Any point on
x−12=y+13=z−14=λ is
(1+2λ,−1+3λ,1+4λ)Any point on x−31=y−k2=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+μ⇒2−2λ+μ=0−1+3λ=k+2μ⇒k+1−3λ+2μ=01+4λ=μ⇒−1−4λ+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^i−5.5^j−5^k).(−5^i+2^j+^k)=−6.
Therefore, the equation of the plane is 5x−2y−z−6=0.