles
Explain why every vector parallel to v (and not the zero vector) is a scalar multiple of v
Assessment Guidance
What to Look For
Look for sketches in which cv lies on a line parallel to v and has the right length, and for magnitudes written as positive numbers. When students give a direction for a negative multiple, check that it differs from the direction of v by 180°, not 90° or a reflection. Ask students who compute ||cv|| from the components to confirm it with |c| ||v||, and ask what happens when c = 0.
02
Classroom Activities
3 Activities
1
Stretch, Shrink, Flip
20 minPairs
Pairs multiply one base vector by 6 scalar cards, draw every result on grid paper and measure it with a ruler. The pattern in their table leads to both parts of the standard: components scale, and length scales by |c|.
Materials and Setup
Base vector: v = ⟨-6, 3⟩, with magnitude √45 = 3√5 ≈ 6.71
6 scalar cards: 3, ½, -1, -2, 0, -⅓
A recording table with columns: c, cv in components, length measured, |c| · ||v||, same or opposite direction
Procedure
Partner A draws v from the origin; Partner B draws each multiple from a different starting point, as in Diagram 1
For each card, compute cv component-wise, then measure the arrow and compare with |c| · 6.71
Which cards reversed the arrow? What do they have in common?
Which card gave an arrow you could not draw? What is its magnitude, and does it have a direction?
Cards -2 and ½ both change the length. Which one also changes the direction?
Modification for Distance Learning
Use a free graphing tool with a slider for c and the vector c⟨-6, 3⟩ drawn from the origin. Students record the magnitude at each card value and describe what happens as c moves through 0.
2
Change of Speed
20 minGroups of 3
Groups receive 4 velocity cards in magnitude and direction form, each with a scalar. They find the new velocity in magnitude and direction form, then check it with components. This practices standard b in context.
Velocity Cards (4)
A cyclist rides at 6 m/s at 40°; the scalar is 1.5 (answer: 9 m/s at 40°)
A ferry moves at 5 m/s at 110°; the scalar is -1 for the return trip (5 m/s at 290°)
A runner jogs at 3.5 m/s at 200°; the scalar is 0.5 for a cool-down (1.75 m/s at 200°)
A delivery robot rolls at 1.2 m/s at 315°; the scalar is -2 (2.4 m/s at 135°)
Procedure
Student 1 finds the new magnitude with |c| times the speed; Student 2 finds the new direction from the sign of c; Student 3 converts the original velocity to components, multiplies by c, and checks that the result matches
For the cyclist, the check is 1.5⟨4.60, 3.86⟩ ≈ ⟨6.89, 5.79⟩, which has magnitude 9 and direction 40°
Rotate roles for each card
Discussion Questions
Why does the runner keep the direction 200° while the ferry changes direction?
What does a scalar between 0 and 1 mean for a velocity? A scalar between -1 and 0?
Could a scalar change the direction by 90°? Why or why not?
Challenge Variation
Give groups only the old and new velocity in magnitude and direction form, for example 8 m/s at 50° and 2 m/s at 230°, and ask them to find the scalar (here -0.25).
3
Multiple or Not?
15 minPairs
Pairs sort 6 cards, each with two vectors, into "scalar multiples" and "not multiples". For each pair of multiples, they find the scalar and say whether it points along or against. This builds the idea that cv is always parallel to v.
Cards (6)
⟨4, -10⟩ and ⟨-2, 5⟩ (the first is -2 times the second)
⟨3, 7⟩ and ⟨9, 21⟩ (the second is 3 times the first)
⟨5, 2⟩ and ⟨10, 5⟩ (not multiples)
⟨-8, 12⟩ and ⟨2, -3⟩ (the first is -4 times the second)
⟨1.5, -3⟩ and ⟨0.5, -1⟩ (the first is 3 times the second)
⟨0, 6⟩ and ⟨0, -2⟩ (the first is -3 times the second)
Procedure
Divide matching components: if both quotients are equal, that quotient is the scalar
Sketch each pair to confirm: multiples lie on parallel lines
For each multiple, write the ratio of the magnitudes and compare it with the absolute value of the scalar
Discussion Questions
Why is ⟨10, 5⟩ not a multiple of ⟨5, 2⟩, even though 10 is twice 5?
For ⟨0, 6⟩ and ⟨0, -2⟩ you cannot divide the x-components. How did you decide?
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: Scaling a Vector
The same vector v = ⟨2, 1⟩ multiplied by 2, 0.5 and -1.5, each arrow drawn from a different starting point so they do not overlap. Every multiple is parallel to v. A positive scalar keeps the direction and a negative scalar reverses it; the absolute value of the scalar sets the length. Red marks the reversed vector.
Diagram 2: Magnitude of cv as the Scalar Changes
For a vector with ||v|| = 5, the magnitude of cv is 5|c|, so the graph is a V drawn to scale. The two marked points are the drone velocities in Example 5. Scalars on the left of 0 give vectors that point against v, scalars on the right give vectors along v, and at c = 0 the result is the zero vector, which has no direction.
04
Homework Assignment
~30 min
HSN.VM.B.5 Homework: Multiplying a Vector by a Scalar
Directions: Show all work and sketch every vector you compute. Give magnitudes in exact form and to two decimal places. Directions are measured counterclockwise from the positive x-axis, between 0° and 360°.
Part 1: Scaling and Components (Problems 1-2)
Let v = ⟨-3, 5⟩. (a) Find 4v, -2v and 0.5v component-wise. (b) Draw v and the three multiples on one grid, starting each from a different point. (c) Which multiples point along v, and which point against it?
Let u = ⟨9, -12⟩. (a) Find ||u||. (b) Use ||cu|| = |c| ||u|| to find ||(2/3)u|| and ||-3u||. (c) Check both answers by finding (2/3)u and -3u in components and computing their magnitudes.
Part 2: Magnitude and Direction (Problems 3-4)
A vector w has magnitude 14 and direction 70°. (a) Find the magnitude and direction of 2.5w. (b) Find the magnitude and direction of -0.5w. (c) Write -0.5w in components, rounded to two decimal places.
Find the scalar c with c⟨4, -2⟩ = ⟨-14, 7⟩. Does c⟨4, -2⟩ point along or against ⟨4, -2⟩, and how many times as long is it? Then explain why no scalar c gives c⟨4, -2⟩ = ⟨8, 4⟩.
Part 3: Applying and Explaining (Problems 5-6)
A boat moves with velocity ⟨5, 12⟩ km/h. It turns back along the same line at 40% of its original speed. (a) Write the new velocity as a scalar multiple of ⟨5, 12⟩ and in components. (b) Find the new speed. (c) Find the directions of the old and new velocities and explain how they are related.
Let v = ⟨a, b⟩ and let c be any real number. (a) Show that ||cv|| = |c| ||v||. Explain where the absolute value comes from. (b) Use the rule to find ||-5⟨1, -2⟩|| exactly. (c) What is cv when c = 0, and why does it have no direction?
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Components
Every component multiplied by c, signs correct
One component or sign error
Scalar applied incorrectly
Sketches
Multiples parallel to v, correct length and sense
Parallel but wrong length or sense in one sketch
No sketches or not parallel
Magnitude and Direction
Magnitudes positive, directions adjusted by 180° exactly when c < 0
One direction or magnitude error
Negative magnitudes or directions unchanged for c < 0
Reasoning
Clear derivation of ||cv|| = |c| ||v|| and explanation of c = 0
Derivation with a gap
No reasoning
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Directions are measured counterclockwise from the positive x-axis. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Find 3⟨-2, 5⟩.
Answer: A
Multiply each component by 3: ⟨3(-2), 3(5)⟩ = ⟨-6, 15⟩. Choice B adds 3 to each component instead of multiplying. Choice C multiplies only the first component, and choice D changes both signs as if the scalar were -3.
Question 2 of 20 · Multiple Choice
Find -2⟨4, -1⟩.
Answer: C
Multiply both components by -2: ⟨-2(4), -2(-1)⟩ = ⟨-8, 2⟩. Choice A forgets that -2 times -1 is positive. Choice B multiplies by 2 instead of -2, and choice D adds -2 to each component.
Question 3 of 20 · Multiple Choice
How does the arrow for -3v compare with the arrow for v?
Answer: B
The length is multiplied by |-3| = 3, and the negative sign reverses the direction. Choice A ignores the sign. Choice C confuses -3 with -1/3. Choice D is wrong because a scalar multiple is always parallel to v; it never turns the arrow by 90°.
Question 4 of 20 · Multiple Choice
A vector v is drawn from (1, 2) to (5, 4). The vector -0.5v is drawn starting at (1, 2). Where does it end?
Answer: D
v = ⟨5 - 1, 4 - 2⟩ = ⟨4, 2⟩, so -0.5v = ⟨-2, -1⟩. Starting at (1, 2), it ends at (1 - 2, 2 - 1) = (-1, 1). Choice A uses 0.5v, which points the wrong way. Choice B gives the components of -0.5v instead of its endpoint, and choice C gives the components of -v.
Question 5 of 20 · Multiple Choice
If ||v|| = 7, what is ||-4v||?
Answer: B
||cv|| = |c| ||v|| = |-4| · 7 = 28. Choice A forgets the absolute value; a magnitude cannot be negative. Choices C and D add the scalar to the magnitude instead of multiplying.
Question 6 of 20 · Multiple Choice
For v = ⟨-8, 15⟩, what is ||3v||?
Answer: A
||v|| = √(64 + 225) = 17, so ||3v|| = 3 · 17 = 51. Check: 3v = ⟨-24, 45⟩ and √(576 + 2025) = √2601 = 51. Choice B is ||v|| without the scalar. Choice C multiplies the sum of the components, -8 + 15 = 7, by 3.
Question 7 of 20 · Multiple Choice
A vector v has direction 60°. What is the direction of -2v?
Answer: C
A negative scalar reverses the direction, so add 180°: 60° + 180° = 240°. Choice A ignores the sign of the scalar. Choice B reflects the vector across the y-axis, and choice D reflects it across the x-axis; neither is the opposite direction.
Question 8 of 20 · Multiple Choice
A vector v has direction 150°. What is the direction of 5v?
Answer: D
A positive scalar changes only the length, so 5v keeps the direction 150°. Choice A adds 180°, which happens only for a negative scalar. Choice B multiplies the angle by 5.
Question 9 of 20 · Multiple Choice
For which scalar c does cv point against v and have twice the magnitude of v?
Answer: B
Pointing against v requires c < 0, and twice the magnitude requires |c| = 2, so c = -2. Choice A keeps the direction. Choice D reverses the vector but halves it.
Question 10 of 20 · Multiple Choice
What is 0v for a nonzero vector v?
Answer: A
Each component is multiplied by 0, so 0v = ⟨0, 0⟩, the zero vector. Its magnitude is |0| ||v|| = 0, and it has no direction, which is why the direction rule only applies when cv is not the zero vector. Choice B confuses multiplying by 0 with multiplying by 1.
Question 11 of 20 · Multiple Choice
u = ⟨-10, 4⟩ and w = ⟨15, -6⟩. Which scalar c gives w = cu?
Answer: C
Divide matching components: 15 / (-10) = -1.5 and -6 / 4 = -1.5. Both quotients agree, so w = -1.5u, which points against u. Choice A drops the sign. Choice B divides the components of u by those of w, giving the scalar for u = cw instead.
Question 12 of 20 · Multiple Choice
v has magnitude 9 and direction 200°. What are the magnitude and direction of -⅓v?
Answer: D
The magnitude is |-⅓| · 9 = 3. The scalar is negative, so the direction is 200° - 180° = 20°. Choice A gives a negative magnitude. Choice B forgets to reverse the direction. Choice C multiplies by 3 instead of dividing.
Question 13 of 20 · Multiple Choice
v goes from (0, 0) to (3, -1). Which vector is parallel to v and points in the same direction?
Answer: B
v = ⟨3, -1⟩ and ⟨6, -2⟩ = 2v, a positive multiple. Choice A is -2v: parallel but pointing the opposite way. Choices C and D are not multiples of v at all; ⟨3, 1⟩ is the reflection of v across the x-axis.
Question 14 of 20 · Multiple Choice
A current has velocity ⟨8, -6⟩ cm/s. A model multiplies the velocity by 0.25. What is the new speed?
Answer: A
The original speed is √(64 + 36) = 10 cm/s, and the new speed is 0.25 · 10 = 2.5 cm/s. Check: 0.25⟨8, -6⟩ = ⟨2, -1.5⟩ and √(4 + 2.25) = 2.5. Choice B divides by 0.25 instead of multiplying. Choice C adds 0.25 to the speed.
Question 15 of 20 · Short Answer
Let v = ⟨-4, 1⟩. Find 4v and -1.5v component-wise, and describe how each arrow compares with v.
4v = ⟨-16, 4⟩: same direction as v, 4 times as long. -1.5v = ⟨6, -1.5⟩: opposite direction, 1.5 times as long. Both are parallel to v.
Question 16 of 20 · Short Answer
Let v = ⟨1, -3⟩. Find ||-4v|| two ways, and explain why the rule is ||cv|| = |c| ||v|| and not c ||v||.
With components: -4v = ⟨-4, 12⟩, so ||-4v|| = √(16 + 144) = √160 = 4√10 ≈ 12.65. With the rule: |-4| · √10 = 4√10. Using c instead of |c| would give -4√10, a negative length, which is impossible. The absolute value comes from √(c²) = |c| when the components are squared.
Question 17 of 20 · Short Answer
v has magnitude 6 and direction 330°. Find the magnitude and direction of -2.5v.
Magnitude: |-2.5| · 6 = 15. The scalar is negative, so the direction is 330° - 180° = 150°.
Question 18 of 20 · Short Answer
Draw v = ⟨1, 2⟩ and 3v from the origin. Use the components to explain why 3v lies along the same line as v and is 3 times as long.
3v = ⟨3, 6⟩. The right triangle under v has legs 1 and 2; the triangle under 3v has legs 3 and 6, a dilation by 3. The triangles are similar, so the arrows make the same angle with the x-axis (both have slope 2) and the hypotenuse is 3 times as long: √45 = 3√5 compared with √5.
Question 19 of 20 · Short Answer
Find the scalar c with c⟨-6, 9⟩ = ⟨4, -6⟩. Does the result point along or against ⟨-6, 9⟩, and how do the magnitudes compare?
From the first components, c = 4 / (-6) = -2/3, and the second components agree: 9 · (-2/3) = -6. So c = -2/3. The result points against ⟨-6, 9⟩ and its magnitude is 2/3 of the original magnitude (√52 compared with √117).
Question 20 of 20 · Short Answer
A tidal current flows with velocity ⟨0.6, -0.8⟩ m/s. At the change of tide it reverses and becomes 1.5 times as fast. Write the new velocity as a scalar multiple and in components, and give its speed and direction.
The new velocity is -1.5⟨0.6, -0.8⟩ = ⟨-0.9, 1.2⟩ m/s. The old speed is √(0.36 + 0.64) = 1 m/s, so the new speed is 1.5 m/s. The old direction is about 306.9°, so the new direction is about 306.9° - 180° = 126.9°.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSN.VM.B.5 mean?
HSN.VM.B.5 asks students to multiply a vector by a scalar. Part a is the picture and the component rule: scale the arrow, reverse it if the scalar is negative, and multiply each component by the scalar. Part b is magnitude and direction: ||cv|| = |c| ||v||, and cv points along v for c > 0 and against v for c < 0.
Is HSN.VM.B.5 part of Precalculus?
Usually, yes. The (+) marks it as an advanced standard, additional mathematics for students who take advanced courses, and it is usually taught in a Precalculus vectors unit together with HSN.VM.B.4 (adding and subtracting vectors).
What is a scalar?
A scalar is a real number, a quantity with size but no direction, such as 3, -0.5 or 2/3. The word comes from "scale": multiplying a vector by a scalar scales its length.
Why is there an absolute value in ||cv|| = |c| ||v||?
Because a magnitude is a length and cannot be negative. When you compute the magnitude from components, c² comes out of the square root as √(c²) = |c|. The sign of c does not disappear: it shows up in the direction instead.
What does the official text mean by ||cv|| = |c|v?
In the official standard, the last v is printed in italic and stands for the magnitude of the vector v. Written with double bars on both sides, the rule is ||cv|| = |c| ||v||. On this page, vectors are bold and magnitudes use double bars.
What happens to the direction of a vector when you multiply by a negative number?
It reverses: the new vector points exactly the opposite way, so its direction angle differs by 180°. The vector stays on a line parallel to the original. Multiplying by a negative number never rotates a vector by any other angle.
What is the difference between multiplying by 2 and by one half?
Both keep the direction because both scalars are positive. Multiplying by 2 doubles the length and multiplying by ½ halves it. A scalar between 0 and 1 shrinks a vector, and a scalar with absolute value greater than 1 stretches it.
What are common mistakes with scalar multiplication of vectors?
Common ones are multiplying only one component, writing a negative magnitude, keeping the same direction angle for a negative scalar, multiplying the direction angle by the scalar, and forgetting that 0v is the zero vector with no direction.
How do you tell whether two vectors are scalar multiples of each other?
Divide matching components. If the quotients are equal, that common quotient is the scalar; if not, the vectors are not multiples and are not parallel. When a component is 0, check that the matching component of the other vector is also 0.
Where is scalar multiplication of vectors used later?
In physics, force equals mass times acceleration, a scalar times a vector, and changing speed without changing direction is a scalar multiple of a velocity. In Common Core it leads to unit vectors, to combinations such as 2u - 3w, and to multiplying matrices by scalars (HSN.VM.C.7).
07
Related Standards
6 standards
These standards connect to HSN.VM.B.5: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSN.VM.A.1Prerequisite
Recognize vectors as having magnitude and direction and use vector notation
