Question

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

• A. $\frac{2}{9}$
• B. $\frac{1}{3}$
• C. $0$
• D. $-1$

Answer 1

The correct option is B $\frac{1}{3}$

$\frac{x-1}{2}=\frac{y+1}{3}=\frac{z-1}{4}=t$

$(x,y,z)\equiv (2t+1,3t-1,4t+1)$

$\frac{x-3}{1}=\frac{y-k}{2}=\frac{z}{1}=m$

$\therefore (x,y,z)\equiv (m+3,2m+k,m)$

$(2t+1,3t-1,4t+1)\equiv (m+3,2m+k,m)$

$2t+1=2m+3$

$4t+!=m$

$2t=-m-3$

$t=\frac{-m-3}{2}$

$-,2m-6+1=m$

$-5=3m$

$m=-\frac{5}{3}$

$-2-1=\frac{-10}{3}+k$

$\frac{10}{3}-3=k$

$k=\frac{1}{3}$