SME
Grade 12 Math · Problem type
A scalar times a vector
Use this for “Find −2u for u = ⟨3, 4⟩ and its magnitude.”
Diagram
u = ⟨3, 4⟩
|u| = 5.
−2u = ⟨−6, −8⟩: 2 times as long, the opposite direction.
This is an example. Type your own values to start a new problem.
k
⟨
u₁
,
u₂
⟩
=
⟨
w₁
,
w₂
⟩
|u|
Magnitude of u
worked out (not typed)
|ku|
Magnitude of ku
calculated
Formulas
w₁
=
k
×
u₁
−6 = −2 × 3
w₂
=
k
×
u₂
−8 = −2 × 4
|u|
=
√
u₁
2
+
u₂
2
5
=
√
3
2
+
4
2
|ku|
=
|
k
|
×
|u|
10 = |−2| × 5
Assumptions
•
k multiplies each component, so ku is |k| times as long.
•
A negative k reverses the direction.
•
Type k = 1 ÷ |u| for the unit vector along u: k = 0.2 turns ⟨3, 4⟩ into ⟨0.6, 0.8⟩.
•
Call the new vector w = ku, with components w₁ and w₂.
Step-by-step
WE KNOW
k = −2, u₁ = 3, u₂ = 4
FIND
x-component of ku (w₁), y-component of ku (w₂), magnitude of u (|u|), magnitude of ku (|ku|)
Step 1 · Find x-component of ku (w₁)
Use w₁ = k × u₁
Multiply
the
x-component
by
k
.
w₁
=
−2
×
3
w₁
=
−6
Step 2 · Find y-component of ku (w₂)
Use w₂ = k × u₂
Multiply
the
y-component
by
k
.
w₂
=
−2
×
4
w₂
=
−8
Step 3 · Find magnitude of u (|u|)
Use |u| = √(u₁² + u₂²)
The Pythagorean theorem on u’s components.
|u|
=
√
3
2
+
4
2
|u|
=
√
9 + 16
|u|
=
√
25
|u| = 5
Step 4 · Find magnitude of ku (|ku|)
Use |ku| = |k| × |u|
Scaling
by
k
scales
the
magnitude
by
|
k
|;
a
negative
k
only
turns
it
around.
|ku| = |−2| × 5
|ku| = 2 × 5
|ku| = 10
Check
Put the numbers back in. Both sides should match.
−6 = −2 × 3 ✓
−8 = −2 × 4 ✓
5
=
√
3
2
+
4
2
✓
10 = |−2| × 5 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
Refresh
Related lessons
Vectors: components, magnitude and dot product
Main lesson
Adding vectors
Use this for “Find u + v for u = ⟨3, 1⟩ and v = ⟨1, 2⟩, and its magnitude.”
The dot product and the angle between vectors
Use this for “Find the angle between ⟨2, 1⟩ and ⟨1, 3⟩” or “Are these vectors perpendicular?”
The resultant of two forces
Use this for “Forces of 30 N at 0° and 40 N at 60° pull on a ring. Find the resultant.”