Question

Equation of a Line in Space of three dimensional geometry.

Answer 1

Here, in this section we shall read about vector and cartesian equation of a line in space.
A line is uniquely defined, if

• it passes through the given point and the given in space.
• it passes through two given points.

Equation of a Line through a given Point A  \((\vec a)\) and Parolielto given Vector \(\vec m.\)

Let L be the line which passes through the point A, whose position vector is \(\vec a\)and is parallel to the vector \(\vec b.\)

Let O be orgin. Then, \(\overrightarrow {OA} = \vec a.\) Let \(\vec r\)be the position vector of an arbitrary point P on the line.

Clearly, for each value of the parameter A, this equation gives the position vector of a point P on the line.
Hence, the vector equation of a line is given by

\(\vec r = \vec a + \lambda \vec b\) .......(1)

Derivation of Cartesian Form from Vector Form:

Let the coordinates of the given point A be (x₁ y₁ z₁) and the direction ratios of the line be a, b, c. Consider the coordinates of any point P be (x, y, z). Then,

\(\vec r \) = = xî + yĵ + zk̂

a = x₁î + y₁ĵ + z₁k̂

∵ Direction ratios of given line are a, b, c and it is parallel to the vector \(\vec b\)

\(\therefore \vec b\) = aî + bĵ + ck̂

Now, vector equation of line is

\(\vec r = \vec a + \lambda \vec b\)

Substituting the values of \(\vec a,\vec b \ and\ \vec r\) in equation (i), we get

xî + yĵ + zk̂ =( x₁î + y₁ĵ + z₁k̂) + λ (ai + bj + ck)

⇒ xî +yĵ +zk̂ = (x₁ + λa)î + (y₁ + kb)ĵ + (z₁ + λc) k̂

Equating the coefficients of î, ĵ and k̂, we get

⇒ x = x₁ + λa, y = y₁ + kb, z = z₁ + λc

These are the parametric equations of the line.

Eliminating the parameter X from (ii), we get the cartesian equation of line, whose direction ratios are a, b, c and which passing through A(x, y, z) as :