The lines x-41-y-1-2-z1 and 9x-1613-9y-17-z-1 are

The correct option is A coplanarGiven lines are x 41=y+1 2=z1 and 9x 1613=9y 17=z 1 |x_1 amp;y_1 amp;z_1 x_2 amp;y_2 amp;z_2 a_2 ...

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Eccentricity

The lines x...

Question

The lines $\frac{x - 4}{1} = \frac{y + 1}{- 2} = \frac{z}{1}$ and $\frac{9 x - 16}{13} = \frac{9 y - 1}{7} = \frac{z}{- 1}$ are

coplanar

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Solution

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Consider lines

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

$L_{2} : \frac{x - 1}{2} = \frac{y - 4}{2} = \frac{z - 5}{1}$

Equation of plane containing these lines is

\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 0}{3} and \frac{x}{- 2} = \frac{y - 2}{- 3} = \frac{z + 1}{- 1}

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

\frac{x - 5}{1} = \frac{y - 7}{2} = \frac{z - 4}{- 1}

\frac{x - 2}{1} = \frac{y - 4}{4} = \frac{z - 6}{7}

\frac{x + 1}{3} = \frac{y + 3}{5} = \frac{z + 5}{7}

\frac{x + 1}{7} = \frac{y + 1}{- 6} = \frac{z + 1}{1}

\frac{x - 3}{1} = \frac{y - 5}{- 2} = \frac{z - 7}{1}

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