Two lines L1 - x-5- y3-z-2 and L2 - x- y-1-Z2-are coplanar- Then - can take values

The correct options are A 1 D 4 #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #10; #1 ...

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Byju's Answer

Standard XII

Mathematics

Condition for Two Lines to be on the Same Plane

Two lines L...

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 options are
A $1$
D $4$

$\frac{x - 5}{0} = \frac{y - 0}{3 - α} = \frac{z - 0}{- 2}$

$\frac{x - α}{0} = \frac{y}{- 1} = \frac{z}{2 - α}$
will be coplanar, if the shortest distance is zero.
$\Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} 5 - α & 0 & 0 0 & 3 - α & - 2 0 & - 1 & 2 - α \end{matrix} ∣ ∣ ∣ ∣ = 0$
$\Rightarrow (5 - α) (α^{2} - 5 α + 4) = 0, α = 1, 4, 5$
Hence, option A and D are correct.

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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 for some

α \in R,

the lines

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

and

L_{2} : \frac{x + 2}{α} = \frac{y + 1}{5 - α} = \frac{z + 1}{1}

coplanar, then the line

L_{2}

passes through the point:

Q. If the lines

x =

\frac{y}{3 - α} = \frac{z}{- 2}

and

x = α

\frac{y}{- 1} = \frac{z}{2 - α}

are coplanar, find the values of

α

Q. The lines

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

and

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

are coplanar. Then the value of

α

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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)

The correct options are A 1 D 4 x−50=y−03−α=z−0−2 x−α0=y−1=z2−α will be coplanar, if the shortest distance is zero.⇒∣∣ ∣∣5−α0003−α−20−12−α∣∣ ∣∣=0⇒(5−α)(α2−5α+4)=0, α=1,4,5Hence, option A and D are correct.

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)