Cross Product

## **Problem**

Consider the two vectors

A

=
⟨
1
,
2
,
3
⟩

{\displaystyle {\boldsymbol {A}}=\langle 1,2,3\rangle }

 and

B

=
⟨
2
,
0
,
−
1
⟩

{\displaystyle {\boldsymbol {B}}=\langle 2,0,-1\rangle }

.

**Part 1:** Calculate the cross product

A
×
B

{\displaystyle {\boldsymbol {A\times B}}}

.

**Part 2:** Determine the norm of

A
×
B

{\displaystyle {\boldsymbol {A\times B}}}

, then calculate angle between

A

{\displaystyle \boldsymbol{A}}

 and

B

{\displaystyle \boldsymbol{B}}

.

**Part 3:** Show that the cross-product

A
×
B

{\displaystyle {\boldsymbol {A\times B}}}

 is orthogonal to

A

{\displaystyle \boldsymbol{A}}

 and

B

{\displaystyle \boldsymbol{B}}

.

## **Solution**

**Part 1**

A
×
B

=

i

j

k

1

2

3

2

0

−
1

{\displaystyle {\boldsymbol {A\times B}}={\begin{vmatrix}{\boldsymbol {i}}&{\boldsymbol {j}}&{\boldsymbol {k}}\\1&2&3\\2&0&-1\end{vmatrix}}}

A
×
B

=

2

3

0

−
1

i

−

1

3

2

−
1

j

1

2

2

0

k

{\displaystyle {\boldsymbol {A\times B}}={\begin{vmatrix}2&3\\0&-1\end{vmatrix}}{\boldsymbol {i}}-{\begin{vmatrix}1&3\\2&-1\end{vmatrix}}{\boldsymbol {j}}+{\begin{vmatrix}1&2\\2&0\end{vmatrix}}{\boldsymbol {k}}}

A
×
B

=
(
2
(
−
1
)
−
3
(
0
)
)

i

−
(
1
(
−
1
)
−
3
(
2
)
)

j

(
1
(
0
)
−
2
(
2
)
)

k

{\displaystyle {\boldsymbol {A\times B}}=(2(-1)-3(0)){\boldsymbol {i}}-(1(-1)-3(2)){\boldsymbol {j}}+(1(0)-2(2)){\boldsymbol {k}}}

A
×
B

=
−
2

i

7

j

−
4

k

{\displaystyle {\boldsymbol {A\times B}}=-2{\boldsymbol {i}}+7{\boldsymbol {j}}-4{\boldsymbol {k}}}

**Part 2**

The norm of the cross product of two vectors is the area of the parallelogram with side lengths of the norms of

A

{\displaystyle \boldsymbol{A}}

 and

B

{\displaystyle \boldsymbol{B}}

.

A
×
B

=

(
−
2

)

2

(
7

)

2

(
−
4

)

2

{\displaystyle |{\boldsymbol {A\times B}}|={\sqrt {(-2)^{2}+(7)^{2}+(-4)^{2}}}}

A
×
B

=

69

{\displaystyle |{\boldsymbol {A\times B}}|={\sqrt {69}}}

To determine the angle

θ

{\displaystyle \theta}

 between the vectors

A

{\displaystyle \boldsymbol{A}}

 and

B

{\displaystyle \boldsymbol{B}}

, use the formula

A
×
B

=

A

B

sin
⁡

θ

.

A

=

(
1

)

2

(
2

)

2

(
3

)

2

=

14

{\displaystyle |{\boldsymbol {A}}|={\sqrt {(1)^{2}+(2)^{2}+(3)^{2}}}={\sqrt {14}}}

B

=

(
2

)

2

(
0

)

2

(
−
1

)

2

=

5

{\displaystyle |{\boldsymbol {B}}|={\sqrt {(2)^{2}+(0)^{2}+(-1)^{2}}}={\sqrt {5}}}

Thus,

sin
⁡

θ

=

69

14

5

{\displaystyle \sin {\theta }={\frac {\sqrt {69}}{{\sqrt {14}}{\sqrt {5}}}}}

# θ

sin

−
1

⁡

(

69

70

)

{\displaystyle \theta =\sin ^{-1}{\left({\frac {\sqrt {69}}{\sqrt {70}}}\right)}}

# θ

83.1354

∘

{\displaystyle \theta =83.1354^{\circ }}

**Part 3**

The fastest way to show that two vectors are orthogonal (perpendicular) is to take the dot-product. If the dot-product is zero, then the two vectors are orthogonal. This is the basis of the right-hand rule.

Let

A
×
B

=

C

=
⟨
−
2
,
7
,
−
4
⟩

{\displaystyle {\boldsymbol {A\times B}}={\boldsymbol {C}}=\langle -2,7,-4\rangle }

,

A

=
⟨
1
,
2
,
3
⟩

{\displaystyle {\boldsymbol {A}}=\langle 1,2,3\rangle }

, and

B

=
⟨
2
,
0
,
−
1
⟩

{\displaystyle {\boldsymbol {B}}=\langle 2,0,-1\rangle }

.

C
⋅
A

=
(
−
2
)
(
1
)
(
7
)
(
2
)
(
−
4
)
(
3
# )
0

{\displaystyle {\boldsymbol {C\cdot A}}=(-2)(1)+(7)(2)+(-4)(3)=0}

C
⋅
B

=
(
−
2
)
(
2
)
(
7
)
(
0
)
(
−
4
)
(
−
1
# )
0

{\displaystyle {\boldsymbol {C\cdot B}}=(-2)(2)+(7)(0)+(-4)(-1)=0}