<p>The planes ${{\mathrm{p}}_{1}}$ and ${{\mathrm{p}}_{2}}$ have equations $\mathbf{r}\cdot \left( \begin{matrix}1  \\ -3  \\  2  \\\end{matrix} \right)=-5$ and $\mathbf{r}\cdot \left( \begin{matrix}  2  \\-1  \\  3  \\\end{matrix} \right)=1$ respectively, and meet in the line ${{l}_{4}}$. Find</p>
<ol class="i-list"><li><div class="list-item-paragraph">the vector equation for ${{l}_{4}}$,</div></li><li><div class="list-item-paragraph">Find the acute angle between ${{\mathrm{p}}_{1}}$ and ${{\mathrm{p}}_{2}}$.</div></li></ol>

The planes ${{\mathrm{p}}_{1}}$ and ${{\mathrm{p}}_{2}}$ have equations $\mathbf{r}\cdot \left( \begin{matrix}1  \\ -3  \\  2  \\\end{matrix} \right)=-5$ and $\mathbf{r}\cdot \left( \begin{matrix}  2  \\-1  \\  3  \\\end{matrix} \right)=1$ respectively, and meet in the line ${{l}_{4}}$. Find

Find the acute angle between ${{\mathrm{p}}_{1}}$ and ${{\mathrm{p}}_{2}}$.

Vectors

International Baccalaureate Higher Level Mathematics

IB HL Math

Unit 13.2 Worked Example 5

\[x = \frac{8}{5} - \frac{7}{5}\lambda, \quad y = \frac{11}{5} + \frac{1}{5}\lambda, \quad z = 0 + \lambda\]

\[l_4: \mathbf{r} = \begin{pmatrix} 8/5 \\ 11/5 \\ 0 \end{pmatrix} + \lambda \begin{pmatrix} -7/5 \\ 1/5 \\ 1 \end{pmatrix}; \quad \lambda \in \mathbb{R}\]

Method II (w/o GC): Find direction vector $\mathbf{d}_4 = \mathbf{n}_1 \times \mathbf{n}_2$

Point on both planes: Let $x = 0$, solve to get $\left(0, \frac{17}{7}, \frac{8}{7}\right)$

\[l_4: \mathbf{r} = \begin{pmatrix} 0 \\ 17/7 \\ 8/7 \end{pmatrix} + \lambda \begin{pmatrix} -7 \\ 1 \\ 5 \end{pmatrix}; \quad \lambda \in \mathbb{R}\]

\[|\mathbf{n}_1 \cdot \mathbf{n}_2| = |\mathbf{n}_1||\mathbf{n}_2|\cos\theta\]

\[\theta = \cos^{-1}\frac{\left|\begin{pmatrix}1\\-3\\2\end{pmatrix} \cdot \begin{pmatrix}2\\-1\\3\end{pmatrix}\right|}{\sqrt{1+9+4}\sqrt{4+1+9}} = \cos^{-1}\left(\frac{2+3+6}{\sqrt{14}\sqrt{14}}\right) = 38.2^\circ \text{ (1 d.p.)}\]

Unit 13.2 Worked Example 5

Unit 13.2 Worked Example 5

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