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Condition for Two Lines to be on the Same Plane

Consider line...

Question

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

A

x - y - 2 = 0

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B

2x - y + 2 = 0

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C

x - y + 7 = 0

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D

None of these

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Solution

The correct option is D
None of these

Equation of plane is

$∣ ∣ ∣ ∣ \begin{matrix} x - x_{1} & y - y_{1} & z - z_{1} l_{1} & m_{1} & n_{1} x_{1} - x_{2} & y_{1} - y_{2} & z_{1} - z_{2} \end{matrix} ∣ ∣ ∣ ∣ = 0$

$∣ ∣ ∣ ∣ ∣ \begin{matrix} x - 2 & y - 3 & z - 4 1 & 1 & \frac{1}{2} 1 & - 1 & - 1 \end{matrix} ∣ ∣ ∣ ∣ ∣ = 0$

$x - 3 y + 4 z - 9 = 0$

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Similar questions

Q. The equation of the plane containing the two lines

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

is
(a) 8x + y − 5z − 7 = 0
(b) 8x + y + 5z − 7 = 0
(c) 8x − y − 5z − 7 = 0
(d) None of these

Q. The equation of plane containing intersecting lines

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

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

Q. The equation of the plane containing the straight line

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

and perpendicular to the plane

x - y + z + 2 = 0

Consider lines

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

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

Value of 'k' so that lines $L_{1}$ and $L_{2}$ are coplanar, is

Equation of the plane containing the straight line $\frac{x}{2} = \frac{y}{3} = \frac{z}{4}$ and perpendicular to the plane containing the staight lines $\frac{x}{3} = \frac{y}{4} = \frac{z}{2}$ and $\frac{x}{4} = \frac{y}{2} = \frac{z}{3}$ is

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