The equation of the plane which contains the line x4-y2-z1 and is perpendicular to the plane containing the lines x-41-y-54-z-92 and x-84-y-62-z-21

Normal vector of the plane containing the lines nbsp; =î amp;ĵ amp;k̂ 1 amp;4 amp;2 4 amp;2 amp;1 =7ĵ 14k̂ So the equation of the plane nbsp; ...

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Equation of Plane Containing Two Lines

The equation ...

Question

The equation of the plane which contains the line $\frac{x}{4} = \frac{y}{2} = \frac{z}{1}$ and is perpendicular to the plane containing the lines $\frac{x - 4}{1} = \frac{y - 5}{4} = \frac{z - 9}{2}$ and $\frac{x + 8}{4} = \frac{y + 6}{2} = \frac{z - 2}{1}$

5 x - 8 y - 4 z = 0

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5 x + 8 y - 4 z = 0

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5 x - 8 y + 4 z = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

5 x + 8 y + 4 z = 0

No worries! We‘ve got your back. Try BYJU‘S free classes today!

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Solution

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Similar questions

\frac{x}{4} = \frac{y}{2} = \frac{z}{1}

\frac{x - 4}{1} = \frac{y - 5}{4} = \frac{z - 9}{2}

\frac{x + 8}{4} = \frac{y + 6}{2} = \frac{z - 2}{1}

\frac{x}{2} = \frac{y}{3} = \frac{z}{4}

\frac{x}{3} = \frac{y}{4} = \frac{z}{2}

\frac{x}{4} = \frac{y}{2} = \frac{z}{3}

\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 + 2}{- 3} = \frac{z}{5}

x - y + z + 2 = 0

\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}

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