Gegeven zijn de punten \(\text{P} (2 , 3) \text{,}\) \(\text{Q} (1 , 0)\) en \(\text{R} (6 , -4) \text{.}\)
3p
Bereken de hoek \(\angle P\kern{-.8pt}Q\kern{-.8pt}R \text{.}\)
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\(\overrightarrow{QP} = \overrightarrow{p} - \overrightarrow{q} = \begin{pmatrix}2 \\ 3\end{pmatrix} - \begin{pmatrix}1 \\ 0\end{pmatrix} = \begin{pmatrix}1 \\ 3\end{pmatrix}\)
en \(\overrightarrow{QR} = \overrightarrow{r} - \overrightarrow{q} = \begin{pmatrix}6 \\ -4\end{pmatrix} - \begin{pmatrix}1 \\ 0\end{pmatrix} = \begin{pmatrix}5 \\ -4\end{pmatrix} \text{.}\)
1p
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\(\cos(\angle P\kern{-.8pt}Q\kern{-.8pt}R) = {\begin{pmatrix}1 \\ 3\end{pmatrix} ⋅ \begin{pmatrix}5 \\ -4\end{pmatrix} \over \begin{vmatrix}\begin{pmatrix}1 \\ 3\end{pmatrix}\end{vmatrix} ⋅ \begin{vmatrix}\begin{pmatrix}5 \\ -4\end{pmatrix}\end{vmatrix}} = {-7 \over \sqrt{10} ⋅ \sqrt{41}} \text{.}\)
1p
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\(\angle P\kern{-.8pt}Q\kern{-.8pt}R = \cos^{-1}({-7 \over \sqrt{10} ⋅ \sqrt{41}}) ≈ 110{,}2\degree\)
1p
Gegeven zijn de lijnen \(k \text{: } \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}6 \\ -5\end{pmatrix} + t ⋅ \begin{pmatrix}3 \\ -2\end{pmatrix}\) en \(l \text{: } \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}3 \\ -1\end{pmatrix} + u ⋅ \begin{pmatrix}4 \\ -7\end{pmatrix} \text{.}\)
2p
Bereken de hoek tussen deze lijnen.
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\(\cos(\angle (k , l)) = {\begin{vmatrix}\begin{pmatrix}3 \\ -2\end{pmatrix} ⋅ \begin{pmatrix}4 \\ -7\end{pmatrix}\end{vmatrix} \over \begin{vmatrix}\begin{pmatrix}3 \\ -2\end{pmatrix}\end{vmatrix} ⋅ \begin{vmatrix}\begin{pmatrix}4 \\ -7\end{pmatrix}\end{vmatrix}} = {26 \over \sqrt{13} ⋅ \sqrt{65}}\)
1p
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\(\angle (k , l) = \cos^{-1}({26 \over \sqrt{13} ⋅ \sqrt{65}}) ≈ 26{,}6\degree\)
1p
Gegeven zijn de vectoren \(\overrightarrow{a} = \begin{pmatrix}5 \\ 1\end{pmatrix}\) en \(\overrightarrow{b} = \begin{pmatrix}3 \\ 4\end{pmatrix} \text{.}\)
2p
Bereken de hoek tussen deze vectoren.
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\(\cos(\angle (\overrightarrow{a} , \overrightarrow{b})) = {\begin{pmatrix}5 \\ 1\end{pmatrix} ⋅ \begin{pmatrix}3 \\ 4\end{pmatrix} \over \begin{vmatrix}\begin{pmatrix}5 \\ 1\end{pmatrix}\end{vmatrix} ⋅ \begin{vmatrix}\begin{pmatrix}3 \\ 4\end{pmatrix}\end{vmatrix}} = {19 \over \sqrt{26} ⋅ \sqrt{25}} \text{.}\)
1p
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\(\angle (\overrightarrow{a} , \overrightarrow{b}) = \cos^{-1}({19 \over \sqrt{26} ⋅ \sqrt{25}}) ≈ 41{,}8\degree\)
1p