Foot of Perpendicular, Image and Distance in 3D

This lesson works through a single Class 12 3D geometry problem. Given the point $P(1,6,3)$ and the line $\tfrac{x}{1}=\tfrac{y-1}{2}=\tfrac{z-2}{3}$, we find the foot of the perpendicular from $P$ to the line, the image of $P$ in the line, and the distance from $P$ to the foot.

A general point on the line

Set the line equal to a parameter $k$:

$\frac{x}{1}=\frac{y-1}{2}=\frac{z-2}{3}=k.$

Reading off each part gives the coordinates of a general point on the line:

$x=k,\qquad y=2k+1,\qquad z=3k+2.$

So the foot of the perpendicular $M$ has the form $M(k,\,2k+1,\,3k+2)$.

Direction ratios of $PM$

With $P(1,6,3)$, the direction ratios of $PM$ are the differences of coordinates:

$\big(k-1,\ (2k+1)-6,\ (3k+2)-3\big)=\big(k-1,\ 2k-5,\ 3k-1\big).$

The given line has direction ratios $(1,2,3)$, the denominators of the symmetric form.

Perpendicularity condition

Since $PM$ is perpendicular to the line, the dot product of the two sets of direction ratios is zero:

$(k-1)\cdot 1+(2k-5)\cdot 2+(3k-1)\cdot 3=0.$

Expanding:

$(k-1)+(4k-10)+(9k-3)=14k-14=0,$

so $k=1$.

Foot of the perpendicular

Substituting $k=1$ into $M(k,\,2k+1,\,3k+2)$:

$M=(1,\ 2(1)+1,\ 3(1)+2)=(1,\ 3,\ 5).$

Image of the point

Let the image be $P'(\alpha,\beta,\gamma)$. The foot $M$ is the midpoint of $PP'$, so by the midpoint formula:

$\left(\frac{\alpha+1}{2},\ \frac{\beta+6}{2},\ \frac{\gamma+3}{2}\right)=(1,3,5).$

$x$ coordinate: $\alpha+1=2$, so $\alpha=1$.

$y$ coordinate: $\beta+6=6$, so $\beta=0$.

$z$ coordinate: $\gamma+3=10$, so $\gamma=7$.

The image is $P'(1,0,7)$.

Distance from the point to the foot

Using the distance formula with $P(1,6,3)$ and $M(1,3,5)$:

$PM=\sqrt{(1-1)^2+(6-3)^2+(3-5)^2}=\sqrt{0+9+4}=\sqrt{13}.$

Key takeaways

• A point on a line in symmetric form can be written with one parameter, here $M(k,\,2k+1,\,3k+2)$.
• The foot of the perpendicular is found by setting the dot product of the direction ratios of $PM$ and the line to zero, which gives $k=1$ and $M(1,3,5)$.
• The image $P'(1,0,7)$ comes from treating the foot as the midpoint of $P$ and $P'$, and the distance from $P$ to the foot is $\sqrt{13}$.