Test Index

UPSC NDA Math Model Paper 5

© examsnet.com

Question : 54 of 120

Marks: +1, -0

The lines

\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}

\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}

are coplanar if the value of

k

is:

1
2
-2
-3

Solution:

We know that

The lines

\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}

and

\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}

are coplanar then

\Rightarrow \begin{vmatrix} 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{vmatrix} =0

=0

Given The lines

\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}

and

\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}

are coplanar

Here,

x_{1}=1, x_{2}=2, y_{1}=4, y_{2}=3, z_{1}=5, z_{2}=4,

a_{1}=k, b_{1}=2, c_{1}=1

a_{2}=1, b_{2}=1, c_{2}=-k

\Rightarrow \begin{vmatrix} 1 & -1 & -1 \\ k & 2 & 1 \\ 1 & 1 & -k \end{vmatrix} =0

\Rightarrow 1(-2k-1)+1(-k^{2}-1)-1(k-2)=0

\Rightarrow -2k-1-k^{2}-1-k+2=0

\Rightarrow -k^{2}-3k=0

\Rightarrow k(k+3)=0

Neglect

k=0

\Rightarrow k=-3

Hence, The lines

\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}

and

\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}

are coplanar if the value of k = - 3

© examsnet.com

Go to Question:

Prev Question

Next Question