Vectors: Question 7

Worked solution

Part (a): Finding $\overrightarrow{MN}$

Since $\overrightarrow{MN} = \overrightarrow{MO} + \overrightarrow{ON} = \overrightarrow{ON} - \overrightarrow{OM}$:

$\overrightarrow{MN} = \begin{pmatrix}5\\9\\9\end{pmatrix} - \begin{pmatrix}2\\5\\-3\end{pmatrix} = \begin{pmatrix}5-2\\9-5\\9-(-3)\end{pmatrix} = \begin{pmatrix}3\\4\\12\end{pmatrix}$

Part (b): Finding $|\overrightarrow{MN}|$

$|\overrightarrow{MN}| = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$

Check by recomputing in a different order: $12^2 + 3^2 + 4^2 = 144 + 9 + 16 = 169$, so $|\overrightarrow{MN}| = \sqrt{169} = 13$ again, the same result.

Part (c): Unit vectors in the direction of $\overrightarrow{MN}$ and $\overrightarrow{NM}$

Dividing $\overrightarrow{MN}$ by its own magnitude gives a unit vector in the same direction:

$\frac{\overrightarrow{MN}}{|\overrightarrow{MN}|} = \frac{1}{13}\begin{pmatrix}3\\4\\12\end{pmatrix} = \frac{3}{13}\mathbf{i} + \frac{4}{13}\mathbf{j} + \frac{12}{13}\mathbf{k}$

Check: the magnitude of this vector must equal $1$: $\left(\frac{3}{13}\right)^2 + \left(\frac{4}{13}\right)^2 + \left(\frac{12}{13}\right)^2 = \frac{9+16+144}{169} = \frac{169}{169} = 1 \checkmark$

Since $\overrightarrow{NM} = -\overrightarrow{MN}$, every component simply changes sign, so the unit vector in the direction of $\overrightarrow{NM}$ is:

$-\frac{3}{13}\mathbf{i} - \frac{4}{13}\mathbf{j} - \frac{12}{13}\mathbf{k}$

Check independently: computing $\overrightarrow{NM} = \overrightarrow{OM} - \overrightarrow{ON} = \begin{pmatrix}2\\5\\-3\end{pmatrix} - \begin{pmatrix}5\\9\\9\end{pmatrix} = \begin{pmatrix}-3\\-4\\-12\end{pmatrix}$ directly, which has the same magnitude $13$ (since squaring removes the signs), gives the same unit vector $-\dfrac{3}{13}\mathbf{i} - \dfrac{4}{13}\mathbf{j} - \dfrac{12}{13}\mathbf{k}$ when divided by $13$, confirming the result.

Final answers

• (a) $\overrightarrow{MN} = \begin{pmatrix}3\\4\\12\end{pmatrix}$
• (b) $|\overrightarrow{MN}| = 13$
• (c) Unit vector for $\overrightarrow{MN}$: $\dfrac{3}{13}\mathbf{i}+\dfrac{4}{13}\mathbf{j}+\dfrac{12}{13}\mathbf{k}$; unit vector for $\overrightarrow{NM}$: $-\dfrac{3}{13}\mathbf{i}-\dfrac{4}{13}\mathbf{j}-\dfrac{12}{13}\mathbf{k}$