Difference between revisions of "Selftest: Dot product"

@@ Line 10: / Line 10: @@
 - 1,5
 + 15
-||The result of the dot product is a ''scalar'' value. In this case the easiest way to compute the dot product is using the angle between the two vectors:<math>\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} =  a b \cos \alpha</math>. The angle is 0° and so the cosine is 1. Further information: see [[Dot product]].
+||The result of the dot product is a ''scalar'' value. In this case the easiest way to compute the dot product is using the angle between the two vectors: <math>\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} =  a b \cos \alpha</math>. The angle is 0° and so the cosine is 1. Further information: see [[Dot product]].
 {'''What is the result of dor product of <math>\vec{\mathbf{a}}</math> and <math>\vec{\mathbf{b}}</math>?'''
@@ Line 19: / Line 19: @@
 - 15
 +  0
-||The result of the dot product is a ''scalar'' value. In this case the easiest way to compute the dot product is using the angle between the two vectors:<math>\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} =  a b \cos \alpha</math>. The angle is 90° and so the cosine is 0. Further information: see [[Dot product]].
+||The result of the dot product is a ''scalar'' value. In this case the easiest way to compute the dot product is using the angle between the two vectors: <math>\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} =  a b \cos \alpha</math>. The angle is 90° and so the cosine is 0. Further information: see [[Dot product]].
@@ Line 31: / Line 31: @@
 + -15
 - 15
-||The result of the dot product is a ''scalar'' value. In this case the easiest way to compute the dot product is using the angle between the two vectors:<math>\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} =  a b \cos \alpha</math>. The angle is 180° and so the cosine is -1. Further information: see [[Dot product]].
+||The result of the dot product is a ''scalar'' value. In this case the easiest way to compute the dot product is using the angle between the two vectors: <math>\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} =  a b \cos \alpha</math>. The angle is 180° and so the cosine is -1. Further information: see [[Dot product]].
@@ Line 38: / Line 38: @@
-{'''Bitte lösen Sie folgende Aufgabe:'''
+{'''Please solve the following exercise:'''
 | type="{}" }
 <math>\begin{pmatrix} 2 \\ 3  \\ 5 \end{pmatrix}\cdot \begin{pmatrix} -3 \\ 6  \\ 4 \end{pmatrix}=</math>{ 32 }
-||Es gibt zwei Möglichkeiten zur Berechnung des Skalarprodukts. Entweder berechnet man es mit Hilfe der Komponentendarstellung:<math>\vec{\mathbf{a}}\cdot\vec{\mathbf{b}} = \begin{pmatrix}a_1 \\ a_2 \\ a_3 \end{pmatrix}\cdot\begin{pmatrix}b_1 \\ b_2 \\ b_3 \end{pmatrix}=a_1b_1+a_2b_2+a_3b_3</math> oder man nutzt den eingeschlossenen Winkel: <math>\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = a \vec{\mathbf{e}}_{a} \cdot \vec{\mathbf{e}}_{b} = a b \cos \alpha </math>.
+||There are two possibilities to compute the dot product. Either the component representation <math>\vec{\mathbf{a}}\cdot\vec{\mathbf{b}} = \begin{pmatrix}a_1 \\ a_2 \\ a_3 \end{pmatrix}\cdot\begin{pmatrix}b_1 \\ b_2 \\ b_3 \end{pmatrix}=a_1b_1+a_2b_2+a_3b_3</math> is used or the angle between the vectors: <math>\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = a \vec{\mathbf{e}}_{a} \cdot \vec{\mathbf{e}}_{b} = a b \cos \alpha </math>.
-{'''Bitte lösen Sie folgende Aufgabe:'''
+{'''Please solve the following exercise:'''
 | type="{}" }
 <math>\begin{pmatrix} 1 \\ 2  \\ 3 \end{pmatrix}\cdot \begin{pmatrix} -4 \\ 1  \\ -6 \end{pmatrix}=</math>{ -20 }
-||Es gibt zwei Möglichkeiten zur Berechnung des Skalarprodukts. Entweder berechnet man es mit Hilfe der Komponentendarstellung:<math>\vec{\mathbf{a}}\cdot\vec{\mathbf{b}} = \begin{pmatrix}a_1 \\ a_2 \\ a_3 \end{pmatrix}\cdot\begin{pmatrix}b_1 \\ b_2 \\ b_3 \end{pmatrix}=a_1b_1+a_2b_2+a_3b_3</math> oder man nutzt den eingeschlossenen Winkel: <math>\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = a \vec{\mathbf{e}}_{a} \cdot \vec{\mathbf{e}}_{b} = ab \cos \alpha </math>.
+||There are two possibilities to compute the dot product. Either the component representation <math>\vec{\mathbf{a}}\cdot\vec{\mathbf{b}} = \begin{pmatrix}a_1 \\ a_2 \\ a_3 \end{pmatrix}\cdot\begin{pmatrix}b_1 \\ b_2 \\ b_3 \end{pmatrix}=a_1b_1+a_2b_2+a_3b_3</math> is used or the angle between the vectors: <math>\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = a \vec{\mathbf{e}}_{a} \cdot \vec{\mathbf{e}}_{b} = a b \cos \alpha </math>.
-{'''Bitte lösen Sie folgende Aufgabe:'''
+{'''Please solve the following exercise:'''
 | type="{}" }
 <math>\begin{pmatrix} 0 \\ 1  \\ 0 \end{pmatrix}\cdot \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}=</math> { 1 }
-||Es gibt zwei Möglichkeiten zur Berechnung des Skalarprodukts. Entweder berechnet man es mit Hilfe der Komponentendarstellung:<math>\vec{\mathbf{a}}\cdot\vec{\mathbf{b}} = \begin{pmatrix}a_1 \\ a_2 \\ a_3 \end{pmatrix}\cdot\begin{pmatrix}b_1 \\ b_2 \\ b_3 \end{pmatrix}=a_1b_1+a_2b_2+a_3b_3</math> oder man nutzt den eingeschlossenen Winkel: <math>\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = a \vec{\mathbf{e}}_{a} \cdot \vec{\mathbf{e}}_{b} = a b \cos \alpha </math>.
+||There are two possibilities to compute the dot product. Either the component representation <math>\vec{\mathbf{a}}\cdot\vec{\mathbf{b}} = \begin{pmatrix}a_1 \\ a_2 \\ a_3 \end{pmatrix}\cdot\begin{pmatrix}b_1 \\ b_2 \\ b_3 \end{pmatrix}=a_1b_1+a_2b_2+a_3b_3</math> is used or the angle between the vectors: <math>\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = a \vec{\mathbf{e}}_{a} \cdot \vec{\mathbf{e}}_{b} = a b \cos \alpha </math>.
 </quiz>
 [[Category:Selftest]]
 [[Category:Vectors]]

Revision as of 16:30, 23 May 2014

← Previous exercise: Simple arithmetic operations

Exercises for chapter {{{chapter}}} | Article: Vector algebra

Next exercise: Cross product →

Point added for a correct answer:
Points for a wrong answer:
Ignore the questions' coefficients:

1. What is the result of dor product of $\vec{\mathbf{a}}$ and $\vec{\mathbf{b}}$ ?

	$\begin{pmatrix} 3 \\ 5 \end{pmatrix}$
	1,5
	15
→	The result of the dot product is a scalar value. In this case the easiest way to compute the dot product is using the angle between the two vectors: $\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = a b \cos \alpha$ . The angle is 0° and so the cosine is 1. Further information: see Dot product.

2. What is the result of dor product of $\vec{\mathbf{a}}$ and $\vec{\mathbf{b}}$ ?

	$\begin{pmatrix} 5 \\ 3 \end{pmatrix}$
	15
	0
→	The result of the dot product is a scalar value. In this case the easiest way to compute the dot product is using the angle between the two vectors: $\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = a b \cos \alpha$ . The angle is 90° and so the cosine is 0. Further information: see Dot product.

3. What is the result of dor product of $\vec{\mathbf{a}}$ and $\vec{\mathbf{b}}$ ?

	$\begin{pmatrix} 5 \\ 3 \end{pmatrix}$
	-15
	15
→	The result of the dot product is a scalar value. In this case the easiest way to compute the dot product is using the angle between the two vectors: $\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = a b \cos \alpha$ . The angle is 180° and so the cosine is -1. Further information: see Dot product.

4. Please solve the following exercise:

\begin{pmatrix} 2 \\ 3 \\ 5 \end{pmatrix}\cdot \begin{pmatrix} -3 \\ 6 \\ 4 \end{pmatrix}=

→ There are two possibilities to compute the dot product. Either the component representation

\vec{\mathbf{a}}\cdot\vec{\mathbf{b}} = \begin{pmatrix}a_1 \\ a_2 \\ a_3 \end{pmatrix}\cdot\begin{pmatrix}b_1 \\ b_2 \\ b_3 \end{pmatrix}=a_1b_1+a_2b_2+a_3b_3

is used or the angle between the vectors:

\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = a \vec{\mathbf{e}}_{a} \cdot \vec{\mathbf{e}}_{b} = a b \cos \alpha

5. Please solve the following exercise:

\begin{pmatrix} 1 \\ 2 \\ 3 \end{pmatrix}\cdot \begin{pmatrix} -4 \\ 1 \\ -6 \end{pmatrix}=

→ There are two possibilities to compute the dot product. Either the component representation

\vec{\mathbf{a}}\cdot\vec{\mathbf{b}} = \begin{pmatrix}a_1 \\ a_2 \\ a_3 \end{pmatrix}\cdot\begin{pmatrix}b_1 \\ b_2 \\ b_3 \end{pmatrix}=a_1b_1+a_2b_2+a_3b_3

is used or the angle between the vectors:

\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = a \vec{\mathbf{e}}_{a} \cdot \vec{\mathbf{e}}_{b} = a b \cos \alpha

6. Please solve the following exercise:

\begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}\cdot \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}=

→ There are two possibilities to compute the dot product. Either the component representation

\vec{\mathbf{a}}\cdot\vec{\mathbf{b}} = \begin{pmatrix}a_1 \\ a_2 \\ a_3 \end{pmatrix}\cdot\begin{pmatrix}b_1 \\ b_2 \\ b_3 \end{pmatrix}=a_1b_1+a_2b_2+a_3b_3

is used or the angle between the vectors:

\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = a \vec{\mathbf{e}}_{a} \cdot \vec{\mathbf{e}}_{b} = a b \cos \alpha

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Difference between revisions of "Selftest: Dot product"

Revision as of 16:30, 23 May 2014

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