Vectoren 101
Gegeven is de vector \(\overrightarrow{a} = \begin{pmatrix}3 \\ 5\end{pmatrix} \text{.}\)
1p
Bereken de lengte van \(\overrightarrow{a} \text{.}\)
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\(\begin{vmatrix}\overrightarrow{a}\end{vmatrix} = \sqrt{3^{2} + 5^{2}} = \sqrt{34} \text{.}\)
1p
Gegeven is de vector \(\overrightarrow{a} = \begin{pmatrix}2 \\ -6\end{pmatrix} \text{.}\)
1p
Bereken de tegengestelde vector van \(\overrightarrow{a} \text{.}\)
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\(-\overrightarrow{a} = \begin{pmatrix}-2 \\ 6\end{pmatrix} \text{.}\)
1p
Gegeven is de vector \(\overrightarrow{a} = \begin{pmatrix}0 \\ 2\end{pmatrix} \text{.}\)
1p
Bereken \(4 ⋅ \overrightarrow{a} \text{.}\)
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\(4 ⋅ \overrightarrow{a} = 4 ⋅ \begin{pmatrix}0 \\ 2\end{pmatrix} = \begin{pmatrix}0 \\ 8\end{pmatrix} \text{.}\)
1p
Gegeven zijn de vectoren \(\overrightarrow{a} = \begin{pmatrix}-2 \\ -1\end{pmatrix}\) en \(\overrightarrow{b} = \begin{pmatrix}-5 \\ 6\end{pmatrix} \text{.}\)
1p
Bereken het inproduct van de vectoren \(\overrightarrow{a}\) en \(\overrightarrow{b} \text{.}\)
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\(\overrightarrow{a} ⋅ \overrightarrow{b} = -2 ⋅ -5 - 1 ⋅ 6 = 4 \text{.}\)
1p
Gegeven zijn de vectoren \(\overrightarrow{a} = \begin{pmatrix}4 \\ -3\end{pmatrix}\) en \(\overrightarrow{b} = \begin{pmatrix}-6 \\ -7\end{pmatrix} \text{.}\)
1p
Bereken \(\overrightarrow{a} + \overrightarrow{b} \text{.}\)
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\(\overrightarrow{a} + \overrightarrow{b} = \begin{pmatrix}4 \\ -3\end{pmatrix} + \begin{pmatrix}-6 \\ -7\end{pmatrix} = \begin{pmatrix}-2 \\ -10\end{pmatrix} \text{.}\)
1p
Gegeven zijn de vectoren \(\overrightarrow{a} = \begin{pmatrix}5 \\ 1\end{pmatrix}\) en \(\overrightarrow{b} = \begin{pmatrix}4 \\ -6\end{pmatrix} \text{.}\)
1p
Bereken \(7 \overrightarrow{a} + 2 \overrightarrow{b} \text{.}\)
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\(7 \overrightarrow{a} + 2 \overrightarrow{b} = 7 \begin{pmatrix}5 \\ 1\end{pmatrix} + 2 \begin{pmatrix}4 \\ -6\end{pmatrix} = \begin{pmatrix}43 \\ -5\end{pmatrix} \text{.}\)
1p
Gegeven zijn de punten \(A (7 , -6)\) en \(B (2 , 5) \text{.}\) Het punt \(C\) is het midden van lijnstuk \(A\kern{-.8pt}B \text{.}\)
1p
Bereken vector \(\overrightarrow{c} \text{.}\)
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\(\overrightarrow{c} = {1 \over 2} ⋅ (\overrightarrow{a} + \overrightarrow{b}) = {1 \over 2} ⋅ (\begin{pmatrix}7 \\ -6\end{pmatrix} + \begin{pmatrix}2 \\ 5\end{pmatrix}) = {1 \over 2} ⋅ \begin{pmatrix}9 \\ -1\end{pmatrix} = \begin{pmatrix}4\frac{1}{2} \\ -\frac{1}{2}\end{pmatrix} \text{.}\)
1p
Gegeven zijn de punten \(A (-3 , -2)\) en \(B (-1 , 7) \text{.}\)
1p
Bereken vector \(\overrightarrow{AB} \text{.}\)
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\(\overrightarrow{AB} = \overrightarrow{b} - \overrightarrow{a} = \begin{pmatrix}-1 \\ 7\end{pmatrix} - \begin{pmatrix}-3 \\ -2\end{pmatrix} = \begin{pmatrix}2 \\ 9\end{pmatrix} \text{.}\)
1p