Worked examples
The question
Question: find the cross product of u = ⟨1, 2, 3⟩ and v = ⟨4, 5, 6⟩.
⟨1, 2, 3⟩ × ⟨4, 5, 6⟩
1 Set up the grid
Write a grid.
On top: i, j, k, for the x, y and z directions.
Under them, u, then v.
u × v = |i, j, k; 1, 2, 3; 4, 5, 6|
2 The x part
Cover the i column.
Four numbers are left: 2, 3, 5, 6.
Write them as a small grid.
Its cross: down 2 times 6, minus up 3 times 5.
x part = |2, 3; 5, 6| = 2 × 6 − 3 × 5
12 minus 15 is −3.
x part = 12 − 15 = −3
3 The y part
Now cover the j column.
Four numbers are left: 1, 3, 4, 6.
Write them as a small grid, with a minus in front: the middle part always gets one.
Its cross: down 1 times 6, minus up 3 times 4.
y part = −|1, 3; 4, 6| = −(1 × 6 − 3 × 4)
Careful!
The y part, under j, always has a minus in front.
Forgetting it is the most common mistake.
Careful: the middle part gets a minus
6 minus 12 is −6.
With the minus in front: 6.
y part = −(6 − 12) = 6
4 The z part
Now cover the k column.
Four numbers are left: 1, 2, 4, 5.
Write them as a small grid.
Its cross: down 1 times 5, minus up 2 times 4.
z part = |1, 2; 4, 5| = 1 × 5 − 2 × 4
5 minus 8 is −3.
z part = 5 − 8 = −3
Put the three parts together: u × v = ⟨−3, 6, −3⟩.
u × v = ⟨−3, 6, −3⟩
5 Check
Let's check.
The answer must be at a square corner to u, so its dot product with u is 0.
It is 0!
⟨−3, 6, −3⟩ • ⟨1, 2, 3⟩ = −3 + 12 − 9 = 0
And with v: 0.
Right!
⟨−3, 6, −3⟩ • ⟨4, 5, 6⟩ = −12 + 30 − 18 = 0
Here is a picture of the directions.
Each arrow is drawn the same length.
u and v lie in this flat piece.
u × v sticks straight out of it.
Let's look back.
First: set up the grid.
Then: the x part.
Then: the y part.
Then: the z part.
Last: check.
Answer u × v = ⟨−3, 6, −3⟩
▶ Watch it solved
The question
Question: find the cross product of u = ⟨−1, 2, 4⟩ and v = ⟨3, 0, −2⟩.
⟨−1, 2, 4⟩ × ⟨3, 0, −2⟩
1 Set up the grid
Write a grid.
On top: i, j, k, for the x, y and z directions.
Under them, u, then v.
u × v = |i, j, k; −1, 2, 4; 3, 0, −2|
2 The x part
Cover the i column.
Four numbers are left: 2, 4, 0, −2.
Write them as a small grid.
Its cross: down 2 times (−2), minus up 4 times 0.
x part = |2, 4; 0, −2| = 2 × (−2) − 4 × 0
−4 minus 0 is −4.
x part = −4 − 0 = −4
3 The y part
Now cover the j column.
Four numbers are left: −1, 4, 3, −2.
Write them as a small grid, with a minus in front: the middle part always gets one.
Its cross: down −1 times (−2), minus up 4 times 3.
y part = −|−1, 4; 3, −2| = −(−1 × (−2) − 4 × 3)
Careful!
The y part, under j, always has a minus in front.
Forgetting it is the most common mistake.
Careful: the middle part gets a minus
2 minus 12 is −10.
With the minus in front: 10.
y part = −(2 − 12) = 10
4 The z part
Now cover the k column.
Four numbers are left: −1, 2, 3, 0.
Write them as a small grid.
Its cross: down −1 times 0, minus up 2 times 3.
z part = |−1, 2; 3, 0| = −1 × 0 − 2 × 3
0 minus 6 is −6.
z part = 0 − 6 = −6
Put the three parts together: u × v = ⟨−4, 10, −6⟩.
u × v = ⟨−4, 10, −6⟩
5 Check
Let's check.
The answer must be at a square corner to u, so its dot product with u is 0.
It is 0!
⟨−4, 10, −6⟩ • ⟨−1, 2, 4⟩ = 4 + 20 − 24 = 0
And with v: 0.
Right!
⟨−4, 10, −6⟩ • ⟨3, 0, −2⟩ = −12 + 0 + 12 = 0
Here is a picture of the directions.
Each arrow is drawn the same length.
u and v lie in this flat piece.
u × v sticks straight out of it.
Let's look back.
First: set up the grid.
Then: the x part.
Then: the y part.
Then: the z part.
Last: check.
Answer u × v = ⟨−4, 10, −6⟩
▶ Watch it solved
The question
Question: find the area of the triangle with corners (1, 0, 0), (0, 2, 0), (0, 0, 3).
P(1, 0, 0), Q(0, 2, 0), R(0, 0, 3)
1 Two sides as vectors
Draw the triangle.
Two sides from P make two vectors.
Each is end minus start.
u goes from P to Q: u = ⟨−1, 2, 0⟩.
u = PQ = ⟨−1, 2, 0⟩
v goes from P to R: v = ⟨−1, 0, 3⟩.
v = PR = ⟨−1, 0, 3⟩
2 Set up the grid
Write a grid.
On top: i, j, k, for the x, y and z directions.
Under them, u, then v.
u × v = |i, j, k; −1, 2, 0; −1, 0, 3|
3 The x part
Cover the i column.
Four numbers are left: 2, 0, 0, 3.
Write them as a small grid.
Its cross: down 2 times 3, minus up 0 times 0.
x part = |2, 0; 0, 3| = 2 × 3 − 0 × 0
6 minus 0 is 6.
x part = 6 − 0 = 6
4 The y part
Now cover the j column.
Four numbers are left: −1, 0, −1, 3.
Write them as a small grid, with a minus in front: the middle part always gets one.
Its cross: down −1 times 3, minus up 0 times (−1).
y part = −|−1, 0; −1, 3| = −(−1 × 3 − 0 × (−1))
Careful!
The y part, under j, always has a minus in front.
Forgetting it is the most common mistake.
Careful: the middle part gets a minus
−3 minus 0 is −3.
With the minus in front: 3.
y part = −(−3 − 0) = 3
5 The z part
Now cover the k column.
Four numbers are left: −1, 2, −1, 0.
Write them as a small grid.
Its cross: down −1 times 0, minus up 2 times (−1).
z part = |−1, 2; −1, 0| = −1 × 0 − 2 × (−1)
0 minus (−2) is 2.
z part = 0 − (−2) = 2
Put the three parts together: u × v = ⟨6, 3, 2⟩.
u × v = ⟨6, 3, 2⟩
6 Find the length
area = ‖u × v‖
The area of the parallelogram is the length of u × v.
Square the parts, add, take the root.
‖u × v‖ = √62 + 32 + 22
6 squared is 36.
3 squared is 9.
2 squared is 4.
‖u × v‖ = √36 + 9 + 4
Add: 36 plus 9 plus 4 is 49.
‖u × v‖ = √49
7 times 7 is 49.
7 × 7 = 49
So ‖u × v‖ = 7.
‖u × v‖ = 7
The triangle is half of the parallelogram.
So we take half.
A = 12 × 7
Half of 7 is 7/2.
A = 72
7 Check
Let's check the cross product: it is at a square corner to u and to v.
Both dot products are 0.
Right!
(u × v) • u = 0, (u × v) • v = 0
And the area is at most half of the two side lengths multiplied, about 3.5.
3.5 fits!
area ≤ ½ × ‖u‖ × ‖v‖ ≈ 3.5
Let's look back.
First: two sides as vectors.
Then: set up the grid.
Then: the x part.
Then: the y part.
Then: the z part.
Then: find the length.
Last: check.
▶ Watch it solved
