Q. 275.0( 1 Vote )

If lines and intersect, then find the value of k and hence find the equation of the plane containing these lines.[CBSE 2015]

Answer :

Given the equation of line are-



any random point on this line is given by (2λ+1, 3λ-1, 4λ+1)


Another line is:



any random point on this line is given by (μ+3, 2μ+k, μ)


At point of intersection of the lines the random coordinates must coincide.


2λ + 1 = μ+3


2λ - μ = 2 …(1)


Also,


4λ + 1 = μ


μ = 4λ + 1 …(2)


Adding equation 1 and 2,we get-


2λ = 4λ + 3




As, 3λ – 1 = 2μ + k


we have the values of λ and μ.


k = 3λ – 2μ – 1


k = 3(-3/2) – 2(-5) – 1


k = 9/2


Now we need to find the equation of the plane containing these 2 lines.


For this we need the normal vector to plain and a point on plane.


For normal we need to take the cross product of direction ratios of line.


Direction ratio of line 1 is (2,3,4)


And direction ratio of line 2 is ( 1,2,1)


the normal vector is given as -




Let be any random vector on pthe lane.



The equation of plane is given by –


where a is any defined vector on planthe e.


As lines lie on a plane, so one of its points can be taken as a poi t on the plane.


(2λ+1, 3λ-1, 4λ+1) will give a point on putting λ = 0


Point is (1, -1, 1)



Hence, the equation of the plane is:



-5x + 2y + z = -5-2+1 = -6


Equation is: -5x+2y+z = -6

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