SME
Grade 12 Math · Problem type
The cross product
Use this for “Find u × v for u = ⟨1, 2, 3⟩ and v = ⟨2, 0, 1⟩, and the areas of the parallelogram and the triangle with sides u and v from one vertex.”
Diagram
u = ⟨1, 2, 3⟩, with length |u| = 3.74.
v = ⟨2, 0, 1⟩, with length |v| = 2.24.
u × v = ⟨2, 5, −4⟩, square to both: (u × v)·u = 0 and (u × v)·v = 0.
The parallelogram’s area is |u × v| = 6.71; the triangle is half: 3.35.
This is an example. Type your own values to start a new problem.
⟨
u₁
,
u₂
,
u₃
⟩
×
⟨
v₁
,
v₂
,
v₃
⟩
=
⟨
w₁
,
w₂
,
w₃
⟩
A
Area of the parallelogram, |u × v|
worked out (not typed) · ≈ 6.7082
AT
Area of the triangle
worked out (not typed) · ≈ 3.3541
Formulas
w₁
=
u₂
×
v₃
−
u₃
×
v₂
2 = 2 × 1 − 3 × 0
w₂
=
u₃
×
v₁
−
u₁
×
v₃
5 = 3 × 2 − 1 × 1
w₃
=
u₁
×
v₂
−
u₂
×
v₁
−4 = 1 × 0 − 2 × 2
A
=
√
w₁
2
+
w₂
2
+
w₃
2
3
√
5
=
√(
2
2
+
5
2
+
(−4)
2
)
AT
=
A
2
3
√
5
2
=
3√
5
2
Assumptions
•
u × v is a vector perpendicular to both u and v: its dot product with each is 0.
•
|u × v| is the area of the parallelogram u and v span; the triangle is half of it.
•
For a triangle PQR, take u = Q − P and v = R − P, the sides from one vertex.
•
Order matters: v × u points the opposite way, −(u × v).
Step-by-step
WE KNOW
u₁ = 1, u₂ = 2, u₃ = 3, v₁ = 2, v₂ = 0, v₃ = 1
FIND
x-component of u × v (w₁), y-component of u × v (w₂), z-component of u × v (w₃), area of the parallelogram, |u × v| (A), area of the triangle (AT)
Step 1 · Find x-component of u × v (w₁)
Use w₁ = u₂ × v₃ − u₃ × v₂
The x-component: cover the x column and take the 2 × 2 determinant of what is left.
w₁
=
2
×
1
−
3
×
0
w₁
=
2
Step 2 · Find y-component of u × v (w₂)
Use w₂ = u₃ × v₁ − u₁ × v₃
The y-component: cover the y column and take the 2 × 2 determinant of what is left, with its sign changed.
w₂
=
3
×
2
−
1
×
1
w₂
=
6
−
1
w₂
=
5
Step 3 · Find z-component of u × v (w₃)
Use w₃ = u₁ × v₂ − u₂ × v₁
The z-component: cover the z column and take the 2 × 2 determinant of what is left.
w₃
=
1
×
0
−
2
×
2
w₃
=
0
−
4
w₃
=
−4
Step 4 · Find area of the parallelogram, |u × v| (A)
Use A = √(w₁² + w₂² + w₃²)
The length of u × v, by the Pythagorean theorem twice. |u × v| = |u||v| sin θ, base times height: the parallelogram’s area.
A
=
√(
2
2
+
5
2
+
(−4)
2
)
A
=
√
4 + 25 + 16
A
=
√
45
A
=
3
√
5
≈
6.7082
Step 5 · Find area of the triangle (AT)
Use AT = A ÷ 2
The diagonal cuts the parallelogram into two equal triangles.
AT
=
3√
5
2
AT
=
3
√
5
2
≈
3.3541
Check
Put the numbers back in. Both sides should match.
2 = 2 × 1 − 3 × 0 ✓
5 = 3 × 2 − 1 × 1 ✓
−4 = 1 × 0 − 2 × 2 ✓
3
√
5
=
√(
2
2
+
5
2
+
(−4)
2
)
✓
3
√
5
2
=
3√
5
2
✓
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Vectors in three dimensions: dot and cross products
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