If the lines (x-2)/1=(y-3)/1=(z-4)/(-k) and (x-1)/k=(y-4)/2=(z-5)/1 are coplanar, then k can have

Correct answer: (A) exactly two values.. For x-x_1a_1=y-y_1b_1=z-z_1c_1 and x-x_2a_2=y-y_2b_2=z-z_2c_2 to be coplanar if x_2-x_1 y_2-y_1 z_2-z_1 a_1 b_1 c_1 a_2 b_2 c_2 =0. ∴ 1 -1 -1 1 1 -k k 2 1 =0⇒k=0, -3 Hence, there are exactly two values of k.

Q53 / 90 · JEE Main 2013 (07 Apr) · Mathematics

If the lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar, then $k$ can have

Correct answer
A
exactly two values.
B
exactly three values.
C
any value.
D
exactly one value.

Solution

For $\frac{x - x_{1}}{a_{1}} = \frac{y - y_{1}}{b_{1}} = \frac{z - z_{1}}{c_{1}}$

and $\frac{x - x_{2}}{a_{2}} = \frac{y - y_{2}}{b_{2}} = \frac{z - z_{2}}{c_{2}}$ to be coplanar if

$|\begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix}| = 0$ .

$∴ |\begin{matrix} 1 & - 1 & - 1 \\ 1 & 1 & - k \\ k & 2 & 1 \end{matrix}| = 0 \Rightarrow k = 0, - 3$

Hence, there are exactly two values of $k$ .

Hence, the correct answer is option: 'A'.

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If the lines x-21=y-31=z-4-k and x-1k=y-42=z-51 are coplanar, then k can have

Solution

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If the lines $\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k}$ and $\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1}$ are coplanar, then $k$ can have