Coordinates in Space

# Projection

## Projection of a point on a line or a plane

### Projection of a point on a line

Consider a point $P$ and a line $AB$. Then, the projection of $P$ on $AB$ is the foot $L$ of the perpendicular drawn from the point $P$ on the line $AB$.

### Projection of a point on a plane

Consider a point $P$ and a plane $CDEF$. Then, the projection of $P$ on the plane $CDEF$ is the foot $L$ of the perpendicular drawn from the point $P$ on the plane $CDEF$. It can also be interpreted as the point $L$ of intersection of the plane through the given point $P$ and perpendicular to a line $AB$ on the plane.

## Projection of a line segment on a line

Consider a line segment $AB$ of a given line and another line $CD$. Then, the projection of $AB$ on $CD$ is the segment $A’B’$ of $CD$ where $A’$ and $B’$ are the projections of $A$ and $B$ on $CD$ respectively.

We can also say that $A’$ and $B’$ are the points in which planes through $A$ and $B$ perpendicular to $CD$ meet the line $CD$.

### The length of the projection of a given line on another line

Consider two lines $AB$ and $CD$. We have to determine the projection of $AB$ on $CD$. From $A$, draw $AA’\perp CD$ and from $B$, draw $BB’\perp CD$ where $A’$ and $B’$ are the projections of the points $A$ and $B$ on $CD$ respectively. Then, $A’B’$ is the length of the projection of $AB$ on $CD$.

From $A$, draw $AN\perp BB’$ so that $AN=A’B’$. Let $\angle NAB=\theta$ i.e. $\theta$ is the angle between $AB$ and $CD$. Then, from right angled $\Delta ANB$, $\cos\theta=\frac{AN}{AB}$ $AN=AB\cos\theta$ $\therefore A’B’=AN\cos\theta$

### Projection of a broken line on a given line

If $A,B,C,…,M,N$ be any number of points in space then the sum of the projection of $AB,BC,…,MN$ on any given line $PQ$ is equal to the projection of the straight line $AN$ on $PQ$.

Let $A’,B’,C’,…, M’,N’$ be the feet of perpendiculars from $A,B,C,…,M,N$ on the line $PQ$. Then the projections of $AB,BC,CD,…,MN$ on the line $PQ$ are $A’B’,B’C’,…,M’N’$ respectively and $A’B’+B’C’+C’D’+…+M’N’=A’N’$ Again the projection of $AN$ on $PQ$ $=A’N’$.

Therefore, the projection of the line joining the end points $AN=A’B’+B’C’+…+M’N’$ $AN=$ algebraic sum of the projection of the broken lines on $PQ$.

### Projection of a line joining two points $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ on another line with direction cosines $l,m,n$

Let $P(x_1,y_1,z_1)$ and $Q(x_2,y_2,z_2)$ be the two points and $AB$ be a line with direction cosines $l,m,n$. Join $P$ and $Q$. Draw $PM$ and $QN$ perpendiculars to $AB$. Then the direction cosines of $PQ$ are $\frac{x_2-x_1}{r},\frac{y_2-y_1}{r},\frac{z_2-z_1}{r}$ where, $r=PQ=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$

If $\theta$ be the angle between $PQ$ and $AB$ then $\cos\theta=\frac{x_2-x_1}{r}l+\frac{y_2-y_1}{r}m+\frac{z_2-z_1}{r}n$ Therefore, the projection of $PQ$ on $AB$ $=MN=PQ\cos\theta$ $=r\left[\frac{x_2-x_1}{r}l+\frac{y_2-y_1}{r}m+\frac{z_2-z_1}{r}n\right]$ $=(x_2-x_1)l+(y_1-y_1)m+(z_2-z_1)n$

Previous: Section Formulae