(5) Finding the magnitude or length
of a vector
This drawing shows you how we set about doing this.
Notice the right-hand coordinate system but with the z-axis vertical
this time.)
The formula for the length OP or |p| is a kind of 3 dimensional
form of Pythagoras' theorem.
Here's how the working would go if we had
a = 4 and b = 3 and c = 12.
Using Pythagoras' theorem in the yellow right-angled triangle we get
OQ = the square root of (16 + 9) = 5
and then, using Pythagoras' theorem in the vertical green right-angled
triangle, we get
OP = the square root of (25 + 144) = 13.
Alternatively, in just one step of working, we have
OP = the square root of (16 + 9 + 144) = 13.
A reminder on unit vectors
Unit vectors are vectors which have unit length. They are
important in many applications. In three dimensions, the vectors i,
j and k which run along the x, y and z axes respectively, all
have unit length.
Sometimes, for example when working with planes, it is necessary to find
a unit vector in the same direction as a given vector.
Suppose you need the unit vector in the same direction
as u = 2i - j + 2k or
u = (2, -1, 2) written as a row vector.
The length or magnitude of u is given by the square root
of (4 + 1 + 4) = 3.
Also, the required vector must be parallel to u.
So we can get the vector we want by just scaling down u by a
factor of 3.
It is 1/3(2i - j + 2k) = 1/3(2, -1, 2) =
(2/3, -1/3, 2/3).
Unit vectors are often written with a little hat on top, so here we
would have found û.
Here's a practical application which uses components and magnitude.
If a body is acted on by three forces P, Q and S
whose lines of action
all pass through the same point, find the resultant force R, and its
magnitude |R| if
P = 3i - 5j + 2k
Q = - 2i + 4j + 4k
S = 3i - 3j - 4k.
P + Q + S = R = (3 - 2 + 3)i +
(- 5 + 4 - 3)j
+ ( 2 + 4 - 4)k = 4i - 4j + 2k
and |R| = the square root of (16 + 16 + 4) = 6.
We might also need to know what angles this resultant makes with the
x, y and z axes. Finding these angles will come as a spin-off from
discovering how we can multiply two vectors together.
But how can we actually do this since the two vectors can have
different directions?
Next
or back to the previous section
or return to the vector homepage.