Question

The straight line $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$ and $\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}$ will intersect provided if k=:

• A. k = {3, –3}
• B. k = {0, –1}
• C. k = {–1, 1}
• D. k = {0, –3}

Answer 1

The correct option is D k = {0, –3}

The two lines $\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}$ and $\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}$ will intersect i.e., coplanar if
$\left|\begin{array}{ccc}1& -1& -1\\ 1& 1& 1\\ k& 2& 1\end{array}\right|=0\Rightarrow k^2+3k=0$ i.e., k =0, -3