The plane containing the line x-1-1-y-2-2-z-3-3 and parallel to the line x-1-y-1-z-4 is

The correct option is C 5x y z = 0 Required equation of plane is x 1 amp; y 2 amp; z 3 1 amp; 2 amp; 3 1 amp; 1 amp; 4 ⇒ 5x y z = 0 ...

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Question

The plane containing the line $\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z - 3}{3}$ and parallel to the line $\frac{x}{1} = \frac{y}{1} = \frac{z}{4}$ is

5x - y + z = 0

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5x + y - z = 0

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5x - y - z = 0

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5x + y + z = 0

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Solution

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Similar questions

\frac{x - 1}{3} = \frac{y + 6}{4} = \frac{z + 1}{2}

\frac{x - 2}{2} = \frac{y - 1}{- 3} = \frac{z + 4}{5}

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

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

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

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

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

x + 2 y + 3 z - 5 = 0 = 3 x + 2 y + z - 5

x - 1 = 2 - y = z - 3

(- 4, 3, 1)

x + 2 y - z - 5 = 0

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

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