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 $\underset{̲}{a}$ and $\underset{̲}{n}$ . Let $\underset{̲}{\stackrel{̂}{n}}$ be a unit vector in the direction of $\underset{̲}{n}$ .

Figure 32

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

$\underset{̲}{a}\cdot \underset{̲}{\stackrel{̂}{n}}=\left|\underset{̲}{a}\right|\left|\stackrel{̂}{\underset{̲}{n}}\right|cos\theta =\left|\underset{̲}{a}\right|cos\theta \left(\text{since }\left|\underset{̲}{\stackrel{̂}{n}}\right|=1\right)$

We conclude that

Example 13

Figure 33 shows a plane containing the point $A$ which has position vector $\underset{̲}{a}$ . The vector $\underset{̲}{\stackrel{̂}{n}}$ 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.

Solution

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