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

2 1 amp;3 4 amp;4 5 1 amp;1 amp; k k amp;2 amp;1 =0 ⇒ 1 amp; 1 amp; 1 1 amp;1 amp; k k amp;2 amp;1 =0 ⇒ 1(1+2k)+1(1+k^2) 1(2 k)=0 ⇒ ...

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Byju's Answer

Standard XI

Mathematics

Condition for Coplanarity of Four Points

If the lines ...

Question

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 coplaner, then $k$ can have :

any value

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exactly one value

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exactly two values

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exactly three values

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Solution

The correct option is C exactly two values
$∣ ∣ ∣ ∣ \begin{matrix} 2 - 1 & 3 - 4 & 4 - 5 1 & 1 & - k k & 2 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$

$\Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} 1 & - 1 & - 1 1 & 1 & - k k & 2 & 1 \end{matrix} ∣ ∣ ∣ ∣ = 0$
$\Rightarrow 1 (1 + 2 k) + 1 (1 + k^{2}) - 1 (2 - k) = 0$
$\Rightarrow k^{2} + 3 k = 0$
$\Rightarrow k = 0, k = - 3$
Hence, $k$ has exactly two values.

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Similar questions

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

The correct option is C exactly two values∣∣ ∣∣2−13−44−511−kk21∣∣ ∣∣=0 ⇒∣∣ ∣∣1−1−111−kk21∣∣ ∣∣=0 ⇒1(1+2k)+1(1+k2)−1(2−k)=0 ⇒k2+3k=0 ⇒k=0, k=−3 Hence, k has exactly two values.

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 coplaner, then $k$ can have :