Question

If the lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{2}{1}$ intersect, then the value of $k$ is

• A. $\frac{7}{2}$
• B. $\frac{9}{2}$
• C. $\frac{-2}{9}$
• D. $\frac{-3}{2}$

Answer 1

Answer: (A)

$\frac{7}{2}$

We know that the condition for the lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$ to intersect is $\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|=0$

Given lines $\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}$ and $\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}$

Here $x_1=1,x_2=3;y_1=-1,y_2=k;z_1=1,z_2=0;a_1=2,b_1=3,c_1=4;a_2=1,b_2=2,c_2=1$

$\Rightarrow \left|\begin{array}{ccc}2& k+1& -1\\ 2& 3& 4\\ 1& 2& 1\end{array}\right|=0$

$\Rightarrow 2(3-8)-(k+1)(2-4)-1(4-3)=0$

$\Rightarrow 2(-5)-(k+1)(-2)-1=0$

$\Rightarrow -10+2k+2-1=0$

$\Rightarrow 2k-7=0$

$\Rightarrow k=\frac{7}{2}$