If the lines $\frac{x - 1}{- 3} = \frac{y - 2}{- 2 k} = \frac{z - 3}{k}$ and $\frac{x - 1}{k} = \frac{y - 2}{1} = \frac{z - 3}{5}$ are perpendicular, find the value of $k$ and hence find the equation of plane containing these lines.

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Solution

The direction vectors of the lines are given by $(- 3, - 2 k, k)$ and $(k, 1, 5)$
Since the two lines are perpendicular, dot product of the direction vectors must yield zero.
$\Rightarrow - 3 k - 2 k + 5 k = 0$
Thus, k can be any real number, except zero.
The point $(1, 2, 3)$ lies on both the lines.
For finding the equation of the place containing the two lines, we find the equation of its normal, which would be the cross product of the two direction vectors.
Assuming $k$ to be $1, (- 3^i - 2^j +^k) \times (^i +^j + 5^k) = - 3^k + 15^j + 2^k - 10^i +^j -^i = - 11^i + 16^j -^k$
The equation of the plane thus would be of the form $- 11 x + 16 y - z + d = 0$
$d$ can be found out by substituting $(1, 2, 3)$ as $(x, y, z)$ since the point lies on the plane as well.
$\Rightarrow - 11 + 32 - 3 + d = 0$
$∴ d = - 18$
The equation of plane is thus $- 11 x + 16 y - z - 18 = 0$

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