The equation of the plane passing through the lines x-4-1-y-3-1-z-2-2 and x-3-1-y-2-4-z-5 is

A(x 4)+b(y 3)+c(z 2)=0 ∴ a+b+2c=0 and a 4b+5c=0 a/5+8=b/2 5=c/ 4 1=k a/13=b/ 3=c/ 5=k Therefore, the required equation of plane is 13x+3y+5z+33=0 ...

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Byju's Answer

Standard XIII

Mathematics

Equation of Plane Containing Two Lines

The equation ...

Question

The equation of the plane passing through the lines $\frac{x - 4}{1} = \frac{y - 3}{1} = \frac{z - 2}{2}$ and $\frac{x - 3}{1} = \frac{y - 2}{- 4} = \frac{z}{5}$ is

11x-y-3z=35

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11x+y-3z=35

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11x-y+3z=35

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None of these

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Solution

The correct option is D None of these
a(x-4)+b(y-3)+c(z-2)=0
$∴$ a+b+2c=0 and a-4b+5c=0
$\frac{a}{5 + 8} = \frac{b}{2 - 5} = \frac{c}{- 4 - 1} = k$
$\frac{a}{13} = \frac{b}{- 3} = \frac{c}{- 5} = k$
Therefore, the required equation of plane is
-13x+3y+5z+33=0

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Similar questions

The equation of the plane passing through the lines $\frac{x - 4}{1} = \frac{y - 3}{1} = \frac{z - 2}{2}$ and $\frac{x - 3}{1} = \frac{y - 2}{- 4} = \frac{z}{5}$ is

Q. Equation of the line passing through the point

(2, 3, 1)

and parallel to the line of the intersection of the plane

x - 2 y - z + 5 = 0 and x + y + 3 z = 6

Q. The equation of the line passing through

(- 4, 3, 1)

, parallel to the plane

x + 2 y - z - 5 = 0

and intersecting the line

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

is:

Q. The equation of the plane passing through the lines

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

and

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

Q. Equation of the plane passing through the point

(3, - 1, 2)

and parallel to the lines

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

and

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

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The equation of the plane passing through the lines x−41=y−31=z−22 and x−31=y−2−4=z5 is

The correct option is D None of thesea(x-4)+b(y-3)+c(z-2)=0 ∴ a+b+2c=0 and a-4b+5c=0 a5+8=b2−5=c−4−1=k a13=b−3=c−5=k Therefore, the required equation of plane is -13x+3y+5z+33=0

The equation of the plane passing through the lines $\frac{x - 4}{1} = \frac{y - 3}{1} = \frac{z - 2}{2}$ and $\frac{x - 3}{1} = \frac{y - 2}{- 4} = \frac{z}{5}$ is

The correct option is D None of these
a(x-4)+b(y-3)+c(z-2)=0
$∴$ a+b+2c=0 and a-4b+5c=0
$\frac{a}{5 + 8} = \frac{b}{2 - 5} = \frac{c}{- 4 - 1} = k$
$\frac{a}{13} = \frac{b}{- 3} = \frac{c}{- 5} = k$
Therefore, the required equation of plane is
-13x+3y+5z+33=0