Vectoren en hoeken
Gegeven zijn de punten \(\text{K} (1 , 3) \text{,}\) \(\text{L} (7 , 5)\) en \(\text{M} (4 , 0) \text{.}\)
3p
Bereken de hoek \(\angle K\kern{-.8pt}L\kern{-.8pt}M \text{.}\)
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\(\overrightarrow{LK} = \overrightarrow{k} - \overrightarrow{l} = \begin{pmatrix}1 \\ 3\end{pmatrix} - \begin{pmatrix}7 \\ 5\end{pmatrix} = \begin{pmatrix}-6 \\ -2\end{pmatrix}\)
en \(\overrightarrow{LM} = \overrightarrow{m} - \overrightarrow{l} = \begin{pmatrix}4 \\ 0\end{pmatrix} - \begin{pmatrix}7 \\ 5\end{pmatrix} = \begin{pmatrix}-3 \\ -5\end{pmatrix} \text{.}\)
1p
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\(\cos(\angle K\kern{-.8pt}L\kern{-.8pt}M) = {\begin{pmatrix}-6 \\ -2\end{pmatrix} ⋅ \begin{pmatrix}-3 \\ -5\end{pmatrix} \over \begin{vmatrix}\begin{pmatrix}-6 \\ -2\end{pmatrix}\end{vmatrix} ⋅ \begin{vmatrix}\begin{pmatrix}-3 \\ -5\end{pmatrix}\end{vmatrix}} = {28 \over \sqrt{40} ⋅ \sqrt{34}} \text{.}\)
1p
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\(\angle K\kern{-.8pt}L\kern{-.8pt}M = \cos^{-1}({28 \over \sqrt{40} ⋅ \sqrt{34}}) ≈ 40{,}6\degree\)
1p
Gegeven zijn de lijnen \(k \text{: } \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}2 \\ 1\end{pmatrix} + t ⋅ \begin{pmatrix}1 \\ 4\end{pmatrix}\) en \(l \text{: } \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}4 \\ 3\end{pmatrix} + u ⋅ \begin{pmatrix}5 \\ 3\end{pmatrix} \text{.}\)
2p
Bereken de hoek tussen deze lijnen.
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\(\cos(\angle (k , l)) = {\begin{vmatrix}\begin{pmatrix}1 \\ 4\end{pmatrix} ⋅ \begin{pmatrix}5 \\ 3\end{pmatrix}\end{vmatrix} \over \begin{vmatrix}\begin{pmatrix}1 \\ 4\end{pmatrix}\end{vmatrix} ⋅ \begin{vmatrix}\begin{pmatrix}5 \\ 3\end{pmatrix}\end{vmatrix}} = {17 \over \sqrt{17} ⋅ \sqrt{34}}\)
1p
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\(\angle (k , l) = \cos^{-1}({17 \over \sqrt{17} ⋅ \sqrt{34}}) = 45{,}0\degree\)
1p
Gegeven zijn de vectoren \(\overrightarrow{a} = \begin{pmatrix}7 \\ -2\end{pmatrix}\) en \(\overrightarrow{b} = \begin{pmatrix}3 \\ 0\end{pmatrix} \text{.}\)
2p
Bereken de hoek tussen deze vectoren.
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\(\cos(\angle (\overrightarrow{a} , \overrightarrow{b})) = {\begin{pmatrix}7 \\ -2\end{pmatrix} ⋅ \begin{pmatrix}3 \\ 0\end{pmatrix} \over \begin{vmatrix}\begin{pmatrix}7 \\ -2\end{pmatrix}\end{vmatrix} ⋅ \begin{vmatrix}\begin{pmatrix}3 \\ 0\end{pmatrix}\end{vmatrix}} = {21 \over \sqrt{53} ⋅ \sqrt{9}} \text{.}\)
1p
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\(\angle (\overrightarrow{a} , \overrightarrow{b}) = \cos^{-1}({21 \over \sqrt{53} ⋅ \sqrt{9}}) ≈ 15{,}9\degree\)
1p