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 and . Let be a unit vector in the direction of .
Study the figure carefully and note that a perpendicular has been drawn from to meet at . The distance is called the projection of onto . Simple trigonometry tells us that the length of the projection is . Now by taking the scalar product of with the unit vector we find
We conclude that
Resolving One Vector Along Another
Figure 33 shows a plane containing the point which has position vector . The vector is a unit vector perpendicular to the plane (such a vector is called a normal vector). Find an expression for the perpendicular distance, , of the plane from the origin.
From the diagram we note that the perpendicular distance of the plane from the origin is the projection of onto and, using Key Point 15, is thus .