Gegeven zijn de punten \(\text{K} (6 , 4) \text{,}\) \(\text{L} (2 , 7)\) en \(\text{M} (1 , -3) \text{.}\)
3p
Bereken de hoek \(\angle K\kern{-.8pt}L\kern{-.8pt}M \text{.}\)
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\(\overrightarrow{LK} = \overrightarrow{k} - \overrightarrow{l} = \begin{pmatrix}6 \\ 4\end{pmatrix} - \begin{pmatrix}2 \\ 7\end{pmatrix} = \begin{pmatrix}4 \\ -3\end{pmatrix}\)
en \(\overrightarrow{LM} = \overrightarrow{m} - \overrightarrow{l} = \begin{pmatrix}1 \\ -3\end{pmatrix} - \begin{pmatrix}2 \\ 7\end{pmatrix} = \begin{pmatrix}-1 \\ -10\end{pmatrix} \text{.}\)
1p
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\(\cos(\angle K\kern{-.8pt}L\kern{-.8pt}M) = {\begin{pmatrix}4 \\ -3\end{pmatrix} ⋅ \begin{pmatrix}-1 \\ -10\end{pmatrix} \over \begin{vmatrix}\begin{pmatrix}4 \\ -3\end{pmatrix}\end{vmatrix} ⋅ \begin{vmatrix}\begin{pmatrix}-1 \\ -10\end{pmatrix}\end{vmatrix}} = {26 \over \sqrt{25} ⋅ \sqrt{101}} \text{.}\)
1p
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\(\angle K\kern{-.8pt}L\kern{-.8pt}M = \cos^{-1}({26 \over \sqrt{25} ⋅ \sqrt{101}}) ≈ 58{,}8\degree\)
1p
Gegeven zijn de lijnen \(k \text{: } \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}2 \\ 6\end{pmatrix} + t ⋅ \begin{pmatrix}3 \\ 1\end{pmatrix}\) en \(l \text{: } \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}0 \\ -3\end{pmatrix} + u ⋅ \begin{pmatrix}5 \\ -6\end{pmatrix} \text{.}\)
2p
Bereken de hoek tussen deze lijnen.
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\(\cos(\angle (k , l)) = {\begin{vmatrix}\begin{pmatrix}3 \\ 1\end{pmatrix} ⋅ \begin{pmatrix}5 \\ -6\end{pmatrix}\end{vmatrix} \over \begin{vmatrix}\begin{pmatrix}3 \\ 1\end{pmatrix}\end{vmatrix} ⋅ \begin{vmatrix}\begin{pmatrix}5 \\ -6\end{pmatrix}\end{vmatrix}} = {9 \over \sqrt{10} ⋅ \sqrt{61}}\)
1p
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\(\angle (k , l) = \cos^{-1}({9 \over \sqrt{10} ⋅ \sqrt{61}}) ≈ 68{,}6\degree\)
1p
Gegeven zijn de vectoren \(\overrightarrow{a} = \begin{pmatrix}3 \\ 0\end{pmatrix}\) en \(\overrightarrow{b} = \begin{pmatrix}7 \\ 4\end{pmatrix} \text{.}\)
2p
Bereken de hoek tussen deze vectoren.
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\(\cos(\angle (\overrightarrow{a} , \overrightarrow{b})) = {\begin{pmatrix}3 \\ 0\end{pmatrix} ⋅ \begin{pmatrix}7 \\ 4\end{pmatrix} \over \begin{vmatrix}\begin{pmatrix}3 \\ 0\end{pmatrix}\end{vmatrix} ⋅ \begin{vmatrix}\begin{pmatrix}7 \\ 4\end{pmatrix}\end{vmatrix}} = {21 \over \sqrt{9} ⋅ \sqrt{65}} \text{.}\)
1p
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\(\angle (\overrightarrow{a} , \overrightarrow{b}) = \cos^{-1}({21 \over \sqrt{9} ⋅ \sqrt{65}}) ≈ 29{,}7\degree\)
1p