<p>The position vectors of two points $A$ and $B$ are $\left( 3\mathbf{i}-\mathbf{j}+4\mathbf{k} \right)$ and $\left( 2\mathbf{i}+\mathbf{j}+2\mathbf{k} \right)$ respectively. Points $A$, $B$ and $C$ are collinear. Given that point $C$ is $6$ units away from $A$ , find the possible position vectors of $C$.</p>

The position vectors of two points $A$ and $B$ are $\left( 3\mathbf{i}-\mathbf{j}+4\mathbf{k} \right)$ and $\left( 2\mathbf{i}+\mathbf{j}+2\mathbf{k} \right)$ respectively. Points $A$, $B$ and $C$ are collinear. Given that point $C$ is $6$ units away from $A$ , find the possible position vectors of $C$.

Vectors

H2 Math

International Baccalaureate Higher Level Mathematics

IB HL Math

Unit 13.1 Worked Example 1

Position vectors: $A = (3\mathbf{i} - \mathbf{j} + 4\mathbf{k})$, $B = (2\mathbf{i} + \mathbf{j} + 2\mathbf{k})$

$\vec{AB} = \begin{pmatrix} 2 \\ 1 \\ 2 \end{pmatrix} - \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} = \begin{pmatrix} -1 \\ 2 \\ -2 \end{pmatrix}$

Unit vector of $\vec{AB} = \displaystyle\frac{1}{\sqrt{1+4+4}} \begin{pmatrix} -1 \\ 2 \\ -2 \end{pmatrix} = \frac{1}{3} \begin{pmatrix} -1 \\ 2 \\ -2 \end{pmatrix}$

1 unit $\rightarrow \displaystyle\frac{1}{3}\begin{pmatrix} -1 \\ 2 \\ -2 \end{pmatrix}$

6 units $\rightarrow 6 \times \displaystyle\frac{1}{3}\begin{pmatrix} -1 \\ 2 \\ -2 \end{pmatrix}$

$\vec{AC} = \pm 6 \times \left[\displaystyle\frac{1}{3}\begin{pmatrix} -1 \\ 2 \\ -2 \end{pmatrix}\right] = \pm 2\begin{pmatrix} -1 \\ 2 \\ -2 \end{pmatrix} = \pm\begin{pmatrix} -2 \\ 4 \\ -4 \end{pmatrix}$

$\vec{OC} = \vec{OA} + \vec{AC} = \begin{pmatrix} 3 \\ -1 \\ 4 \end{pmatrix} \pm \begin{pmatrix} -2 \\ 4 \\ -4 \end{pmatrix}$

$= \begin{pmatrix} 5 \\ -5 \\ 8 \end{pmatrix}$ or $\begin{pmatrix} 1 \\ 3 \\ 0 \end{pmatrix}$

Unit 13.1 Worked Example 1

Unit 13.1 Worked Example 1

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