Question

Equation of a Line Passing Through Two Given Points of three dimensional geometry.

Answer 1

Vector Form :

Let  \(\vec a \ and \ \vec b\)be the position vectors of two points A(x₁ y₁ z₁) and B(x₂, y₂, z₂), respectively that are lying on a line L.

If O be origin, then \(\overrightarrow {OA} = \vec a \ and \ \overrightarrow {OB} = \vec b.\)

\(\overrightarrow {AB}\) = (Position vector of B) - (Position vector of A)

= \(\vec b \ - \ \vec a\)

Let \(\vec r\)be the position vector of an arbitrary point P(x, y, z) on line L, then \(\overrightarrow {OP} = \vec r\)

\(\overrightarrow {AP} = \vec r - \vec a\)

Since, \(\overrightarrow {AP} = \overrightarrow {AB}\) are collinear vectors,

Thus, vector equation of line L, passing throughpoints \(A(\vec a) \ and \ B(\vec b)\) is 

Derivation of Cartesian Form from Vector Form:

We have \(\vec r\)= xî + yĵ + zk̂

\(\vec a\)= x₁î + y₁ĵ + z₁k̂

\(\vec b\)= x₂î + y₂ĵ + z₂k̂

Substituting these values in equation (2), we get

xî + yĵ + zk̂ = x₁î + y₁ĵ + z₁k̂ + λ(x₂ - x₁)î + (y₂ - y₁)ĵ + (z₂ - z₁)k̂]

Eliminating the like coefficients of î, ĵ, k̂, we get

x = x₁ + λ(x₂ - x₁),

⇒ x - x₁ = λ(x₂ - x₁);

y = y₁ + λ(y₂ - y₁)

⇒ y - y₁ = λ(y₂ - y₁);

z = z₁ + λ(z₂ - z₁)

⇒ z - z₁ = λ(z₂ - z₁)

On eliminating λ, we get

Which is the equation of line in cartesian form passing through the given points.

Angle Between Two Lines

(i) Let L₁ and L₂ be two lines passing through origin whose directionratios are a₁ b₁ c₁ and a₂, b₂, c₂ respectively. Let P and Q be two points on L₁ and L₂ respectively and θ be the acute angle between OP and OQ.

The angle θ between them is given by

The angle between the lines in terms of sin θ is given by

Remarks:

If the lines L₁ and L₂ do not pass through the origin, we may take lines L₁ and L₂ which are parallel to L₁ and L₂ respestively and pass through the origin.

(ii) Let l₁ m₁ n₁ and l₂, m₂, n₂ be the direction-cosines of line OP and OQ respectively. If θ is the angle between these lines, then

cos θ = |l₁l₂ + m₁m₂ + n₁n₂|

(Since, l₁² + m₁² + n₁² = 1 = l₂² + m₂² + n₂²)

and sin θ = 

⇒ l₁l₂ + m₁m₂ + n₁n₂ = 0

If a₁ b₁ c₁ and a₂, b₂, c₂ are the direction-ratios of the lines, then the lines are mutually perpendicular if a₁a₂ + b₁b₂ + c₁c₂ = 0

(iv) Two parallel lines : If l₁ m₁ n₁ and l₂, m₂, n₂ are the direction-cosines of two lines and the lines are mutually parallel, then 0 = 0°

sin θ = sin 0° = 0

When a₁ b₁ c₁ and a₂, b₂, c₂ are the direction-ratios of the lines then they are mutually parallel if

Now, we find the angle between two lines when their equations are given. If θ is the acute angle between the lines

where, a₁ b₁ c₁ and a₂, b₂, c₂ are the direction ratios of the lines (1) and (2), respectively, then