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]
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)
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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