If the lines x-5- y 3-z-2 and x- y-1-z 2- are coplanar- find the values of -

The lines nbsp;x x1a1=y y1b1=z z1c1 nbsp;and nbsp;x x2a2=y y2b2=z z2c2 nbsp;are coplanar if nbsp;x2 x1y2 y1z2 z1a1b1c1a2b2c2=0. The given lines nb ...

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Byju's Answer

Standard X

Mathematics

Equation of a Line Perpendicular to a Given Line

If the lines ...

Question

If the lines $x =$ 5, $\frac{y}{3 - α} = \frac{z}{- 2}$ and $x = α$ , $\frac{y}{- 1} = \frac{z}{2 - α}$ are coplanar, find the values of $α$ .

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Solution

The lines $\frac{x - x_{1}}{a_{1}} = \frac{y - y_{1}}{b_{1}} = \frac{z - z_{1}}{c_{1}}$ and $\frac{x - x_{2}}{a_{2}} = \frac{y - y_{2}}{b_{2}} = \frac{z - z_{2}}{c_{2}}$ are coplanar if $|\begin{matrix} x_{2} - x_{1} & y_{2} - y_{1} & z_{2} - z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{matrix}| = 0$ .

The given lines $\frac{x - 5}{0} = \frac{y}{3 - α} = \frac{z}{- 2}$ and $\frac{x - α}{0} = \frac{y}{- 1} = \frac{z}{2 - α}$ are coplanar.

$∴ |\begin{matrix} α - 5 & 0 - 0 & 0 - 0 \\ 0 & 3 - α & - 2 \\ 0 & - 1 & 2 - α \end{matrix}| = 0 \Rightarrow |\begin{matrix} α - 5 & 0 & 0 \\ 0 & 3 - α & - 2 \\ 0 & - 1 & 2 - α \end{matrix}| = 0 \Rightarrow (α - 5) [(3 - α) \times (2 - α) - 2] - 0 + 0 = 0 \Rightarrow (α - 5) (α^{2} - 5 α + 4) = 0 \Rightarrow (α - 5) (α - 1) (α - 4) = 0$
$\Rightarrow α - 1 = 0 or α - 4 = 0 or α - 5 = 0 \Rightarrow α = 1 or α = 4 or α = 5$
Thus, the values of $α$ are 1, 4 and 5.

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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)