The magnitude of a vector

The magnitude of a vector u⃗\vec u, written ∣u⃗∣\vert \vec{u} \vert, is the length of the arrow that this vector represents.

If the vector has the components

u⃗=(xyz)\vec{u} =\left(\begin{array}{r} x\\y\\z \end{array}\right)

we can apply the theorem of Pythagoras twice to find the following formula for the magnitude:

∣u⃗∣=x2+y2+z2\boxed{\vert\vec u\vert = \sqrt{x^2+y^2+z^2}}

A more detailed derivation of this formula is given in the figure below, for the vector with the components

u⃗=(234)\vec{u} =\left(\begin{array}{r} 2\\3\\4 \end{array}\right)

In this case, the magnitude is

∣u⃗∣=22+32+42=29\vert \vec{u}\vert = \sqrt{2^2+3^2+4^2}=\sqrt{29}
Exercise 1

  1. Determine the magnitude of the vectors w⃗=(−11−1)\vec{w}=\left(\begin{array}{r} -1\\ 1\\ -1 \end{array}\right) and s⃗=(1.50−2)\vec{s}=\left(\begin{array}{r} 1.5\\ 0\\ -2 \end{array}\right)

  2. Vector u⃗\vec{u} has the xx-component 22 and the yy-component −3-3. Determine the zz-component such that the magnitude of the vector is 3232 .

  3. Determine the distance between the points A(2∣1∣0)A(2 \vert 1\vert 0) and B(−7∣2∣5)B(-7 \vert 2 \vert 5 ).

Solution

  1. ∣w⃗∣=3,∣s⃗∣=6.25=2.5\vert \vec{w}\vert =\sqrt{3}, \vert \vec{s}\vert =\sqrt{6.25}=2.5

  2. ∣w⃗∣=22+(−3)2+z2=32→22+(−3)2+z2=322=1024→z=1011=±31.796...\vert \vec{w}\vert =\sqrt{2^2+(-3)^2+z^2}=32 \rightarrow 2^2+(-3)^2+z^2=32^2=1024 \rightarrow z=\sqrt{1011}=\pm 31.796...

  3. The arrow from AA to BB has the components AB→=(−915)\overrightarrow{AB}=\left(\begin{array}{r} -9\\ 1\\ 5 \end{array}\right). The distance between AA and BB is the length of this arrow: ∣AB→∣=(−9)2+12+52=107\vert \overrightarrow{AB}\vert =\sqrt{(-9)^2+1^2+5^2}=\sqrt{107}..