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…
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 :
-
$\frac{11}{2}$
-
$-\frac{11}{2}$
-
$\frac{9}{2}$
-
$-\frac{9}{2}$
Solution
Two given planes are coplanar, if
$
\begin{aligned}
& \left|\begin{array}{ccc}
-2-(-1) & k-1 & 0-(-1) \\
2 & 1 & 3 \\
2 & 3 & 4
\end{array}\right|=0 \\
\Rightarrow & \left|\begin{array}{ccc}
-1 & k-1 & 1 \\
2 & 1 & 3 \\
2 & 3 & 4
\end{array}\right|=0 \\
\Rightarrow & (-1)(4-9)-(k-1)(8-6)+6-2=0 \\
\Rightarrow & k=\frac{11}{2}
\end{aligned}
$
Asked in: JEE Main 2013 (09 Apr Online)
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