3D vectors: lines and planes

<ol class="abc-list" start="2"><li><div class="list-item-paragraph">By expanding $\left( \mathbf{b}-\mathbf{c} \right)\,\centerdot \,\left( \mathbf{b}-\mathbf{c} \right)$, simplify</div><div class="list-item-paragraph" style="text-align: center">${{\left| \mathbf{b} \right|}^{2}}+{{\left| \mathbf{c} \right|}^{2}}-\left( \mathbf{b}-\mathbf{c} \right)\,\centerdot \,\left( \mathbf{b}-\mathbf{c} \right)$</div><div class="list-item-paragraph">By taking $\mathbf{b}=\overrightarrow{AC}$ and $\mathbf{c}=\overrightarrow{AB}$, deduce the cosine formula for triangle $ABC$.</div><p style="text-align: right;">[4]</p></li></ol>

By expanding $\left( \mathbf{b}-\mathbf{c} \right)\,\centerdot \,\left( \mathbf{b}-\mathbf{c} \right)$, simplify

${{\left| \mathbf{b} \right|}^{2}}+{{\left| \mathbf{c} \right|}^{2}}-\left( \mathbf{b}-\mathbf{c} \right)\,\centerdot \,\left( \mathbf{b}-\mathbf{c} \right)$

By taking $\mathbf{b}=\overrightarrow{AC}$ and $\mathbf{c}=\overrightarrow{AB}$, deduce the cosine formula for triangle $ABC$.

Default solution

$(\mathbf{b}-\mathbf{s}) \cdot (\mathbf{b}-\mathbf{s}) = \mathbf{b} \cdot \mathbf{b} + \mathbf{b}(-\mathbf{s}) + (-\mathbf{s}) \cdot \mathbf{b} - \mathbf{s} \cdot (-\mathbf{s})$
$(\mathbf{b}-\mathbf{s}) \cdot (\mathbf{b}-\mathbf{s}) = |\mathbf{b}|^2 - \mathbf{b} \cdot \mathbf{s} - \mathbf{s} \cdot \mathbf{b} + |\mathbf{s}|^2$
$(\mathbf{b}-\mathbf{s}) \cdot (\mathbf{b}-\mathbf{s}) = |\mathbf{b}|^2 - 2\mathbf{b} \cdot \mathbf{s} + |\mathbf{s}|^2$
$2\mathbf{b} \cdot \mathbf{s} = |\mathbf{b}|^2 + |\mathbf{s}|^2 - (\mathbf{b}-\mathbf{s}) \cdot (\mathbf{b}-\mathbf{s})$

$|\mathbf{b}|^2 + |\mathbf{s}|^2 - (\mathbf{b}-\mathbf{s}) \cdot (\mathbf{b}-\mathbf{s}) = 2\mathbf{b} \cdot \mathbf{s}$ --- (1)

$AB = |\mathbf{s}|$, $AC = |\mathbf{b}|$, $\mathbf{b} \cdot \mathbf{s} = |\mathbf{b}||\mathbf{s}|\cos\theta$

$\mathbf{s} + \mathbf{b} = \overrightarrow{BC}$
$(\mathbf{b}-\mathbf{s}) = \overrightarrow{BC} \Rightarrow (\mathbf{b}-\mathbf{s}) \cdot (\mathbf{b}-\mathbf{s}) = |\overrightarrow{BC}|^2$

Sub $AB = |\mathbf{s}|$, $AC = |\mathbf{b}|$, $\mathbf{b} \cdot \mathbf{s} = |\mathbf{b}||\mathbf{s}|\cos\theta$,
$(\mathbf{b}-\mathbf{s}) \cdot (\mathbf{b}-\mathbf{s}) = BC^2$ into (1)

$AC^2 + AB^2 - BC^2 = 2|\mathbf{b}||\mathbf{s}|\cos\theta$

$BC^2 = AC^2 + AB^2 - 2 \cdot AC \cdot AB \cos\theta$ (shown)

N2001 P2 Q15

N2001 P2 Q15

Video Solution:

Default solution