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Question

If lines x12=y+13=z14 and x31=yk2=z1 intersect, then find the value of k and hence find the equation of the plane containing these lines.

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Solution

Any point on x12=y+13=z14=λ is (1+2λ,1+3λ,1+4λ)
Any point on x31=yk2=z1=μ is
(3+μ,k+2μ,μ).
For the two lines to intersect, the following three equations must be satisfied simultaneously:
1+2λ=3+μ22λ+μ=01+3λ=k+2μk+13λ+2μ=01+4λ=μ14λ+μ=0
Solving the above equations, we get, λ=1.5,μ=5,k=4.5
the point of intersection is therefore, (2,5.5,5).
The normal to the plane containing the lines is obtained by the cross product of the vectors:
(2^i+3^j+4^k)×(^i+2^j+^k)=5^i+2^j+^k.
Since the plane contains both the lines it must contain the point of intersection as well, so in order to find the constant d in the equation of the plane r.n=d, where r is any point on the plane and n is the normal vector to the plane, we substitute the point of intersection for r in the equation:
d=(2^i5.5^j5^k).(5^i+2^j+^k)=6.
Therefore, the equation of the plane is 5x2yz6=0.

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