SME
Linear Algebra · Problem type
Projection onto a line
Use this for “Find the projection of u = ⟨3, 1⟩ onto v = ⟨1, 1⟩ and the part of u orthogonal to v.”
Diagram
u = ⟨3, 1⟩
v = ⟨1, 1⟩
u·v = 3 × 1 + 1 × 1 = 4.
v·v
=
2,
so
projᵥ
u
=
(
4
2
)v
=
⟨2,
2⟩.
u − projᵥ u = ⟨1, −1⟩; (u − projᵥ u)·v = 0: square to v.
This is an example. Type your own values to start a new problem.
u
=
⟨
uₓ
,
uᵧ
⟩,
v
=
⟨
vₓ
,
vᵧ
⟩
projᵥ
u
=
⟨
pₓ
,
pᵧ
⟩
u
−
projᵥ
u
=
⟨
qₓ
,
qᵧ
⟩
u·v
Dot product u·v
worked out (not typed)
v·v
Dot product v·v
worked out (not typed)
k
Factor u·v ÷ v·v
worked out (not typed)
Formulas
u·v = uₓ × vₓ + uᵧ × vᵧ
4 = 3 × 1 + 1 × 1
v·v
=
vₓ²
+
vᵧ
2
2
=
1
2
+
1
2
k
=
u·v
v·v
2
=
4
2
pₓ
=
k
×
vₓ
2 = 2 × 1
pᵧ
=
k
×
vᵧ
2 = 2 × 1
qₓ = uₓ − pₓ
1 = 3 − 2
qᵧ = uᵧ − pᵧ
−1 = 1 − 2
Assumptions
•
projᵥ u is the multiple of v closest to u, so what is left, u − projᵥ u, is orthogonal to v.
•
The factor k = u·v ÷ v·v is negative when the angle between u and v is more than 90°.
•
v is not the zero vector, so v·v > 0.
Step-by-step
WE KNOW
uₓ = 3, uᵧ = 1, vₓ = 1, vᵧ = 1
FIND
dot product u·v (u·v), dot product v·v (v·v), factor u·v ÷ v·v (k), projection, x-component (pₓ), projection, y-component (pᵧ), part orthogonal to v, x-component (qₓ), part orthogonal to v, y-component (qᵧ)
Step 1 · Find dot product u·v (u·v)
Use u·v = uₓ × vₓ + uᵧ × vᵧ
Multiply matching components and add.
u·v = 3 × 1 + 1 × 1
u·v = 4
Step 2 · Find dot product v·v (v·v)
Use v·v = vₓ² + vᵧ²
v·v is the square of v’s length.
v·v
=
1
2
+
1
2
v·v = 1 + 1
v·v = 2
Step 3 · Find factor u·v ÷ v·v (k)
Use k = (u·v) ÷ (v·v)
How
many
v’s
long
the
projection
is.
Then
(u
−
kv)·v
=
u·v
−
k
(v·v)
=
0.
k
=
4
2
k
=
2
Step 4 · Find projection, x-component (pₓ)
Use pₓ = k × vₓ
The
projection
is
k
times
v:
scale
each
component
of
v
by
k
.
pₓ = 2 × 1
pₓ = 2
Step 5 · Find projection, y-component (pᵧ)
Use pᵧ = k × vᵧ
Scale the other component of v the same way.
pᵧ = 2 × 1
pᵧ = 2
Step 6 · Find part orthogonal to v, x-component (qₓ)
Use qₓ = uₓ − pₓ
What is left of u once the projection is taken off.
qₓ = 3 − 2
qₓ = 1
Step 7 · Find part orthogonal to v, y-component (qᵧ)
Use qᵧ = uᵧ − pᵧ
The other component of what is left. It is orthogonal to v: its dot product with v is 0.
qᵧ = 1 − 2
Check: 1 × 1 + (−1) × 1 = 0
qᵧ = −1
Check
Put the numbers back in. Both sides should match.
4 = 3 × 1 + 1 × 1 ✓
2
=
1
2
+
1
2
✓
2
=
4
2
✓
2 = 2 × 1 ✓
2 = 2 × 1 ✓
1 = 3 − 2 ✓
−1 = 1 − 2 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.