Class 12 Maths Three Dimensional Geometry Equation of a line in Space

Equation of a line in Space

A line is uniquely determined if:

  • It passes through a given point and has direction, or
  • It passes through two given points.

 

Line through a given point and parallel to a given vector:-

 

 

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.

Let the coordinates of the any point P be (x, y, z).

Consider r-> = xî +yĵ +zk̂, a=x1î+y1ĵ+z1k̂and b->= (a î +bĵ +ck̂) line vector L->=a î +bĵ +ck̂.

Equation of line: - r->= P-> + λL->.(equation(1))

Using equation (1) x1î+y1ĵ+z1k̂= x1î+y1ĵ+z1k̂+λ (aî +bĵ +ck̂).

Therefore, x=x1+λa => (x-x1)/(a)=λ

y=y1+λb =>  (y-y1)/(b) =λ

z=z1+λc => (z-z1)/(c)=λ

Therefore Equation of  line in Cartesian form is given by :-

(x-x1)/ (a) = (y-y1)/ (b) = (z-z1)/(c)(By eliminatingλ).

Note: - If l, m and n are direction cosines of the line then the equation of line is given as:

=(x–x1)/l =(y-y1)/m= (z-z1)/n

Problem:-

Find the equation of the line which passes through the point (1, 2, and 3) and is parallel tothe vector 3î+2ĵ -2k̂.

Answer:-

It is given that the line passes through the point A (1, 2, and 3). Therefore, the position vector through A is a->=î+2ĵ +3k̂ and b-> =3î +2ĵ -2k̂.

It is known that the line which passes through point A and parallel to b-> is given by  r->=a->+ λb->

=>r-> =î +2ĵ +3k̂+λ (3î +2ĵ -2k̂).

This is the required equation of the line.

Problem:-

Find the equation of the line in vector and in Cartesian form that passes through the point with position vector 2î-ĵ +4k̂ and is in the direction î +2ĵ -k̂?

Answer:-

It is given that the line passes through the point with position vector a->=2î -ĵ +4k̂(equation (1))

b-> = î +2ĵ -k̂(equation (2))

It is known that a line through a point with position vector a⃗and parallel tob ⃗ is givenby the equation, r-> = a-> + λb ->.

=>r-> =2î -ĵ +4k̂+ λ(î +2ĵ -k̂)

This is the required equation of the line in vector form.

r-> = x î +yĵ +zk̂=>(λ+2)î+ (2λ-1)ĵ+ (-λ+4)k̂

Eliminating λ, we obtain the Cartesian form equation as:

(x-2)/ (1) =(y+2)/ (2) = (z-4)/ (-1)

This is the required equation of the given line in Cartesian form.

Problem:-

Find the Cartesian equation of the line which passes through the point(−2, 4, −5) and parallel to the line given by (x+3)/ (3) =(y-4)/ (5) = (z+8)/ (6)?

Answer:-

It is given that the line passes through the point (−2, 4, −5) and is parallel to (x+3)/ (3) =(y-4)/ (5) = (z+8)/ (6)

The direction ratios of the line,(x+3)/ (3) =(y-4)/ (5) = (z+8)/ (6), are 3, 5, and 6.

The required line is parallel to (x+3)/ (3) =(y-4)/ (5) = (z+8)/ (6).

Therefore, its direction ratios are 3k, 5k, and 6k, where k ≠ 0

It is known that the equation of the line through the point (x1, y1, z1) and with direction ratios, a, b, c, is given by (x-x1)/ (a) =(y-y1)/ (b) = (z-z1)/(c)

Therefore the equation of the required line is (x+2)/ (3k) =(y-4)/ (5k) = (z+5)/ (6k)

=>(x+3)/ (3) = k

=> (y-4)/ (5) = k

=> (z+8)/ (6) =k.

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