If the straight lines (x-1)/(2)=(y+1)/(k)=(z)/(2) and (x+1)/(5)=(y+1)/(2)=(z)/(k) are coplanar, then the plane (s) containing these two lines is (are)

Correct answer: B. y+z=-1 Given that lines are coplanar. | 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 array |=| array ccc 2 & 0 & 0 2 & k & 2 5 & 2 & k array |=0 k=± 2 For k=2 , equation of the plane is given by | array ccc x-1 & y+1 & z 2 & 2 & 2 5 & 2 & 2 array |=0 y-z+1=0 For k=-2 , equation of the plane is given by | array ccc x-1 & y+1 & z 2 & -2 & 2 5 & 2 & -2 array |=0 y+z+1=0

JEE Advanced 2012Paper 2

Exam: JEE Advanced
Year: 2012
Paper: Paper 2
Subject: Mathematics
Chapter: Three Dimensional Geometry

JEE Advanced 2012 · Paper 2 · Mathematics · Three Dimensional Geometry

If the straight lines (x-1)/(2)=(y+1)/(k)=(z)/(2) and...

If the straight lines

\frac{x-1}{2}=\frac{y+1}{k}=\frac{z}{2}

and

\frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}

are coplanar, then the plane (s) containing these two lines is (are)

Options

A $y+2 z=-1$
B $y+z=-1$
C $y-z=-1$
D $y-2 z=-1$

Correct answer

B. $y+z=-1$

Step-by-step solution

Given that lines are coplanar.

\therefore\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|=\left|

\begin{array}{ccc} 2 & 0 & 0 \\ 2 & k & 2 \\ 5 & 2 & k \end{array}

\right|=0 \Rightarrow k=\pm 2

For

k=2

, equation of the plane is given by \left|

\begin{array}{ccc}x-1 & y+1 & z \\ 2 & 2 & 2 \\ 5 & 2 & 2\end{array}

\right|=0 \Rightarrow y-z+1=0

For

k=-2

, equation of the plane is given by \left|

\begin{array}{ccc}x-1 & y+1 & z \\ 2 & -2 & 2 \\ 5 & 2 & -2\end{array}

\right|=0 \Rightarrow y+z+1=0

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