Vectoren en hoeken
Gegeven zijn de punten \(\text{P} (1 , -2) \text{,}\) \(\text{Q} (5 , 0)\) en \(\text{R} (3 , 7) \text{.}\)
3p
Bereken de hoek \(\angle R\kern{-.8pt}P\kern{-.8pt}Q \text{.}\)
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\(\overrightarrow{PR} = \overrightarrow{r} - \overrightarrow{p} = \begin{pmatrix}3 \\ 7\end{pmatrix} - \begin{pmatrix}1 \\ -2\end{pmatrix} = \begin{pmatrix}2 \\ 9\end{pmatrix}\)
en \(\overrightarrow{PQ} = \overrightarrow{q} - \overrightarrow{p} = \begin{pmatrix}5 \\ 0\end{pmatrix} - \begin{pmatrix}1 \\ -2\end{pmatrix} = \begin{pmatrix}4 \\ 2\end{pmatrix} \text{.}\)
1p
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\(\cos(\angle R\kern{-.8pt}P\kern{-.8pt}Q) = {\begin{pmatrix}2 \\ 9\end{pmatrix} ⋅ \begin{pmatrix}4 \\ 2\end{pmatrix} \over \begin{vmatrix}\begin{pmatrix}2 \\ 9\end{pmatrix}\end{vmatrix} ⋅ \begin{vmatrix}\begin{pmatrix}4 \\ 2\end{pmatrix}\end{vmatrix}} = {26 \over \sqrt{85} ⋅ \sqrt{20}} \text{.}\)
1p
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\(\angle R\kern{-.8pt}P\kern{-.8pt}Q = \cos^{-1}({26 \over \sqrt{85} ⋅ \sqrt{20}}) ≈ 50{,}9\degree\)
1p
Gegeven zijn de lijnen \(k \text{: } \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}7 \\ -1\end{pmatrix} + t ⋅ \begin{pmatrix}3 \\ 0\end{pmatrix}\) en \(l \text{: } \begin{pmatrix}x \\ y\end{pmatrix} = \begin{pmatrix}3 \\ -4\end{pmatrix} + u ⋅ \begin{pmatrix}4 \\ -7\end{pmatrix} \text{.}\)
2p
Bereken de hoek tussen deze lijnen.
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\(\cos(\angle (k , l)) = {\begin{vmatrix}\begin{pmatrix}3 \\ 0\end{pmatrix} ⋅ \begin{pmatrix}4 \\ -7\end{pmatrix}\end{vmatrix} \over \begin{vmatrix}\begin{pmatrix}3 \\ 0\end{pmatrix}\end{vmatrix} ⋅ \begin{vmatrix}\begin{pmatrix}4 \\ -7\end{pmatrix}\end{vmatrix}} = {12 \over \sqrt{9} ⋅ \sqrt{65}}\)
1p
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\(\angle (k , l) = \cos^{-1}({12 \over \sqrt{9} ⋅ \sqrt{65}}) ≈ 60{,}3\degree\)
1p
Gegeven zijn de vectoren \(\overrightarrow{a} = \begin{pmatrix}0 \\ -2\end{pmatrix}\) en \(\overrightarrow{b} = \begin{pmatrix}4 \\ -6\end{pmatrix} \text{.}\)
2p
Bereken de hoek tussen deze vectoren.
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\(\cos(\angle (\overrightarrow{a} , \overrightarrow{b})) = {\begin{pmatrix}0 \\ -2\end{pmatrix} ⋅ \begin{pmatrix}4 \\ -6\end{pmatrix} \over \begin{vmatrix}\begin{pmatrix}0 \\ -2\end{pmatrix}\end{vmatrix} ⋅ \begin{vmatrix}\begin{pmatrix}4 \\ -6\end{pmatrix}\end{vmatrix}} = {12 \over \sqrt{4} ⋅ \sqrt{52}} \text{.}\)
1p
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\(\angle (\overrightarrow{a} , \overrightarrow{b}) = \cos^{-1}({12 \over \sqrt{4} ⋅ \sqrt{52}}) ≈ 33{,}7\degree\)
1p