### 3 Resolving one vector along another

The scalar product can be used to find the component of a vector in the direction of another vector. Consider Figure 32 which shows two arbitrary vectors
$\underset{\u0332}{a}$
and
$\underset{\u0332}{n}$
. Let
$\underset{\u0332}{\stackrel{\u0302}{n}}$
be a
**
unit vector
**
in the direction of
$\underset{\u0332}{n}$
.

**
Figure 32
**

Study the figure carefully and note that a perpendicular has been drawn from
$P$
to meet
$\underset{\u0332}{n}$
at
$Q$
. The distance
$OQ$
is called the
**
projection
**
of
$\underset{\u0332}{a}$
onto
$\underset{\u0332}{n}$
. Simple trigonometry tells us that the length of the projection is
$\left|\underset{\u0332}{a}\right|cos\theta $
. Now by taking the scalar product of
$\underset{\u0332}{a}$
with the unit vector
$\underset{\u0332}{\stackrel{\u0302}{n}}$
we find

$\phantom{\rule{2em}{0ex}}\underset{\u0332}{a}\cdot \underset{\u0332}{\stackrel{\u0302}{n}}=\left|\underset{\u0332}{a}\right|\phantom{\rule{0.3em}{0ex}}\left|\stackrel{\u0302}{\underset{\u0332}{n}}\right|\phantom{\rule{0.3em}{0ex}}cos\theta =\left|\underset{\u0332}{a}\right|cos\theta \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\left(\text{since}\left|\underset{\u0332}{\stackrel{\u0302}{n}}\right|=1\right)$

We conclude that

##### Key Point 15

**
Resolving One Vector Along Another
**

##### Example 13

Figure 33 shows a plane containing the point
$A$
which has position vector
$\underset{\u0332}{a}$
. The vector
$\underset{\u0332}{\stackrel{\u0302}{n}}$
is a unit vector perpendicular to the plane (such a vector is called a
**
normal
**
vector). Find an expression for the perpendicular distance,
$\ell $
, of the plane from the origin.

**
Figure 33
**

##### Solution

From the diagram we note that the perpendicular distance $\ell $ of the plane from the origin is the projection of $\underset{\u0332}{a}$ onto $\underset{\u0332}{\stackrel{\u0302}{n}}$ and, using Key Point 15, is thus $\underset{\u0332}{a}\cdot \stackrel{\u0302}{\underset{\u0332}{n}}$ .