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)

Problem:-

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->=--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 - -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)

Problem:-

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)

Problem:-

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.

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