3D Geometry: Foot of Perpendicular, Image, and Distance

This lesson works through one 3D coordinate geometry question: given the point $P(0, 2, 3)$ and the line $\dfrac{x+3}{5} = \dfrac{y-1}{2} = \dfrac{z+4}{3}$, we find the foot of the perpendicular from $P$ to the line, the image of $P$ in the line, and the length of the perpendicular.

A general point on the line

Set each ratio equal to a parameter $\lambda$:

$\frac{x+3}{5} = \frac{y-1}{2} = \frac{z+4}{3} = \lambda$

Solving each part gives a general point on the line in terms of $\lambda$:

$x = 5\lambda - 3, \qquad y = 2\lambda + 1, \qquad z = 3\lambda - 4$

Let $M(5\lambda - 3,\; 2\lambda + 1,\; 3\lambda - 4)$ be the foot of the perpendicular from $P$ to the line.

Direction ratios and the perpendicularity condition

The direction ratios of $PM$ come from subtracting the coordinates of $P(0, 2, 3)$ from those of $M$:

$PM:\quad (5\lambda - 3,\; 2\lambda - 1,\; 3\lambda - 7)$

The direction ratios of the line are the denominators:

$AB:\quad (5,\; 2,\; 3)$

Since $PM$ is perpendicular to the line, the sum of the products of corresponding direction ratios is zero:

$5(5\lambda - 3) + 2(2\lambda - 1) + 3(3\lambda - 7) = 0$

Expanding:

$25\lambda - 15 + 4\lambda - 2 + 9\lambda - 21 = 0$

$38\lambda - 38 = 0 \quad\Longrightarrow\quad \lambda = 1$

The foot of the perpendicular

Substitute $\lambda = 1$ back into $M$:

$M = (5(1) - 3,\; 2(1) + 1,\; 3(1) - 4) = (2,\; 3,\; -1)$

So the foot of the perpendicular is $M(2, 3, -1)$.

Image of the point

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

$(2,\; 3,\; -1) = \left(\frac{\alpha + 0}{2},\; \frac{\beta + 2}{2},\; \frac{\gamma + 3}{2}\right)$

Equating each coordinate:

x: $\;2 = \dfrac{\alpha}{2} \Rightarrow \alpha = 4$

y: $\;3 = \dfrac{\beta + 2}{2} \Rightarrow 6 = \beta + 2 \Rightarrow \beta = 4$

z: $\;-1 = \dfrac{\gamma + 3}{2} \Rightarrow -2 = \gamma + 3 \Rightarrow \gamma = -5$

So the image is $P'(4, 4, -5)$.

Length of the perpendicular

The length of the perpendicular is the distance $PM$ between $P(0, 2, 3)$ and $M(2, 3, -1)$:

$PM = \sqrt{(0 - 2)^2 + (2 - 3)^2 + (3 - (-1))^2}$

$PM = \sqrt{(-2)^2 + (-1)^2 + 4^2} = \sqrt{4 + 1 + 16} = \sqrt{21}\ \text{units}$

Key takeaways

• A general point on a line in symmetric form is found by setting the ratios equal to a parameter $\lambda$.
• The foot of the perpendicular comes from the condition that the direction ratios of $PM$ and the line satisfy $a_1 a_2 + b_1 b_2 + c_1 c_2 = 0$.
• The image of a point in a line is found by treating the foot of the perpendicular as the midpoint of the point and its image.
• The length of the perpendicular is the distance between the point and the foot, here $\sqrt{21}$ units.