Area of the triangle with corners (1, 1, 1), (2, 3, 1), (1, 2, 3) Step by Step

Step by step

The question

Question: find the area of the triangle with corners (1, 1, 1), (2, 3, 1), (1, 2, 3).

P(1, 1, 1), Q(2, 3, 1), R(1, 2, 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 = ⟨0, 1, 2⟩.

v = PR = ⟨0, 1, 2⟩

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; 0, 1, 2|

3 The x part

Cover the i column.

Four numbers are left: 2, 0, 1, 2.

Write them as a small grid.

Its cross: down 2 times 2, minus up 0 times 1.

x part = |2, 0; 1, 2| = 2 × 2 − 0 × 1

4 minus 0 is 4.

x part = 4 − 0 = 4

4 The y part

Now cover the j column.

Four numbers are left: 1, 0, 0, 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 0 times 0.

y part = −|1, 0; 0, 2| = −(1 × 2 − 0 × 0)

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 0 is 2.

With the minus in front: −2.

y part = −(2 − 0) = −2

5 The z part

Now cover the k column.

Four numbers are left: 1, 2, 0, 1.

Write them as a small grid.

Its cross: down 1 times 1, minus up 2 times 0.

z part = |1, 2; 0, 1| = 1 × 1 − 2 × 0

1 minus 0 is 1.

z part = 1 − 0 = 1

Put the three parts together: u × v = ⟨4, −2, 1⟩.

u × v = ⟨4, −2, 1⟩

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‖ = √4² + (−2)² + 1²

4 squared is 16.

(−2) squared is 4.

1 squared is 1.

Minus times minus is plus.

‖u × v‖ = √16 + 4 + 1

Add: 16 plus 4 plus 1 is 21.

‖u × v‖ = √21

21 is not a square number.

So √21 is the exact length.

‖u × v‖ = √21

The triangle is half of the parallelogram.

So we take half.

A = 12 × √21

So the exact area is √21/2.

A = √212

With a calculator, that is about 2.3.

A ≈ 2.3

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 2.5.

2.3 fits!

area ≤ ½ × ‖u‖ × ‖v‖ ≈ 2.5

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.

Answer