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IB Mathematics: Applications and Interpretation HL · AHL 4: Vectors · AHL 4.20
Mixed practice: scalar and vector products, lines and distances Problem set B Answers
Skills
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- (a) 119°
- (b) 83.4°
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- (a) 65.4°
- (b) 134°
- (c) 2.65
- (d) \(20 \sqrt{2}\)
- (e) 0.927 rad
- (f) 1.12 rad
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- (a) 35.4°
- (b) 25.2°
- (c) \(\mathbf{d}_1\cdot\mathbf{d}_2=0\)
- (d) \(\mathbf{r}=\begin{pmatrix}-1\\4\\3\end{pmatrix}+\lambda \begin{pmatrix}2\\-3\\6\end{pmatrix}\); 31.0°
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- (a) \(6\mathbf{i} - 4\mathbf{j} + 7\mathbf{k}\) (or any multiple of it)
- (b) \(\dfrac{1}{\sqrt{3}}\begin{pmatrix}1\\-1\\-1\end{pmatrix}\) (or its negative)
- (c) t = 5.5
- (d) t = -2 or t = 2.5
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- (a) \(\mathbf{a}\times\mathbf{b}\) is perpendicular to both \(\mathbf{a}\) and \(\mathbf{b}\), and the scalar product of perpendicular vectors is 0
- (b) 0
- (c) Both sides equal 294
- (d) \(\mathbf{p}=\begin{pmatrix}3\\-5\\-4\end{pmatrix}\), \(\mathbf{q}=\begin{pmatrix}-3\\5\\4\end{pmatrix}\); \(\mathbf{q}=-\mathbf{p}\)
Shapes in 3D
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- (a) 58.5°, 71.5°, 50.1°
- (b) 8.96
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- (a) \(\overrightarrow{CA}\cdot\overrightarrow{CB}=0\)
- (b) A: 36.7°, B: 53.3°
- (c) 3.35
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- (a) (3, 2, 1)
- (b) 8.66
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- (a) (3, 3, 4)
- (b) \(|\overrightarrow{AB}|=|\overrightarrow{AD}|=3\)
- (c) \(\overrightarrow{AC}\cdot\overrightarrow{BD}=0\)
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- (a) C(1, 7, 1), F(1, 2, 4), G(1, 7, 4), H(-1, 7, 4)
- (b) 9.50
In context
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- (a) \(\begin{pmatrix}\frac{5}{2}\\-6\end{pmatrix}\)
- (b) -1.25 m/s
Stretch
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- (a) \(\overrightarrow{OB}=\lambda\overrightarrow{OA}\)
- (b) \(\lambda=2\)
- (c) \(\sqrt{5}\)
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- (a) M(1, -2, 7)
- (b) Q is on \(l\) when \(\lambda=-1\)
- (c) R(5, -10, 15)
jmaths.xyz · AI HL · AHL 4.20 Mixed practice: scalar and vector products, lines and distances