If the lines x-k1 - y-22 - z-33 and x-13 - y-22 - z-31 are co-planar- then the value of k is

∴ 1 amp;2 amp;3 3 amp;2 amp;1 k+1 amp;4 amp; 6 =0 ⇒ 1(12 4) 2(18 k 1)+3(12 2k 2)=0 ⇒ 8 34+2k+30 6k=0 ⇒ 4k=4 ∴ k=1 ...

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Slope Intercept Form of a Line

If the lines ...

Question

If the lines $\frac{x - k}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and $\frac{x + 1}{3} = \frac{y + 2}{2} = \frac{z + 3}{1}$ are co-planar, then the value of $k$ is

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

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

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x i + j + k

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i + j + z k

x \neq 1

y \neq 1

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\frac{x}{1 - x} + \frac{y}{1 - y} + \frac{z}{1 - z}

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

\frac{x}{1} = \frac{y}{2} = \frac{z}{k}

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

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

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\frac{x + 3}{- 3} = \frac{y - 1}{1} = \frac{z - k}{5}

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

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