Two lines L1-x-5-y3-z-2 and L2-x-y-1-z2- are coplanar- Then - can take values

The first equation of line is expressed as, L_1:x 50=y3 α=z 2 The second equation of line is expressed as, L_2:x α0=y 1=z2 α If two lines are coplanar, then ...

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Standard XII

Mathematics

Skew Lines

Two lines L1:...

Question

Two lines $L_{1} : x = 5, \frac{y}{3 - α} = \frac{z}{- 2}$ and $L_{2} : x = α, \frac{y}{- 1} = \frac{z}{2 - α}$ are coplanar. Then $α$ can take value(s)

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Solution

The correct option is D $4$
The first equation of line is expressed as,
$L_{1} : \frac{x - 5}{0} = \frac{y}{3 - α} = \frac{z}{- 2}$

The first equation of line is expressed as,
$L_{2} : \frac{x - α}{0} = \frac{y}{- 1} = \frac{z}{2 - α}$

If two lines are coplanar, then
$∣ ∣ ∣ ∣ \begin{matrix} 5 - α & 0 & 0 0 & 3 - α & - 2 0 & - 1 & 2 - α \end{matrix} ∣ ∣ ∣ ∣ = 0 \Rightarrow (5 - α) [(α - 2) (α - 3) - 2] = 0 \Rightarrow α = 5, 4, 1$
The values of $α$ are $5, 4, 1$

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Similar questions

Two lines $L_{1}$ : $x = 5, \frac{y}{3 - α} = \frac{z}{- 2}$ and $L_{2}$ : $x = α, \frac{y}{- 1} = \frac{Z}{2 - α}$ are coplanar. Then $α$ can take value(s)

Q. If the angle between the line

x = \frac{y - 1}{2} = \frac{z - 3}{λ}

and the plane

x + 2 y + 3 z = 4

{cos}^{- 1} (\sqrt{5 / 14})

then

λ

Q. If the angle bwteen a line

x = \frac{y - 1}{2} = \frac{z - 3}{λ}

and normal to the plane

x + 2 y + 3 z = 4

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, then possible value(s) of

λ

is/are

Two lines $L_{1} : x = 5, \frac{y}{3 - α} = \frac{z}{- 2}$ and $L_{2} : x = α, \frac{y}{- 1} = \frac{z}{2 - α}$ are coplanar. Then, $α$ can take value(s)

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Two lines L1:x=5,y3−α=z−2 and L2:x=α,y−1=z2−α are coplanar. Then α can take value(s)

Two lines $L_{1} : x = 5, \frac{y}{3 - α} = \frac{z}{- 2}$ and $L_{2} : x = α, \frac{y}{- 1} = \frac{z}{2 - α}$ are coplanar. Then $α$ can take value(s)