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133 - FML 3 - 18.1W e.g. 39.1

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Wize University Linear Algebra Textbook > Products of Vectors

Cross Product and Area

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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}.




More Cross Product and Area Questions:
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}}}.
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).