If the lines x-11-y2-z3 and xk-1-yk-zk2 are coplanar- then k equals

1 amp;0 nbsp; amp;0 1 nbsp; nbsp; amp;2 nbsp; amp;3 k 1 nbsp; nbsp; amp;k nbsp; amp;k^2 nbsp; =0 ⇒ 2k^2 3k=0⇒ k=0,k=32 ...

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Byju's Answer

Standard XII

Mathematics

Condition for Coplanarity of Four Points

If the lines ...

Question

If the lines $\frac{x - 1}{1} = \frac{y}{2} = \frac{z}{3}$ and $\frac{x}{k - 1} = \frac{y}{k} = \frac{z}{k^{2}}$ are coplanar, then k equals

k = 0

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k = 1

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k = 2

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k = \frac{3}{2}

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Solution

The correct option is D $k = \frac{3}{2}$
$∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 1 & 2 & 3 k - 1 & k & k^{2} \end{matrix} ∣ ∣ ∣ ∣ = 0$

$\Rightarrow 2 k^{2} - 3 k = 0 \Rightarrow k = 0, k = \frac{3}{2}$

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If the lines x−11=y2=z3 and xk−1=yk=zk2 are coplanar, then k equals

The correct option is D k=32∣∣ ∣∣100123k−1kk2∣∣ ∣∣=0 ⇒2k2−3k=0⇒k=0,k=32

If the lines $\frac{x - 1}{1} = \frac{y}{2} = \frac{z}{3}$ and $\frac{x}{k - 1} = \frac{y}{k} = \frac{z}{k^{2}}$ are coplanar, then k equals

The correct option is D $k = \frac{3}{2}$
$∣ ∣ ∣ ∣ \begin{matrix} 1 & 0 & 0 1 & 2 & 3 k - 1 & k & k^{2} \end{matrix} ∣ ∣ ∣ ∣ = 0$

$\Rightarrow 2 k^{2} - 3 k = 0 \Rightarrow k = 0, k = \frac{3}{2}$