Question

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

• A. $\frac{11}{2}$
• B. $-\frac{11}{2}$
• C. $\frac{9}{2}$
• D. $-\frac{9}{2}$

Answer 1

The correct option is A $\frac{11}{2}$

To find: value of k

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

$x=2m-1$; $y=m+1$; $z=3m-1$

$\frac{x+2}{2}=\frac{y-k}{3}=\frac{z}{4}=n$

$x=2n-2;z=4n;y=3n+k$

$2m-1=2n-2$

$m=\frac{2n-1}{2}$

Also, $3m-1=4n$

$\frac{3(2n-1)}{2}-1=4n$

$6n-5=8n$

$n=\frac{-5}{2}$

$m+1=3n+k$

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

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

$k=\frac{1-4n}{2}$

$=\frac{1+10}{2}$

$k=\frac{11}{2}$.