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=x_{1} î +y_{1}ĵ +z_{1}k̂ and b ^{->}= (x_{2} î +y_{2}ĵ +z_{2}k̂)

Substituting the values in equation r=a+λ(b^{->}-a^{->}) ,Therefore

x î +yĵ +zk̂=x_{1} î +y_{1}ĵ +z_{1}k̂+ λ((x_{2}-x_{1})î+(y_{2}-y_{1})ĵ +(z_{2}-z_{1})k̂)

Equating the coefficients of î**,**ĵ**,**k̂we get,

x=x_{1}+ λ(x_{2}-x_{1}); y=y_{1}+λ(y_{2}-y_{1});z=z_{1}+λ(z_{2}-z_{1})

On eliminating λ, we get,

Therefore equation of the line in Cartesian form will be :-

**(x-x _{1})/(x_{2}-x_{1}) = (y-y_{1})/(y_{2}-y_{1}) =(z-z_{1})/(z_{2}-z_{1})**

__Problem:-__

Find the vector and the Cartesian equations of the line that passes through the points (3, −2, −5), (3, −2, 6).

__Answer:-__

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-x_{1})/ (a) =(y-y_{1})/ (b) = (z-z_{1})/(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).

__Answer:-__

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 (x_{1}, y_{1}, z_{1}) and direction ratios a, b, c is given by,

(x-x_{1})/ (a) =(y-y_{1})/ (b) = (z-z_{1})/(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.

__Answer:-__

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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