|Class 12 Maths Three Dimensional Geometry||Equation of a line passing through two given points|
Equation of a line passing through two given points
Derivation of Cartesian form from vector form
Consider r-> = x î +yĵ +zk̂, a=x1 î +y1ĵ +z1k̂ and b ->= (x2 î +y2ĵ +z2k̂)
Substituting the values in equation r=a+λ(b->-a->) ,Therefore
x î +yĵ +zk̂=x1 î +y1ĵ +z1k̂+ λ((x2-x1)î+(y2-y1)ĵ +(z2-z1)k̂)
Equating the coefficients of î,ĵ,k̂we get,
x=x1+ λ(x2-x1); y=y1+λ(y2-y1);z=z1+λ(z2-z1)
On eliminating λ, we get,
Therefore equation of the line in Cartesian form will be :-
(x-x1)/(x2-x1) = (y-y1)/(y2-y1) =(z-z1)/(z2-z1)
Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).
Let the line passing through the points, P (3, −2, −5) and Q (3, −2, 6), be PQ.
Since PQ passes through P (3, −2, −5), its position vector is given by,a->=3î-2ĵ-5k̂
The direction ratios of PQ are given by,
(3 − 3) = 0, (−2 + 2) = 0, (6 + 5) = 11
The equation of the vector in the direction of PQ is: -b->=0.î-0ĵ+11k̂=11k̂
The equation of PQ in vector form is given by,r-> = a-> + λb->,whereλ∈ R.
=>r->= (3î -2ĵ -5k̂) +11λk̂
The equation of PQ in Cartesian form is=(x-x1)/ (a) =(y-y1)/ (b) = (z-z1)/(c)
=(x-3)/ (0) =(y+2)/ (0) = (z+5)/ (11)
Find the vector and the Cartesian equations of the lines that pass through the origin and (5, −2, 3).
The required line passes through the origin. Therefore, its position vector is given by,
a->=0 (equation (1))
The direction ratios of the line through origin and (5, −2, 3) are
(5 − 0) = 5, (−2 − 0) = −2, (3 − 0) = 3
The line is parallel to the vector given by the equation, b->= 5î-2ĵ +3k̂
The equation of the line in vector form through a point with position vectors a-> and parallel to
b->= r->=0 +λ (5î -2ĵ +3k̂)
r->=λ(5î -2ĵ +3k̂)
The equation of the line through the point (x1, y1, z1) and direction ratios a, b, c is given by,
(x-x1)/ (a) =(y-y1)/ (b) = (z-z1)/(c)
Therefore, the equation of the required line in the Cartesian form is:-
(x-0)/ (5) =(y-0)/ (-2) = (z-0)/ (3)
=>(x/5) =(y/-2)/ (z/3)
The Cartesian equation of a line is (x-5)/ (3) =(y+4)/ (7) = (z-6)/ (2). Write its vector form.
The Cartesian equation of the line is (x-5)/ (3) =(y+4)/ (7) = (z-6)/ (2) (equation (1)).
The given line passes through the point (5, −4, 6). The position vector of this point is a->=5î-4ĵ +6k̂.
Also, the direction ratios of the given line are 3, 7, and 2.
This means that the line is in the direction of vector, b-> =3î+7ĵ +2k̂.
It is known that the line through position vector a⃗ and in the direction of the vector b-> is given by the equation, r->= a-> + λb->, whereλ∈ R.
=>r-> = (5î -4ĵ +6k̂) + λ (3î +7ĵ +2k̂)
This is the required equation of the given line in vector form.