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Chapter 10 · Vector Algebra

Exercise 10.4 · Q9: Area of the triangle A(1,1,2)A(1,1,2)⁠, B(2,3,5)B(2,3,5)⁠, C(1,5,5)C(1,5,5)

Exercise 10.4Area of a triangle by the vector product · Section 10.6.3
The question

Find the area of the triangle with vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).

Exam answer AB→=(2−1)i^+(3−1)j^+(5−2)k^=i^+2j^+3k^AC→=(1−1)i^+(5−1)j^+(5−2)k^=4j^+3k^\begin{aligned} \overrightarrow{AB} &= (2 - 1)\hat i + (3 - 1)\hat j \\ &\quad + (5 - 2)\hat k = \hat i + 2\hat j + 3\hat k \\ \overrightarrow{AC} &= (1 - 1)\hat i + (5 - 1)\hat j \\ &\quad + (5 - 2)\hat k = 4\hat j + 3\hat k \end{aligned}

Area of △ABC=12∣AB→×AC→∣\triangle ABC = \frac12|\overrightarrow{AB}\times\overrightarrow{AC}|⁠.

AB→×AC→=∣i^j^k^123043∣=i^(6−12)−j^(3−0)+k^(4−0)=−6i^−3j^+4k^\begin{aligned} \overrightarrow{AB}\times\overrightarrow{AC} &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix} \\ &= \hat i(6 - 12) - \hat j(3 - 0) \\ &\quad + \hat k(4 - 0) \\ &= -6\hat i - 3\hat j + 4\hat k \end{aligned} ∣AB→×AC→∣=36+9+16=61\begin{aligned} |\overrightarrow{AB}\times\overrightarrow{AC}| &= \sqrt{36 + 9 + 16} \\ &= \sqrt{61} \end{aligned}

∴\therefore the required area is 1261\dfrac12\sqrt{61} sq. units.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

If two sides of a triangle from the same vertex are the vectors a⃗\vec a and b⃗\vec b⁠, the triangle is half of the parallelogram on a⃗\vec a and b⃗\vec b⁠, so its area is

12∣a⃗×b⃗∣.\tfrac12|\vec a\times\vec b|.

Here the two sides from A are AB→\overrightarrow{AB} and AC→\overrightarrow{AC}⁠.

Given, to find, and the plan
Given
The vertices A(1, 1, 2), B(2, 3, 5) and C(1, 5, 5).
To find
The area of triangle ABC.
Plan
(1) Find AB→\overrightarrow{AB} and AC→\overrightarrow{AC}⁠. (2) Use area =12∣AB→×AC→∣= \frac12|\overrightarrow{AB}\times\overrightarrow{AC}|⁠. (3) Compute the vector product and check it. (4) Take its length and halve it. (5) Check with the scalar product, and with another vertex.

Basics you need

1Vector joining two points

P1P2→=(x2−x1)i^\overrightarrow{P_1P_2} = (x_2 - x_1)\hat i +(y2−y1)j^+(z2−z1)k^+ (y_2 - y_1)\hat j + (z_2 - z_1)\hat k (Section 10.5.2). Basics: three numbers fix a point

2Area of a triangle

If a⃗\vec a and b⃗\vec b are adjacent sides of a triangle, its area is 12∣a⃗×b⃗∣\frac12|\vec a\times\vec b| (Section 10.6.3, Observation 8). The book works the same pattern in Example 24.

3The determinant form of a⃗×b⃗\vec a\times\vec b

i^,j^,k^\hat i, \hat j, \hat k in the top row, then the components of a⃗\vec a and of b⃗\vec b⁠; expand along the top row with signs +,−,++, -, +⁠. Basics: a determinant measures a signed area

4Magnitude from components

∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

Solution

  1. Two sides from the vertex A

    AB→=(2−1)i^+(3−1)j^+(5−2)k^=i^+2j^+3k^,AC→=(1−1)i^+(5−1)j^+(5−2)k^=4j^+3k^.\begin{aligned} \overrightarrow{AB} &= (2 - 1)\hat i + (3 - 1)\hat j + (5 - 2)\hat k \\ &= \hat i + 2\hat j + 3\hat k, \\ \overrightarrow{AC} &= (1 - 1)\hat i + (5 - 1)\hat j + (5 - 2)\hat k \\ &= 4\hat j + 3\hat k. \end{aligned}
    The formula needs two sides that start at the same vertex. Each side vector is terminal point minus initial point. AC→\overrightarrow{AC} has no i^\hat i part, because A and C have the same x-coordinate; that 0 goes into the determinant.
    P1P2→=OP2→−OP1→\overrightarrow{P_1P_2} = \overrightarrow{OP_2} - \overrightarrow{OP_1} (Section 10.5.2). Basics: three numbers fix a point

  2. Why the area is 12∣AB→×AC→∣\frac12|\overrightarrow{AB}\times\overrightarrow{AC}|

    Take AB as the base. If θ\theta is the angle at A, the height from C to the line AB is ∣AC→∣sin⁡θ|\overrightarrow{AC}|\sin\theta⁠. So

    area=12×∣AB→∣×∣AC→∣sin⁡θ=12∣AB→×AC→∣.\begin{aligned} \text{area} &= \tfrac12 \times |\overrightarrow{AB}| \times |\overrightarrow{AC}|\sin\theta \\ &= \tfrac12|\overrightarrow{AB}\times\overrightarrow{AC}|. \end{aligned}
    The length of a vector product is ∣a⃗∣∣b⃗∣sin⁡θ|\vec a||\vec b|\sin\theta⁠, which is base times height of the parallelogram on a⃗\vec a and b⃗\vec b⁠; the triangle is exactly half of that parallelogram.
    Area of a triangle =12∣a⃗×b⃗∣= \frac12|\vec a\times\vec b| (Section 10.6.3, Observation 8). Lesson: the cross product

  3. The vector product

    AB→×AC→=∣i^j^k^123043∣=i^(6−12)−j^(3−0)+k^(4−0)=−6i^−3j^+4k^.\begin{aligned} \overrightarrow{AB}\times\overrightarrow{AC} &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 2 & 3 \\ 0 & 4 & 3 \end{vmatrix} \\ &= \hat i(6 - 12) - \hat j(3 - 0) \\ &\quad + \hat k(4 - 0) \\ &= -6\hat i - 3\hat j + 4\hat k. \end{aligned}
    The minors are (2)(3)−(3)(4)=6−12(2)(3) - (3)(4) = 6 - 12⁠, (1)(3)−(3)(0)=3−0(1)(3) - (3)(0) = 3 - 0 and (1)(4)−(2)(0)=4−0(1)(4) - (2)(0) = 4 - 0⁠; the j^\hat j term takes a minus sign, giving −3j^-3\hat j⁠. As a check, (−6i^−3j^+4k^)⋅(i^+2j^+3k^)=−6−6+12=0(-6\hat i - 3\hat j + 4\hat k)\cdot(\hat i + 2\hat j + 3\hat k) = -6 - 6 + 12 = 0 and (−6i^−3j^+4k^)⋅(4j^+3k^)=−12+12=0(-6\hat i - 3\hat j + 4\hat k)\cdot(4\hat j + 3\hat k) = -12 + 12 = 0⁠.
    Expansion along the top row, signs +,−,++, -, +⁠. Basics: a determinant measures a signed area

  4. Its length, and the area

    ∣AB→×AC→∣=(−6)2+(−3)2+42=36+9+16=61,\begin{aligned} |\overrightarrow{AB}\times\overrightarrow{AC}| &= \sqrt{(-6)^2 + (-3)^2 + 4^2} \\ &= \sqrt{36 + 9 + 16} = \sqrt{61}, \end{aligned} area of △ABC=1261 sq. units.\text{area of } \triangle ABC = \tfrac12\sqrt{61}\ \text{sq. units}.
    Squaring removes the signs. 61 is prime, so 61\sqrt{61} does not simplify. Halving is the step that turns the parallelogram into the triangle.
    ∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

  5. Two independent checks

    By the scalar product: ∣AB→∣2=1+4+9=14|\overrightarrow{AB}|^2 = 1 + 4 + 9 = 14⁠, ∣AC→∣2=16+9=25|\overrightarrow{AC}|^2 = 16 + 9 = 25⁠, AB→⋅AC→=0+8+9=17\overrightarrow{AB}\cdot\overrightarrow{AC} = 0 + 8 + 9 = 17⁠, so

    ∣AB→×AC→∣2=∣AB→∣2∣AC→∣2−(AB→⋅AC→)2=14×25−172=350−289=61.\begin{aligned} &|\overrightarrow{AB}\times\overrightarrow{AC}|^2 \\ &= |\overrightarrow{AB}|^2|\overrightarrow{AC}|^2 \\ &\quad - (\overrightarrow{AB}\cdot\overrightarrow{AC})^2 \\ &= 14 \times 25 - 17^2 \\ &= 350 - 289 = 61. \end{aligned}

    From the vertex B: BA→=−i^−2j^−3k^\overrightarrow{BA} = -\hat i - 2\hat j - 3\hat k and BC→=−i^+2j^\overrightarrow{BC} = -\hat i + 2\hat j⁠, and BA→×BC→=6i^+3j^−4k^\overrightarrow{BA}\times\overrightarrow{BC} = 6\hat i + 3\hat j - 4\hat k⁠, again of length 61\sqrt{61}⁠.

    A(1, 1, 2) B(2, 3, 5) C(1, 5, 5) √14 5 √5 height area = ½|AB × AC| = √61/2 ≈ 3.91 AB · AC = 17
    Triangle ABC drawn in its own plane, true to scale: AB=14\text{AB} = \sqrt{14}⁠, AC=5\text{AC} = 5⁠, BC=5\text{BC} = \sqrt5⁠. The height from C to the line AB meets it beyond B, because the angle at B is obtuse; the area is 1261≈3.91\frac12\sqrt{61} \approx 3.91⁠.
    The first check uses only scalar products, so it is independent of the determinant. The second uses two other sides; any two sides from one vertex give the same area, and here the vector product comes out as the negative of the first one, with the same length.
    ∣a⃗×b⃗∣2=∣a⃗∣2∣b⃗∣2−(a⃗⋅b⃗)2|\vec a\times\vec b|^2 = |\vec a|^2|\vec b|^2 - (\vec a\cdot\vec b)^2⁠, from sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 (Sections 10.6.1 and 10.6.3). Lesson: Examples 22–25
Result

The area of triangle ABC is 1261\frac12\sqrt{61} sq. units.

(Textbook answer: 612\frac{\sqrt{61}}{2}⁠.)

Check it with numbers

61≈7.810250\sqrt{61} \approx 7.810250⁠, so the area is 1261≈3.905125\frac12\sqrt{61} \approx 3.905125⁠.

Base and height: ∣AB→∣=14≈3.741657|\overrightarrow{AB}| = \sqrt{14} \approx 3.741657 and the height from C is 6114≈2.087377\frac{\sqrt{61}}{\sqrt{14}} \approx 2.087377⁠; 12×3.741657×2.087377≈3.905124\frac12 \times 3.741657 \times 2.087377 \approx 3.905124⁠, the same area up to the rounding of the two factors (exactly, 12×14×6114=1261\frac12 \times \sqrt{14} \times \frac{\sqrt{61}}{\sqrt{14}} = \frac12\sqrt{61}⁠).

The angle at A: cos⁡A=17350≈0.908688\cos A = \frac{17}{\sqrt{350}} \approx 0.908688⁠, so A≈24.68∘A \approx 24.68^\circ⁠.

Common mistakes
  1. Forgetting the 12\frac12⁠. 61\sqrt{61} is the area of the parallelogram on AB→\overrightarrow{AB} and AC→\overrightarrow{AC}⁠, twice the triangle.
  2. Crossing the position vectors. OA→×OB→\overrightarrow{OA}\times\overrightarrow{OB} belongs to the triangle OAB; the sides of ABC are AB→\overrightarrow{AB} and AC→\overrightarrow{AC}⁠.
  3. Losing the 0 in AC→\overrightarrow{AC}⁠. AC→=0i^+4j^+3k^\overrightarrow{AC} = 0\hat i + 4\hat j + 3\hat k⁠; its bottom row in the determinant is 0,4,30, 4, 3⁠, not 4,3,04, 3, 0⁠.
What to take away
  1. Area of a triangle ABC =12∣AB→×AC→∣= \frac12|\overrightarrow{AB}\times\overrightarrow{AC}|⁠, using two sides from one vertex.
  2. Check the vector product by dotting it with both sides: both must give 0.
  3. ∣a⃗∣2∣b⃗∣2−(a⃗⋅b⃗)2|\vec a|^2|\vec b|^2 - (\vec a\cdot\vec b)^2 gives the same squared length without a determinant.

Every fact used, and where it comes from

FactUsed inSource
Vector joining two pointsSteps 1, 5Section 10.5.2 (Lesson: components)
Area of a triangle =12∣a⃗×b⃗∣= \frac12|\vec a\times\vec b|⁠; ∣a⃗×b⃗∣=∣a⃗∣∣b⃗∣sin⁡θ|\vec a\times\vec b| = |\vec a||\vec b|\sin\thetaStep 2Section 10.6.3, Definition 3 and Observation 8, Fig 10.26
a⃗×b⃗\vec a\times\vec b as a determinant in componentsSteps 3, 5Section 10.6.3 (Basics: a determinant measures a signed area)
∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}Steps 4, 5Section 10.5.1
Scalar product in components; zero for perpendicular vectorsSteps 3, 5Section 10.6.1, Observation 2
The question and the answerResultExercise 10.4 Q9; NCERT answer key (Lesson: the cross product, exercises)
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