Question

The lines $\frac{x-1}{a}=\frac{y-2}{3}=\frac{z-3}{4}$; $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$ are coplanar then, the value of $a$ is:

• A. $1$
• B. $2$
• C. $-1$
• D. $-2$

Answer 1

The correct option is B $2$

Given lines are $\frac{x-1}{a}=\frac{y-2}{3}=\frac{z-3}{4}$; $\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}$
Points on the two lines respectively are $(x_1,y_1,z_1)=(1,2,3)$ and $(x_2,y_2,z_2)=(2,3,4)$
The corresponding $D.R^{\prime }s$ are $(a_1,b_1,c_1)=(a,3,4)$ and $(a_2,b_2,c_2)=(3,4,5)$
For the lines to be coplanar, $D=\left|\begin{array}{ccc}{x}_1-x_2& y_1-y_2& z_1-z_2\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|=0$
$\Rightarrow \left|\begin{array}{ccc}-1& -1& -1\\ a& 3& 4\\ 3& 4& 5\end{array}\right|=0$
$\Rightarrow -1(15-16)+1(5a-12)-1(4a-9)=0$
$\Rightarrow 1+5a-12+9-4a=0$
$\Rightarrow -2+a=0\Rightarrow a=2$