Magnitude and direction of a vector

Vectors

A-level Maths (9MA0)

ORIGINAL

Pythagoras for the magnitude; for the direction, sketch the vector so the inverse tan gives an angle in the right quadrant.

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Question

\mathbf{p} = -5\mathbf{i} + 12\mathbf{j}

. (a) Find

|\mathbf{p}|

. (b) Find the angle between

\mathbf{p}

and the unit vector

\mathbf{i}

, to 1 decimal place. (c) Find a unit vector in the direction of

\mathbf{p}

Every step worked, with the reasoning.

1
$|\mathbf{p}| = \sqrt{(-5)^2 + 12^2} = \sqrt{169} = 13$
The magnitude is the length of the vector: Pythagoras on its components.
2
$\mathbf{p}$ goes 5 left and 12 up, so it points into the second quadrant.
Sketch the components before using inverse tan — the calculator only returns an angle between $-90^\circ$ and $90^\circ$ .
3
Angle with the negative $x$ -axis: $\tan^{-1}\left(\dfrac{12}{5}\right) = 67.380\ldots^\circ$
Use the sizes of the components as the sides of the right-angled triangle.
4
Angle with $\mathbf{i}$ : $180^\circ - 67.380\ldots^\circ = 112.619\ldots^\circ$
$\mathbf{i}$ points along the positive $x$ -axis, so subtract from $180^\circ$ .
5
Unit vector $= \dfrac{\mathbf{p}}{|\mathbf{p}|} = \dfrac{1}{13}(-5\mathbf{i} + 12\mathbf{j}) = -\dfrac{5}{13}\mathbf{i} + \dfrac{12}{13}\mathbf{j}$
Divide each component by the magnitude so the length becomes 1.

Answer: (a) $13$ (b) $112.6^\circ$ (c) $-\dfrac{5}{13}\mathbf{i} + \dfrac{12}{13}\mathbf{j}$