<ol class="i-list"><li><div class="list-item-paragraph">Find a vector equation of the line through the points $A$ and $B$ with position vectors $\mathrm{7}\mathbf{i}\,+\,8\mathbf{j}\,+\,9\mathbf{k}$ and $-\mathbf{i}-8\mathbf{j}+\mathbf{k}$ respectively.</div><p style="text-align: right;">[3]</p></li><li><div class="list-item-paragraph">The perpendicular to this line from the point $C$ with position vector $\mathbf{i}+8\mathbf{j}+3\mathbf{k}$ meets the line at point $N$. Find the position vector of $N$ and the ratio $AN:NB$.</div><p style="text-align: right;">[5]</p></li><li><div class="list-item-paragraph">Find the Cartesian equation of the line which is a reflection of the line $AC$ in the line $AB$.</div><p style="text-align: right;">[4]</p></li></ol>

Find a vector equation of the line through the points $A$ and $B$ with position vectors $\mathrm{7}\mathbf{i}\,+\,8\mathbf{j}\,+\,9\mathbf{k}$ and $-\mathbf{i}-8\mathbf{j}+\mathbf{k}$ respectively.

The perpendicular to this line from the point $C$ with position vector $\mathbf{i}+8\mathbf{j}+3\mathbf{k}$ meets the line at point $N$. Find the position vector of $N$ and the ratio $AN:NB$.

Find the Cartesian equation of the line which is a reflection of the line $AC$ in the line $AB$.

Vectors - General

Vectors

International Baccalaureate Higher Level Mathematics

IB HL Math

N2012 P1 Q9

$A\begin{pmatrix} 7 \\ 8 \\ 9 \end{pmatrix}, \quad B\begin{pmatrix} -1 \\ -1 \\ 8 \end{pmatrix}$

$\vec{AB} = \begin{pmatrix} -8 \\ -16 \\ -8 \end{pmatrix} \div (-8) \Rightarrow \mathbf{d} = \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$

$l_{AB}: \mathbf{r} = \begin{pmatrix} -1 \\ -1 \\ 8 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}; \quad \lambda \in \mathbb{R}$

For some $\lambda \in \mathbb{R}$, $\vec{ON} = \begin{pmatrix} -1 \\ -1 \\ 8 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix}$

Since $\angle CNB = 90^\circ$, $\vec{CN} \cdot \mathbf{d} = 0$

$\left[ \begin{pmatrix} -1 \\ -1 \\ 8 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} - \begin{pmatrix} 8 \\ 3 \\ 5 \end{pmatrix} \right] \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 0$

$\left[ \begin{pmatrix} -9 \\ -4 \\ 3 \end{pmatrix} + \lambda \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} \right] \cdot \begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = 0$

$(-9 - 4 + 3) + \lambda(1 + 1 + 1) = 0$

$6\lambda = 36 \Rightarrow \lambda = 6$

$\vec{ON} = \begin{pmatrix} -1 \\ -1 \\ 8 \end{pmatrix} + 6\begin{pmatrix} 1 \\ 1 \\ 1 \end{pmatrix} = \begin{pmatrix} 5 \\ 4 \\ 7 \end{pmatrix}$

$\therefore N\begin{pmatrix} 5 \\ 4 \\ 7 \end{pmatrix}$

$AN = |\vec{AN}| = \left| \begin{pmatrix} 5 \\ 4 \\ 7 \end{pmatrix} - \begin{pmatrix} 7 \\ 8 \\ 9 \end{pmatrix} \right| = \left| \begin{pmatrix} -2 \\ -4 \\ -2 \end{pmatrix} \right| = \sqrt{24} = 2\sqrt{6}$

$NB = |\vec{NB}| = \left| \begin{pmatrix} -1 \\ -1 \\ 8 \end{pmatrix} - \begin{pmatrix} 5 \\ 4 \\ 7 \end{pmatrix} \right| = \left| \begin{pmatrix} -6 \\ -5 \\ 1 \end{pmatrix} \right| = \sqrt{216} = 6\sqrt{6}$

$AN : NB = 2\sqrt{6} : 6\sqrt{6} = 1 : 3$

By Midpoint Theorem, $\vec{ON} = \dfrac{\vec{OC} + \vec{OC'}}{2}$

$\vec{OC'} = 2\vec{ON} - \vec{OC} = 2\begin{pmatrix} 5 \\ 4 \\ 7 \end{pmatrix} - \begin{pmatrix} 8 \\ 3 \\ 5 \end{pmatrix} = \begin{pmatrix} 2 \\ 5 \\ 9 \end{pmatrix}$

Direction of line $AC': \vec{AC'} = \begin{pmatrix} 2 \\ 5 \\ 9 \end{pmatrix} - \begin{pmatrix} 7 \\ 8 \\ 9 \end{pmatrix} = \begin{pmatrix} -2 \\ -2 \\ 2 \end{pmatrix} \div (-2) \Rightarrow \begin{pmatrix} 1 \\ -1 \\ -4 \end{pmatrix}$

$l_{AC'}: \mathbf{r} = \begin{pmatrix} 7 \\ 8 \\ 9 \end{pmatrix} + \mu \begin{pmatrix} 1 \\ -4 \\ -1 \end{pmatrix}; \quad \mu \in \mathbb{R}$

$y = 8 - 4\mu \Rightarrow \mu = \dfrac{y - 8}{-4}$

$l_{AC'}: x - 7 = \dfrac{y - 8}{-4} = 9 - z$

N2012 P1 Q9

N2012 P1 Q9

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