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Wize University Linear Algebra Textbook > Products of Vectors

Cross Product and Area

Cross Product PropertiesPrevious Section
Volume of a Parallelepiped (Triple Product)Next Section
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Cross Product and Area

Parallelograms

Area▱=∥BA→×BC→∥\boxed{\quad \text{Area}_\text{▱} = \lVert \overrightarrow{BA} \times \overrightarrow{BC} \rVert \quad}
Find the cross product of two adjacent vectors (starting from the same point) in R3\reals^3, then take the norm of the resulting vector.

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Triangles

Area△=12∥BA→×BC→∥\boxed{\quad \text{Area}_\text{△} = \frac{1}{2} \lVert \overrightarrow{BA} \times \overrightarrow{BC} \rVert \quad}
Find the cross product of the two vectors in R3\reals^3, then take half of the norm of the resulting vector.


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Example: Cross Product and Area

Find the area of the triangle that has vertices 𝑃(1,2,3)𝑃(1,2,3), Q(−1,0,2)Q(−1,0,2), and 𝑅(5,1,1)𝑅(5,1,1).
Each axis is measured in millimetres; don't forget units!


PQ→=[−102]−[123]=[−2−2−1]\overrightarrow{PQ} = \begin{bmatrix} -1\\ 0\\ 2\\ \end{bmatrix} - \begin{bmatrix} 1\\ 2\\ 3\\ \end{bmatrix} =\begin{bmatrix} -2\\ -2\\ -1\\ \end{bmatrix}

PR→=[511]−[123]=[4−1−2]\overrightarrow{PR} = \begin{bmatrix} 5\\ 1\\ 1\\ \end{bmatrix} - \begin{bmatrix} 1\\ 2\\ 3\\ \end{bmatrix} = \begin{bmatrix} 4\\ -1\\ -2\\ \end{bmatrix}

PQ→×PR→=[(−2)(−2) − (−1)(−1)(−1)(4) − (−2)(−2)(−2)(−1) − (−2)(4)]=[3−810]\overrightarrow{PQ} \times \overrightarrow{PR} = \begin{bmatrix} (−2)(−2) \ −\ (−1)(−1)\\[0.5em] (−1)(4) \ −\ (−2)(−2)\\[0.5em] (−2)(−1) \ −\ (−2)(4)\\ \end{bmatrix} = \begin{bmatrix} 3\\ -8\\ 10\\ \end{bmatrix}

We can now calculate the area of the triangle:

Area△=12∥PQ→×PR→∥=1232+(−8)2+102=12173≈6.576 mm3\begin{aligned} \text{Area}_\text{△} &=\frac{1}{2}\lVert\overrightarrow{PQ} \times \overrightarrow{PR}\rVert\\[1em] &= \frac{1}{2}\sqrt{3^2+(−8)^2+10^2}\\[1em] &= \boxed{\frac{1}{2}\sqrt{173} \approx 6.576 \ \rm{mm}^3} \end{aligned}

Practice: Cross Product and Area

Given the points A(−3,−1,0)A(-3,-1,0), B(1,2,2)B(1,2,2), C(3,1,2)C(3,1,2), answer the following questions.

Extra Practice
Practice: Cross Product and Areas
Find the area of the triangle with vertices M(−3,−1,0)M(-3,-1,0), N(3,1,−1)N(3,1,-1), and P(−2,−1,0)P(-2,-1,0)
Practice Question: Application of Cross Product
Practice Question: Application of Cross Product
Find the area of the triangle that has edges defined by 𝑢⃗=(−1,0,2)𝑢⃗ = (−1,0,2) and 𝑣=(3,−3,0)𝑣=(3,−3,0)
133 - FML 3 - 18.1W e.g. 28
Find the area of the parallelogram with vertices at (2,−3), (0,2), (13,−14),\bcb{(2,-3),~(0,2),~(13,-14),} and (11,−9)\bcb{(11,-9)}.
133 - FML 3 - 18.1W e.g. 27
Find the area of the triangle defined by the vectors [57]\bcb{\boldsymbol{ \begin{bmatrix} 5 \\ 7 \end{bmatrix}}} and [31]\bcb{\boldsymbol{ \begin{bmatrix} 3 \\ 1 \end{bmatrix}}}.
133 - FML 3 - 18.1W e.g. 39.1
Consider the points P1=(1,4,1)\bcb{P_1 = (1,4,1)}, P2=(3,4,−1)\bcb{P_2 = (3,4,-1)}, P3=(2,4,−2)\bcb{P_3 = (2,4,-2)}, P4=(0,2,0)\bcb{P_4 = (0,2,0)}.
Find the area of the triangle with vertices P1, P2, P3\bcb{P_1,\, P_2,\, P_3}.
Practice: Cross Product (2)
Let X=(0, 0, 0), Y=(1, 1, 0), & Z=(0, 1, 1).X=(0,~0,~0),~Y=(1,~1,~0),~\&~Z=(0,~1,~1)..
What is the area of the triangle XYZ?
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 39.1_$\tkcth{Mock F}\tkct{?}$_$\tkct{no vid / soln }$
Consider the points P1=(1,4,1)\bcb{P_1 = (1,4,1)}, P2=(3,4,−1)\bcb{P_2 = (3,4,-1)}, P3=(2,4,−2)\bcb{P_3 = (2,4,-2)}, P4=(0,2,0)\bcb{P_4 = (0,2,0)}.
Find the area of the triangle with vertices P1, P2, P3\bcb{P_1,\, P_2,\, P_3}.
A parallelogram in R3R^3 has vertices A(−1,1,2)A\left(-1,1,2\right), B(1,−2,8)B\left(1,-2,8\right) and C=(3,−1,2)C=\left(3,-1,2\right).
Additional practice problems--vector products
Additional Practice Problems--Vector Products
1. If u⃗=(−2,3,−1),  v⃗=(−1,−2,3)\vec{u}=\left(-2,3,-1\right),\ \ \vec{v}=\left(-1,-2,3\right), find
a) u⃗ ∙ v⃗\vec{u}\ \bullet\ \vec{v}
Find the area of the parallelogram defined by the vectors u→=(1,3,0)\overrightarrow{u}=\left(1,3,0\right)and v→=(−2,0,1)\overrightarrow{v}=\left(-2,0,1\right).
Cross Product and Area
Find the area of the parallelogram ABCD that has vertices A(1,2,5)A(1,2,5), B(−3,0,2)B(-3,0,2), C(−3,−5,−2)C(-3,-5,-2) and D(1,−3,1)D(1,-3,1).