A position vector starts at the chosen origin. If OA= a and OB= b, then AB= b- a. Subtract initial position from final position. For v= u v, | v|=sqrt(u2+v2). Magnitude is a non-negative scalar, not a vector. The unit vector in the direction of a nonzero v is v| v|. The zero vector has no direction and cannot be normalised. Example: A(1,-2), B(4,2). AB= 3 4, | AB|=5, and the unit vector from A towards B is 35 45. A vector of magnitude 10 in this direction is 10 35 45 = 6 8. Check its magnitude and direction separately.

A position vector starts at the chosen origin. If \(\overrightarrow{OA}=\mathbf a\) and \(\overrightarrow{OB}=\mathbf b\), then \(\overrightarrow{AB}=\mathbf b-\mathbf a\). Subtract initial position from final position.

For \(\mathbf v=\begin{pmatrix}u\\v\end{pmatrix}\), \(|\mathbf v|=\sqrt{u^2+v^2}\). Magnitude is a non-negative scalar, not a vector.

The unit vector in the direction of a nonzero \(\mathbf v\) is \(\dfrac{\mathbf v}{|\mathbf v|}\). The zero vector has no direction and cannot be normalised.

Example: \(A(1,-2)\), \(B(4,2)\). \(\overrightarrow{AB}=\begin{pmatrix}3\\4\end{pmatrix}\), \(|\overrightarrow{AB}|=5\), and the unit vector from \(A\) towards \(B\) is \(\begin{pmatrix}\frac35\\\frac45\end{pmatrix}\).

A vector of magnitude \(10\) in this direction is \(10\begin{pmatrix}\frac35\\\frac45\end{pmatrix}=\begin{pmatrix}6\\8\end{pmatrix}\). Check its magnitude and direction separately.

Position, displacement and unit vectors

Position, displacement and unit vectors