SME
Linear Algebra · Problem type
Volume of a box from a determinant
Use this for “Find the volume of the box with edges ⟨2, 0, 0⟩, ⟨1, 3, 0⟩ and ⟨0, 1, 4⟩.”
Diagram
u = ⟨2, 0, 0⟩, with length |u| = 2.
v = ⟨1, 3, 0⟩, with length |v| = 3.16.
w = ⟨0, 1, 4⟩, with length |w| = 4.12.
u·(v × w) = 24, so the box’s volume is |24| = 24.
This is an example. Type your own values to start a new problem.
u
=
⟨
u₁
,
u₂
,
u₃
⟩
v
=
⟨
v₁
,
v₂
,
v₃
⟩
w
=
⟨
w₁
,
w₂
,
w₃
⟩
u · (v × w)
Triple product
worked out (not typed)
V
Volume of the box
worked out (not typed)
Formulas
u
·
(v
×
w)
=
det
[[
u₁
,
u₂
,
u₃
],
[
v₁
,
v₂
,
v₃
],
[
w₁
,
w₂
,
w₃
]]
24 = det [[2, 0, 0], [1, 3, 0], [0, 1, 4]]
V
=
|u
·
(v
×
w)|
24 = |24|
Assumptions
•
The box (a parallelepiped) has edges u, v and w from one corner.
•
Its volume is |u · (v × w)|, the size of det of the matrix with rows u, v and w.
•
det = 0 means a flat box: the three edges lie in one plane.
Step-by-step
WE KNOW
u₁ = 2, u₂ = 0, u₃ = 0, v₁ = 1, v₂ = 3, v₃ = 0, w₁ = 0, w₂ = 1, w₃ = 4
FIND
triple product (u · (v × w)), volume of the box (V)
Step 1 · Find triple product (u · (v × w))
Use u · (v × w) = det [[u₁, u₂, u₃], [v₁, v₂, v₃], [w₁, w₂, w₃]]
The triple product is the det with u, v and w as the rows: first v × w, then its dot product with u.
u · (v × w) = det [[2, 0, 0], [1, 3, 0], [0, 1, 4]]
⟨1, 3, 0⟩ × ⟨0, 1, 4⟩ = ⟨3 × 4 − 0 × 1, 0 × 0 − 1 × 4, 1 × 1 − 3 × 0⟩ = ⟨12, −4, 1⟩
⟨2, 0, 0⟩ · ⟨12, −4, 1⟩ = 2 × 12 + 0 × (−4) + 0 × 1 = 24
u · (v × w) = 24
Step 2 · Find volume of the box (V)
Use V = |u · (v × w)|
Volume is never negative: take the size. A negative triple product only means u, v, w turn the other way.
V
=
|24|
V
=
24
Check
Put the numbers back in. Both sides should match.
24 = det [[2, 0, 0], [1, 3, 0], [0, 1, 4]] ✓
24 = |24| ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.