The lines x-3-3-y-1-1-z-5-5 and x-1-1-y-2-z-5-5 are coplanar- Then the value of - isA- -1B- 2C- 5D- 1

The correct option is B 2 nbsp; nbsp; nbsp;Given lines are coplanar. ⇒ 1 ( 3) amp; nbsp; 2 1 amp; 5 5 3 amp;1 amp;5 1 amp;α amp;5 = ...

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Byju's Answer

Standard XII

Mathematics

Condition for Coplanarity of Four Points

The lines x+3...

Question

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 $α$ is

- 1

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Solution

The correct option is B $2$
Given lines are coplanar.
$\Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} - 1 - (- 3) & 2 - 1 & 5 - 5 - 3 & 1 & 5 - 1 & α & 5 \end{matrix} ∣ ∣ ∣ ∣ = 0$

$\Rightarrow ∣ ∣ ∣ ∣ \begin{matrix} 2 & 1 & 0 - 3 & 1 & 5 - 1 & α & 5 \end{matrix} ∣ ∣ ∣ ∣ = 0$

$\Rightarrow α = 2$

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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. Show that the lines

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

and

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

are coplanar.

Q. If the lines

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

and

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

are coplanar for

K = α

and

K = β

(α \neq β)

, then

α + β + 6

is equal to

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

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The lines x+3−3=y−11=z−55 and x+1−1=y−2α=z−55 are coplanar. Then the value of α is

The correct option is B 2 Given lines are coplanar. ⇒∣∣ ∣∣−1−(−3)2−15−5−315−1α5∣∣ ∣∣=0 ⇒∣∣ ∣∣210−315−1α5∣∣ ∣∣=0 ​​​​​​​ ⇒α=2

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 $α$ is