Magnitude, direction and position vectorsAQA A-Level Maths: Revision notes

The magnitude (length, or modulus) of $\mathbf a=x\mathbf i+y\mathbf j$ is found with Pythagoras: $|\mathbf a|=\sqrt{x^2+y^2}.$ For $\mathbf p=5\mathbf i-12\mathbf j$, $|\mathbf p|=\sqrt{25+144}=13$. A unit vector has magnitude $1$; the unit vector in the direction of $\mathbf a$ is $\frac{\mathbf a}{|\mathbf a|}$, so for $\mathbf p$ it is $\frac{5}{13}\mathbf i-\frac{12}{13}\mathbf j$. Always square each component first, so negatives become positive.

The direction of a vector is the angle $\theta$ it makes with the positive $x$-direction, usually measured anticlockwise. Find the acute reference angle from $\tan^{-1}\left|\frac yx\right|$, then use the quadrant from the signs of $x$ and $y$:

• $x>0,y>0$: $\theta$ equals the reference angle.
• $x<0,y>0$: $\theta=180^\circ-$ reference angle.
• $x<0,y<0$: $\theta=180^\circ+$ reference angle.
• $x>0,y<0$: $\theta=360^\circ-$ reference angle (or the negative angle $-$ reference angle). For $5\mathbf i-12\mathbf j$ the reference angle is $\tan^{-1}\frac{12}{5}=67.4^\circ$, in the fourth quadrant, so $\theta=292.6^\circ$.

A vector of magnitude $r$ and direction $\theta$ has components $\mathbf a=r\cos\theta\,\mathbf i+r\sin\theta\,\mathbf j.$ For magnitude $8$ and direction $150^\circ$: $8\cos150^\circ=-4\sqrt3$ and $8\sin150^\circ=4$, so $\mathbf a=-4\sqrt3\,\mathbf i+4\mathbf j$. Going the other way, use $r=\sqrt{x^2+y^2}$ and the quadrant rule for $\theta$. A sketch of the vector with its components as a right-angled triangle makes the signs clear.

The position vector of a point $A$ is $\overrightarrow{OA}=\mathbf a$, the vector from the origin $O$ to $A$. The components of $\mathbf a$ are the coordinates of $A$, so the point $(2,3)$ has position vector $2\mathbf i+3\mathbf j$. A position vector is fixed by its end point, unlike a general vector, which can be drawn anywhere.

To go from $A$ to $B$ you go back to $O$ and then out to $B$: $\overrightarrow{AB}=\overrightarrow{AO}+\overrightarrow{OB}=\mathbf b-\mathbf a$. Remember end minus start. The distance between the points is the magnitude $|\mathbf b-\mathbf a|$. Example: $\mathbf a=2\mathbf i+3\mathbf j$, $\mathbf b=8\mathbf i-5\mathbf j$. $\overrightarrow{AB}=6\mathbf i-8\mathbf j$, so $AB=\sqrt{36+64}=10$. Combining these ideas lets you test shapes: for $A(1,1)$, $B(5,3)$, $C(3,7)$, $AB^2=BC^2=20$ and $AC^2=40$, so the triangle is isosceles with a right angle at $B$.