SME
Grade 12 Math · Problem type
Volume from the triple product
Use this for “Find the volume of the box that u = ⟨1, 2, 0⟩, v = ⟨0, 1, 3⟩ and w = ⟨2, 0, 1⟩ span.”
Diagram
u = ⟨1, 2, 0⟩, with length |u| = 2.24.
v = ⟨0, 1, 3⟩, with length |v| = 3.16.
w = ⟨2, 0, 1⟩, with length |w| = 2.24.
u·(v × w) = 13, so the box’s volume is |13| = 13.
This is an example. Type your own values to start a new problem.
=
T
V
Volume
worked out (not typed)
Formulas
T
=
u₁
×
(
v₂
×
w₃
−
v₃
×
w₂
)
−
u₂
×
(
v₁
×
w₃
−
v₃
×
w₁
)
+
u₃
×
(
v₁
×
w₂
−
v₂
×
w₁
)
13 = 1 × (1 × 1 − 3 × 0) − 2 × (0 × 1 − 3 × 2) + 0 × (0 × 0 − 1 × 2)
V
=
|
T
|
13 = |13|
Assumptions
•
u · (v × w) is the 3 × 3 determinant with rows u, v and w.
•
The volume of the slanted box (parallelepiped) the three vectors span is its absolute value.
•
A triple product of 0 means the three vectors lie in one plane.
Step-by-step
WE KNOW
u₁ = 1, u₂ = 2, u₃ = 0, v₁ = 0, v₂ = 1, v₃ = 3, w₁ = 2, w₂ = 0, w₃ = 1
FIND
triple product u · (v × w) (T), volume (V)
Step 1 · Find triple product u · (v × w): T
Use T = u₁ × (v₂ × w₃ − v₃ × w₂) − u₂ × (v₁ × w₃ − v₃ × w₁) + u₃ × (v₁ × w₂ − v₂ × w₁)
Expand the determinant along u’s row: each entry times the 2 × 2 determinant left, signs +, −, +.
T
=
1
×
(1
×
1
−
3
×
0)
−
2
×
(0
×
1
−
3
×
2)
+
0
×
(0
×
0
−
1
×
2)
T
=
1
−
3
×
0
−
2
×
(0
−
6)
+
0
×
(0
−
2)
T
=
1
−
3
×
0
−
2
×
(−6)
+
0
×
(−2)
T
=
1
−
(−12)
T
=
13
Step 2 · Find volume (V)
Use V = |T|
A volume is never negative: the sign only says which way the three vectors turn.
V
=
|13|
V
=
13
Check
Put the numbers back in. Both sides should match.
13 = 1 × (1 × 1 − 3 × 0) − 2 × (0 × 1 − 3 × 2) + 0 × (0 × 0 − 1 × 2) ✓
13 = |13| ✓
Topic check
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Vectors in three dimensions: dot and cross products
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Use this for “Find the distance between P(1, 2, 3) and Q(3, 5, 9) and the midpoint of PQ.”