<p>The coordinates of the points $A$, $B$ and $C$ are given by $A\,(0,\,2,\,4)$, $B\,(4,\,6,\,11)$ and $C\,(8,\,1,\,0)$.</p>
<ol class="i-list"><li><div class="list-item-paragraph">Show that the triangle with vertices $A$, $B$ and $C$ is an isosceles right-angled triangle.</div></li><li><div class="list-item-paragraph">Find the point $D$ in the same plane as $A$, $B$ and $C$ such that $BCD$ is an equilateral triangle.</div></li></ol>

The coordinates of the points $A$, $B$ and $C$ are given by $A\,(0,\,2,\,4)$, $B\,(4,\,6,\,11)$ and $C\,(8,\,1,\,0)$.

Show that the triangle with vertices $A$, $B$ and $C$ is an isosceles right-angled triangle.

Find the point $D$ in the same plane as $A$, $B$ and $C$ such that $BCD$ is an equilateral triangle.

Vectors - General

Vectors

International Baccalaureate Higher Level Mathematics

IB HL Math

Finding a point to form an equilateral triangle given a line segment in 3D vectors

Given: $A\begin{pmatrix} 0 \\ 2 \\ 4 \end{pmatrix}$, $B\begin{pmatrix} 4 \\ 6 \\ 11 \end{pmatrix}$, $C\begin{pmatrix} 8 \\ 1 \\ 0 \end{pmatrix}$

For an equilateral triangle, all angles are $60^\circ$.

Point $D$ lies on line $AM$ (the perpendicular from $A$ through midpoint $M$ of $BC$).

$D'$ is the reflection of $D$ below the line $BC$.

$|\overrightarrow{CD}| = |\overrightarrow{BC}|$

For some $\lambda \in \mathbb{R}$, $\overrightarrow{OD} = l_{AM}$

Midpoint of $BC$, $M = \dfrac{\begin{pmatrix} 4 \\ 6 \\ 11 \end{pmatrix} + \begin{pmatrix} 8 \\ 1 \\ 0 \end{pmatrix}}{2} = \frac{1}{2}\begin{pmatrix} 12 \\ 7 \\ 11 \end{pmatrix}$

$|\overrightarrow{BC}| = \left| \begin{pmatrix} 4 \\ -5 \\ -11 \end{pmatrix} \right| = \sqrt{4^2 + 5^2 + 11^2} = \sqrt{162}$

$\overrightarrow{AM} = \frac{1}{2}\begin{pmatrix} 12 \\ 7 \\ 11 \end{pmatrix} - \begin{pmatrix} 0 \\ 2 \\ 4 \end{pmatrix} = \begin{pmatrix} 6 \\ \frac{3}{2} \\ \frac{3}{2} \end{pmatrix} \xrightarrow{\times 2} \begin{pmatrix} 12 \\ 3 \\ 3 \end{pmatrix} \xrightarrow{\div 3} \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix}$

$l_{AM}: \mathbf{r} = \begin{pmatrix} 0 \\ 2 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix}$; $\lambda \in \mathbb{R}$

For some $\lambda \in \mathbb{R}$, $\overrightarrow{OD} = \begin{pmatrix} 0 \\ 2 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix}$

$\left| \begin{pmatrix} 0 \\ 2 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix} - \begin{pmatrix} 8 \\ 1 \\ 0 \end{pmatrix} \right| = \sqrt{162}$

$\left| \begin{pmatrix} -8 \\ 1 \\ 4 \end{pmatrix} + \lambda \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix} \right| = \sqrt{162}$

$\sqrt{(4\lambda - 8)^2 + (\lambda + 1)^2 + (\lambda + 4)^2} = \sqrt{162}$

$16\lambda^2 - 64\lambda + 64 + \lambda^2 + 2\lambda + 1 + \lambda^2 + 8\lambda + 16 = 162$

$\div 9: \quad 2\lambda^2 - 6\lambda - 9 = 0$

$\lambda = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(2)(-9)}}{2(2)} = \dfrac{6 \pm \sqrt{108}}{4}$

$= \dfrac{6 \pm \sqrt{9 \cdot 4 \cdot 3}}{4} = \dfrac{6 \pm 6\sqrt{3}}{4} = \dfrac{3 \pm 3\sqrt{3}}{2}$

$\therefore \overrightarrow{OD} = \begin{pmatrix} 0 \\ 2 \\ 4 \end{pmatrix} + \left( \dfrac{3 + 3\sqrt{3}}{2} \right) \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix}$

or $\overrightarrow{OD} = \begin{pmatrix} 0 \\ 2 \\ 4 \end{pmatrix} + \left( \dfrac{3 - 3\sqrt{3}}{2} \right) \begin{pmatrix} 4 \\ 1 \\ 1 \end{pmatrix}$

Finding a point to form an equilateral triangle given a line segment in 3D vectors

Finding a point to form an equilateral triangle given a line segment in 3D vectors

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