The lines x -2-1- y -3-1- z -4- k and x -1- k - y -4-2- z -5-1 are coplanar ifA- k -1 or -1B- k -0 or -3C- k -3 or -3D- k -0 or -1

The correct option is B k = 0 or 3x 21 = y 31 = z 4 k ...(1) nbsp; nbsp; x 1k = y 42 = z 51 nbsp;...(2) nbsp; The lines (1) and (2) ...

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Byju's Answer

Standard XII

Mathematics

Condition for Coplanarity of Four Points

The lines x -...

Question

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 if

k = 1 or - 1

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k = 0 or - 3

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k = 3 or - 3

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k = 0 or - 1

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Solution

The correct option is B $k = 0 or - 3$
$\frac{x - 2}{1} = \frac{y - 3}{1} = \frac{z - 4}{- k} . . . (1)$
$\frac{x - 1}{k} = \frac{y - 4}{2} = \frac{z - 5}{1} . . . (2)$

The lines $(1)$ and $(2)$ are coplanar if
$∣ ∣ ∣ ∣ \begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} l_{1} & m_{1} & n_{1} l_{2} & m_{1} & n_{2} \end{matrix} ∣ ∣ ∣ ∣ = 0$

$\Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} 1 & - 1 & - 1 1 & 1 & - k k & 2 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$
$\Rightarrow k^{2} + 3 k = 0$
$\Rightarrow k = 0 or - 3$

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The lines x−21=y−31=z−4−k and x−1k=y−42=z−51 are coplanar if

The correct option is B k=0 or −3x−21=y−31=z−4−k ...(1) x−1k=y−42=z−51 ...(2) The lines (1) and (2) are coplanar if ∣∣ ∣∣x2−x1y2−y1z2−z1l1m1n1l2m1n2∣∣ ∣∣=0 ⇒∣∣ ∣∣1−1−111−kk21∣∣ ∣∣=0 ⇒k2+3k=0 ⇒k=0 or −3

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 if