Question

The number of real values of $k$ for which the line $\frac{x-1}{4}=\frac{y+1}{3}=\frac{z}{k}$ and $\frac{x}{1}=\frac{y-k}{3}=\frac{z-1}{-2}$ are coplanar, is

• A. $2$
• B. $1$
• C. $3$
• D. $0$

Answer 1

The correct option is A $2$

Direction ratio of lines are $(4,3,k)$ and $(1,3,-2)$ respectively. Now direction ration of a line passing through the points on the two lines will be $(1,-1-k,0-1)$
As these are coplanar, so scalar dot product of these three must be $0$,
$\left|\begin{array}{ccc}1& 1-k& -1\\ 4& 3& k\\ 1& 3& -2\end{array}\right|$ $=$ $0$
On expanding, we get
$1(-6-3k)-(1-k)(-8-k)-1(12-3)=0$
$\Rightarrow -k^2-10k-7=0$
$\Rightarrow k^2+10k+7=0$
$\Rightarrow D={10}^2-4\times 7>0$

$\Rightarrow D=100-28>0$

$\Rightarrow D=72>0$
So it has two real roots.