Any point on \(\frac{x - 1}2 = \frac{y + 1}3 = \frac {z-1}4=\) λ is (1 + 2λ, −1 + 3λ, 1 + 4λ)
Any point on \(\frac{x - 3}1 = \frac{y - k}2 = \frac z1 =\) 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 = 0
1 + 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:
\((2\hat i + 3\hat j + 4\hat k) \times (\hat i + 2\hat j + \hat k) = -5\hat i + 2\hat j + \hat k\)
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 \(\vec r.\vec n = d\), where \(\vec r\) is any point on the plane and \(\vec n\) is the normal vector to the plane, we substitute the point of intersection for \(\vec r\) in the equation:
\(d = (-2 \hat i - 5.5\hat j - 5 \hat k). (-5\hat i + 2\hat j + \hat k) = -6\).
Therefore, the equation of the plane is 5x − 2y − z − 6 = 0.
