5. Vectors

a. Definition, Magnitude and Direction

4. Direction

A vector \(\vec v\) has a magnitude and a direction. The magnitude is its length \[ |\vec v|=\sqrt{ {v_1}^2+{v_2}^2} \] What do we want to mean by its direction? We could mean the slope \(m\) of a line parallel to the vector, but this would not say if the vector points right or left and would not work for vertical vectors. Alternatively, we could mean its polar angle \(\theta\) in polar coordinates. This works in two dimensions but does not easily generalize to higher dimensions. Further, these ideas depend on the coordinate system and change if we rotate the axes.

We would like to have a notion of direction which is independent of the coordinate system and generalizes to higher dimensions. This is the unit vector.

The Direction Unit Vector

Recall that the unit vector in the direction of a vector \(\vec v\) is \(\hat v=\dfrac{\vec v}{|\vec v|}\). This is what is left when you eliminate (divide out by) the magnitude. So we define:

The direction of a vector \(\vec v\) is its unit vector \(\hat v\): \[ \hat{v}=\dfrac{\vec v}{|\vec v|} =\left\langle\dfrac{v_1}{|\vec v|},\dfrac{v_2}{|\vec v|}\right\rangle \]

If we know this unit vector, we know the direction of \(\vec v\) but we know nothing about its magnitude.

This definition sounds circular. It simply says that the vector \(\hat v\) is simultaneously the unit vector in the direction of \(\vec v\) and the direction of \(\vec v\).

Find the unit vector in the direction opposite to \(\vec v=\left\langle-3,4\right\rangle\).

The magnitude of \(\vec v=\left\langle-3,4\right\rangle\) is: \[ |\vec v|=\sqrt{(-3)^2+4^2}=\sqrt{25}=5 \] So the direction of \(\vec v\) is the unit vector: \[ \hat v=\dfrac{\vec v}{|\vec v|} =\left\langle\dfrac{-3}{5},\dfrac{4}{5}\right\rangle \] Therefore, the unit vector in the direction opposite to \(\vec v\) is: \[ \hat w=-\hat v =\left\langle\dfrac{3}{5},\dfrac{-4}{5}\right\rangle \]

Find the vector of length \(6\) in the same direction as \(\vec v=\left\langle6,-6\right\rangle\).

\( \vec w=\left\langle3\sqrt{2},-3\sqrt{2}\right\rangle\)

The magnitude of \(\vec v=\left\langle6,-6\right\rangle\) is: \[ |\vec v|=\sqrt{6^2+(-6)^2}=\sqrt{72}=6\sqrt{2} \] So the direction of \(\vec v\) is the unit vector: \[ \hat v=\dfrac{\vec v}{|\vec v|} =\left\langle\dfrac{6}{6\sqrt{2}},\dfrac{-6}{6\sqrt{2}}\right\rangle =\left\langle\dfrac{1}{\sqrt{2}},\dfrac{-1}{\sqrt{2}}\right\rangle \] Since \(\hat v\) has a length of \(1\), a vector of length \(6\) in the same direction is: \[ \vec w=6\hat v =\left\langle\dfrac{6}{\sqrt{2}},\dfrac{-6}{\sqrt{2}}\right\rangle =\left\langle3\sqrt{2},-3\sqrt{2}\right\rangle \]

We check by computing the length of \(\vec w\): \[ |\vec w|=\sqrt{(3\sqrt{2})^2+(-3\sqrt{2})^2}=\sqrt{18+18}=6 \]

Direction Angles and Direction Cosines (Optional)

If there is a coordinate system, then giving the direction of a vector as a unit vector is equivalent to giving the direction angles of the vector.

The direction angles of a vector \(\vec v\) are the angles, \(\alpha_1\) and \(\alpha_2\), between \(\vec v\) (when located at the origin) and each of the positive coordinate axes. The direction cosines are the cosines of these angles: \(\cos\alpha_1\) and \(\cos\alpha_2\).

Notice that \(\alpha_1\) is the polar angle \(\alpha_1=\theta\) and \(\alpha_2\) is its complement \(\alpha_2=\dfrac{\pi}{2}-\theta\).

The components of the direction of \(\vec v\) are the direction cosines. \[ \hat v =\left\langle\cos\alpha_1,\cos\alpha_2\right\rangle \] So, if you know the direction angles or the direction cosines, you know the direction of the vector and vice versa.

Look at the diagram. We drop a perpendicular from the tip of \(\vec v\) to each of the coordinate axes. In the right triangle formed by \(\vec v\), an axis and the perpendicular, the hypotenuse is \(|\vec v|\) and the adjacent side is the component, i.e. the side adjacent to \(\alpha_1\) is \(v_1\) and the side adjacent to \(\alpha_2\) is \(v_2\). So \(\cos\alpha_1=\dfrac{v_1}{|\vec v|}\) and \(\cos\alpha_2=\dfrac{v_2}{|\vec v|}\). If you put these together to form a vector, you get \[ \left\langle\cos\alpha_1,\cos\alpha_2\right\rangle =\left\langle\dfrac{v_1}{|\vec v|},\dfrac{v_2}{|\vec v|}\right\rangle =\hat v \] which is the unit vector direction of \(\vec v\).

The components of the direction of \(\vec v\) are the direction cosines not the direction sines.

5. Vectors

a. Definition, Magnitude and Direction

4. Direction

The Direction Unit Vector

Direction Angles and Direction Cosines (Optional)

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