Two lines L 1 : x = 5 , y 3 - α = z - 2 and L 2 : x = α , y - 1 = z 2 - α are coplanar. Then α can take value(s)

x - 5 0 = - y α - 3 = z - 2 x - α 0 = y - 1 = z 2 - α 5 - α 0 0 0 3 - α - 2 0 - 1 2 - α = 0 5 - α 3 - α 2 - α - 2 = 0 α - 5 α 2 - 5 α + 6 - 2 = 0 α - 5 α 2 - 5 α + 4 = 0 α = 1 , 4 , 5

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Mathematics - Vector Question with Solution | TestHub

MathematicsVectorVector equation of a line, Vector equation of the angle bisectorsEasy2 minPYQ_2013

MathematicsEasymultiple choice

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)

Options:(select one or more)

Answer:

A, D

Solution:

$\frac{x - 5}{0} = \frac{- y}{α - 3} = \frac{z}{- 2}$
$\frac{x - α}{0} = \frac{y}{- 1} = \frac{z}{2 - α}$
$|\begin{matrix} 5 - α & 0 & 0 \\ 0 & 3 - α & - 2 \\ 0 & - 1 & 2 - α \end{matrix}| = 0$
$(5 - α) ((3 - α) (2 - α) - 2) = 0$
$(α - 5)$ $(α^{2} - 5 α + 6 - 2) = 0$
$(α - 5) (α^{2} - 5 α + 4) = 0$
$α = 1, 4, 5$

Stream:JEE_ADVSubject:MathematicsTopic:VectorSubtopic:Vector equation of a line, Vector equation of the angle bisectors

⏱ 2mℹ️ Source: PYQ_2013

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