\(\newcommand{L}[1]{\| #1 \|}\newcommand{VL}[1]{\L{ \vec{#1} }}\newcommand{R}[1]{\operatorname{Re}\,(#1)}\newcommand{I}[1]{\operatorname{Im}\, (#1)}\)
Vector projection¶
This here page follows the discussion in this Khan academy video on projection. Please watch that video for a nice presentation of the mathematics on this page.
For the video and this page, you will need the definitions and mathematics from Vectors and dot products.
Start¶
Consider two vectors \(\vec{w}\) and \(\vec{v}\).
We can scale \(\vec{v}\) with a scalar \(c\). By choosing the correct \(c\) we can create any vector on the infinite length dotted line in the diagram. \(c \vec{v}\) defines this infinite line.
We’re going to find the projection of \(\vec{w}\) onto \(\vec{v}\), written as:
The projection of \(\vec{w}\) onto \(\vec{v}\) is a vector on the line \(c \vec{v}\). Specifically it is \(c \vec{v}\) such that the line joining \(\vec{w}\) and \(c \vec{v}\) is perpendicular to \(\vec{v}\).
Why is it called projection?¶
Imagine a light source, parallel to \(\vec{v}\), above \(\vec{w}\). The light would cast rays perpendicular to \(\vec{v}\).
\(\mathrm{proj}_\vec{v}\vec{w}\) is the shadow cast by \(\vec{w}\) on the line defined by \(\vec{v}\).
Calculating the projection¶
The vector connecting \(\vec{w}\) and \(c \vec{v}\) is \(\vec{w} - c \vec{v}\).
We want to find \(c\) such that \(\vec{w} - c \vec{v}\) is perpendicular to \(\vec{v}\).
Two perpendicular vectors have vector dot product of zero, so:
By distribution over addition of dot products:
Because \(\VL{v} = \sqrt(\vec{v} \cdot \vec{v})\):
So:
We can also write the projection in terms of the unit vector defined by \(\vec{v}\):
\(\frac{\vec{w} \cdot \vec{v}}{\VL{v}}\) is called the scalar projection of \(\vec{w}\) onto \(\vec{v}\).