Question

If the lines \(\frac{x-k}{1}=\frac{y-2}{2}=\frac{z-3}{3}\) and \(\frac{x+1}{3}=\frac{y+2}{2}=\frac{z+3}{1}\) are co-planar, then the value of \(k\) is _____________.

Hint:

Coplanar lines have a vanishing scalar triple product of the joining vector and the two direction vectors.

Answer 1

Correct Answer: 1

The direction vectors are not parallel, so the lines are coplanar iff they intersect. A point on line 1: \((k+\lambda,\,2+2\lambda,\,3+3\lambda)\); on line 2: \((-1+3\mu,\,-2+2\mu,\,-3+\mu)\).
\(y\): \(2+2\lambda=-2+2\mu\Rightarrow\lambda=\mu-2\). \(z\): \(3+3\lambda=-3+\mu\Rightarrow3+3\mu-6=-3+\mu\Rightarrow\mu=0,\ \lambda=-2\).
\(x\): \(k+\lambda=-1+3\mu\Rightarrow k-2=-1\Rightarrow k=1\).
Answer: 1