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).

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