Q. 274.8( 5 Votes )

Find the equation

Answer :

The two given lines are:




Let a, b, c be the direction Ratios of the normal to the plane containing the line (1).


Also, as plane contains line (1), so it passes through (1, 4, 4)


Therefore, equation of plane is -


a (x –1) + b(y – 4)+c(z – 4) = 0 …(3)


We know that if a1, b1, c1 are direction Ratios of a plane and a2, b2, c2 are direction Ratios of a line parallel to it then,


a1a2 + b1b2 + c1c2 = 0


As, required plane is parallel to line (1),


3a+2b – 2c = 0 … (4)


Also, as required plane is parallel to line (2),


2a – 4b + 1c = 0 … (5)


Solving (4) and (5),




Let,


Putting, a = 6λ, b = λ7, c = 16λ in (3), we get


6λ(x –1) + 7λ(y – 4) + 16λ(z – 4) = 0


λ((6x –6) + (7y – 28) + (16z – 64)) = 0


6x + 7y +16z – 6 – 28 – 64 = 0,


6x + 7y +16z – 98 = 0,


which is the required equation of the plane.


Since line (2) is parallel to required plane


SD between two lines = Perpendicular distance of the point


(–1, 1 –2) from the plane.


Distance between a point (x’, y’, z’) from a line


ax + by + cz – d = 0 is -






Therefore,


OR


Equation of given line,



Also, (1) is coplanar with the line determined by the planes


x + 2y + 3z – 8 = 0 … (2)


and


2x + 3y + 4z – 11 = 0 … (3)


So, need to show there exists a plane passing through intersection of planes (2) and (3) containing the line (1).


Any plane passing through intersection of two planes P1 and P2 is


P1 + kP2 = 0


Where k is some constant,


So, equation of the plane passing through the intersection of planes (2) and (3) is -


(x + 2y + 3z – 8) + k (2x + 3y + 4z – 11) = 0 … (4)


As, (4) passes through (-1, -1, -1) so, substituting the coordinates of the point (–1, –1, –1) in (4), we get,


(–1–2–3–8) + k (–2–3–4–11) = 0


(-14) + k (-20) = 0



Putting in (4) we get,



10(x + 2y + 3z – 8) -7(2x + 3y + 4z – 11) = 0


10x + 20y + 30z – 80 – 14x + 21y + 28z – 77 = 0


4x + y – 2z + 3 = 0


So, plane passing through intersection of planes (2) and (3) is-


4x + y – 2z + 3 =0 … (5)


Now we find value of a1a2 + b1b2 + c1c2 where a1, b1, c1 are


Direction Ratios of the line (1) and a2, b2, c2 are Direction Ratios of the normal to the plane (5)


From equation (1) – (a1, b1, c1) = (1, 2, 3) and


From equation (5) – (a2, b2, c2) = (4, 1, -2)


So, a1a2 + b1b2 + c1c2 = (1)(4) + (2)(1) + (3) (-2)


= 4 + 2 – 6


= 0


i.e., a1a2 + b1b2 + c1c2 = 0


which implies line (1) lies in plane (5)


Hence the two lines are coplanar and the equation of the plane containing them is 4x + y – 2z + 3 = 0


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