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Practice: Cross Product (2)
Related Topics
Wize University Linear Algebra Textbook > Products of Vectors
Cross Product and Area
3 Activities
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).
X
=
(
0
,
0
,
0
)
,
Y
=
(
1
,
1
,
0
)
,
&
Z
=
(
0
,
1
,
1
)
.
.
What is the area of the triangle XYZ?
Answer
I don't know
Check Submission
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)
M
(
−
3
,
−
1
,
0
)
,
N
(
3
,
1
,
−
1
)
N(3,1,-1)
N
(
3
,
1
,
−
1
)
, and
P
(
−
2
,
−
1
,
0
)
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
𝑢
⃗
=
(
−
1
,
0
,
2
)
𝑢⃗ = (−1,0,2)
u
⃗
=
(
−
1
,
0
,
2
)
and
𝑣
=
(
3
,
−
3
,
0
)
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),}
(
2
,
−
3
)
,
(
0
,
2
)
,
(
13
,
−
14
)
,
and
(
11
,
−
9
)
\bcb{(11,-9)}
(
11
,
−
9
)
.
133 - FML 3 - 18.1W e.g. 27
Find the area of the triangle defined by the vectors
[
5
7
]
\bcb{\boldsymbol{ \begin{bmatrix} 5 \\ 7 \end{bmatrix}}}
[
5
7
]
and
[
3
1
]
\bcb{\boldsymbol{ \begin{bmatrix} 3 \\ 1 \end{bmatrix}}}
[
3
1
]
.
133 - FML 3 - 18.1W e.g. 39.1
Consider the points
P
1
=
(
1
,
4
,
1
)
\bcb{P_1 = (1,4,1)}
P
1
=
(
1
,
4
,
1
)
,
P
2
=
(
3
,
4
,
−
1
)
\bcb{P_2 = (3,4,-1)}
P
2
=
(
3
,
4
,
−
1
)
,
P
3
=
(
2
,
4
,
−
2
)
\bcb{P_3 = (2,4,-2)}
P
3
=
(
2
,
4
,
−
2
)
,
P
4
=
(
0
,
2
,
0
)
\bcb{P_4 = (0,2,0)}
P
4
=
(
0
,
2
,
0
)
.
Find the area of the triangle with vertices
P
1
,
P
2
,
P
3
\bcb{P_1,\, P_2,\, P_3}
P
1
,
P
2
,
P
3
.
$\tkct{cut from 19.4F}$ Mid $\tkco{ S}$ | 133 - FML 3 - 18.1W e.g. 39.1_$\tkcth{Mock F}\tkct{?}$_$\tkct{no vid / soln }$
Consider the points
P
1
=
(
1
,
4
,
1
)
\bcb{P_1 = (1,4,1)}
P
1
=
(
1
,
4
,
1
)
,
P
2
=
(
3
,
4
,
−
1
)
\bcb{P_2 = (3,4,-1)}
P
2
=
(
3
,
4
,
−
1
)
,
P
3
=
(
2
,
4
,
−
2
)
\bcb{P_3 = (2,4,-2)}
P
3
=
(
2
,
4
,
−
2
)
,
P
4
=
(
0
,
2
,
0
)
\bcb{P_4 = (0,2,0)}
P
4
=
(
0
,
2
,
0
)
.
Find the area of the triangle with vertices
P
1
,
P
2
,
P
3
\bcb{P_1,\, P_2,\, P_3}
P
1
,
P
2
,
P
3
.
A parallelogram in
R
3
R^3
R
3
has vertices
A
(
−
1
,
1
,
2
)
A\left(-1,1,2\right)
A
(
−
1
,
1
,
2
)
,
B
(
1
,
−
2
,
8
)
B\left(1,-2,8\right)
B
(
1
,
−
2
,
8
)
and
C
=
(
3
,
−
1
,
2
)
C=\left(3,-1,2\right)
C
=
(
3
,
−
1
,
2
)
.
Additional practice problems--vector products
Additional Practice Problems--Vector Products
1. If
u
⃗
=
(
−
2
,
3
,
−
1
)
,
v
⃗
=
(
−
1
,
−
2
,
3
)
\vec{u}=\left(-2,3,-1\right),\ \ \vec{v}=\left(-1,-2,3\right)
u
=
(
−
2
,
3
,
−
1
)
,
v
=
(
−
1
,
−
2
,
3
)
, find
a)
u
⃗
∙
v
⃗
\vec{u}\ \bullet\ \vec{v}
u
∙
v
Find the area of the parallelogram defined by the vectors
u
→
=
(
1
,
3
,
0
)
\overrightarrow{u}=\left(1,3,0\right)
u
=
(
1
,
3
,
0
)
and
v
→
=
(
−
2
,
0
,
1
)
\overrightarrow{v}=\left(-2,0,1\right)
v
=
(
−
2
,
0
,
1
)
.
Cross Product and Area
Find the area of the parallelogram ABCD that has vertices
A
(
1
,
2
,
5
)
A(1,2,5)
A
(
1
,
2
,
5
)
,
B
(
−
3
,
0
,
2
)
B(-3,0,2)
B
(
−
3
,
0
,
2
)
,
C
(
−
3
,
−
5
,
−
2
)
C(-3,-5,-2)
C
(
−
3
,
−
5
,
−
2
)
and
D
(
1
,
−
3
,
1
)
D(1,-3,1)
D
(
1
,
−
3
,
1
)
.