# Perpendicular Distance of a Point From a Line in 3d

Here you will learn how to find perpendicular distance of a point from a line in 3d in both vector form and cartesian form.

Let’s begin –

## Perpendicular Distance of a Point From a Line in 3d

#### (a) Cartesian Form

Algorithm :

Let P$$(\alpha, \beta, \gamma)$$ be the given point, and let the given line be

$$x – x_1\over a$$ = $$y – y_1\over b$$ = $$z – z_1\over c$$

1). Write the coordinates of a general point on the given line. The coordinates of general point on the line are ($$x_1 + a\lambda$$, $$y_1 + b\lambda$$, $$z_1 + c\lambda$$), where $$\lambda$$ is a parameter. Assume that this point L is the foot of the perpendicular drawn from P on the given line.

2). Write direction ratios of PL.

3). Apply the condition of perpedidularity of the given line and PL.

4). Obtain the value of $$\lambda$$  from step 3.

5). Substitute $$\lambda$$ in ($$x_1 + a\lambda$$, $$y_1 + b\lambda$$, $$z_1 + c\lambda$$) to obtain the coordinates of L.

6). Obtain PL by using distance formula.

#### (b) Vector Form

Algorithm :

Let P($$\vec{\alpha})$$ be the given point, and let $$\vec{r}$$ = $$\vec{a}$$ + $$\lambda \vec{b}$$ be the given line be

1). Write the position vector of a general point on the given line. The position vector of a general point on $$\vec{r}$$ = $$\vec{a}$$ + $$\lambda \vec{b}$$  is $$\vec{a}$$ + $$\lambda \vec{b}$$, where $$\lambda$$ is a parameter. Assume that this point L is required foot of the perpendicular from P on the given line.

2). Obtain $$\vec{PL}$$ = Position vector of L – Position Vector of P = $$\vec{a}$$ + $$\lambda \vec{b}$$ – $$\vec{\alpha}$$.

3). Put $$\vec{PL}$$.$$\vec{b}$$ = 0 i.e. ($$\vec{a}$$ + $$\lambda \vec{b}$$ – $$\vec{\alpha}$$).$$\vec{b}$$ = 0 to obtain the value of $$\lambda$$.

4). Substitute the value of $$\lambda$$ in $$\vec{r}$$ = $$\vec{a}$$ + $$\lambda \vec{b}$$ to obtain the position vector of L.

5). Find |$$\vec{PL}$$| to obtain the required length of the perpendicular.

Example : Find the foot of the perpendicular from the point (0, 2, 3) on the line $$x + 3\over 5$$ = $$y – 1\over 2$$ = $$z + 4\over 3$$.

Solution : Let L be the foot of the perpendicular drawn from the point P (0, 2, 3) to the given line.

The coordinates of a general point on the line $$x + 3\over 5$$ = $$y – 1\over 2$$ = $$z + 4\over 3$$ is given by

$$x + 3\over 5$$ = $$y – 1\over 2$$ = $$z + 4\over 3$$ = $$\lambda$$

or, x = $$5\lambda – 3$$, y = $$2\lambda + 1$$, z = $$3\lambda – 4$$.

Let the coordinates of L be ($$5\lambda – 3$$, $$2\lambda + 1$$, $$3\lambda – 4$$). Therefore, direction ratios of PL are proportional to

$$5\lambda – 3$$ – 0, $$2\lambda + 1$$ – 2, $$3\lambda – 4$$ – 3 i.e. $$5\lambda – 3$$, $$2\lambda – 1$$, $$3\lambda – 7$$.

Direction ratios of the given line are proportional to 5, 2, 3.

But, PL is perpendicular to the given line.

$$\therefore$$  5($$5\lambda – 3$$) + 2($$2\lambda – 1$$) + 3($$3\lambda – 7$$) = 0 $$\implies$$ $$\lambda$$ = 1

Putting $$\lambda$$ = 1 in ($$5\lambda – 3$$, $$2\lambda + 1$$, $$3\lambda – 4$$), the coordinates of L are (2, 3, -1).