Question

Two lines ${L_1}:x - 5 = \dfrac{y}{{3 - \alpha }} = \dfrac{z}{{ - 2}}$ and ${L_2}:x - \alpha = \dfrac{y}{{ - 1}} = \dfrac{z}{{2 - \alpha }}$ are coplanar. Then $\alpha$ can take value(s)?

Answer 1

Coplanarity of two lines can be expressed using determinants. The Cartesian coordinates and direction ratios are obtained by writing the given equations in the standard form. Solving we get possible value(s) for$\alpha$.

Formula used: Two lines ${L_1}:\dfrac{{x - {x_1}}}{{{a_1}}} = \dfrac{{y - {y_1}}}{{{b_1}}} = \dfrac{{{z_{}} - {z_1}}}{{{c_1}}}$ and ${L_2}:\dfrac{{x - {x_2}}}{{{a_2}}} = \dfrac{{y - {y_2}}}{{{b_2}}} = \dfrac{{{z_{}} - {z_2}}}{{{c_2}}}$ are coplanar (lie on the same plane) if and only if \det \left( {\begin{array}{*{20}{c}} {{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{array}} \right) = 0
Here, ${x_i}$,${y_i}$,${z_i}$ are the Cartesian coordinates (of a point in the line) and ${a_i},{b_i},{c_i}$ are the direction ratios of the line and $\det$ represents determinant of the particular matrix.
\det \left( {\begin{array}{*{20}{c}} {{a_{11}}}&{{a_{12}}}&{{a_{13}}} \\\ {{a_{21}}}&{{a_{22}}}&{{a_{23}}} \\\ {{a_{31}}}&{{a_{32}}}&{{a_{33}}} \end{array}} \right) = {a_{11}}[{a_{22}}{a_{33}} - {a_{32}}{a_{23}}] - {a_{12}}[{a_{21}}{a_{33}} - {a_{31}}{a_{23}}] + {a_{13}}[{a_{21}}{a_{32}} - {a_{31}}{a_{22}}]
(using first row)
“if and only if” means it is a necessary and sufficient condition.

Complete step-by-step answer:
Given that
Two lines ${L_1}:x - 5 = \dfrac{y}{{3 - \alpha }} = \dfrac{z}{{ - 2}}$ and ${L_2}:x - \alpha = \dfrac{y}{{ - 1}} = \dfrac{z}{{2 - \alpha }}$ are coplanar.
Line ${L_1}:x - 5 = \dfrac{y}{{3 - \alpha }} = \dfrac{z}{{ - 2}}$ can be written as
${L_1}:\dfrac{{x - 5}}{1} = \dfrac{{y - 0}}{{3 - \alpha }} = \dfrac{{z - 0}}{{ - 2}}$
Now it is in the form ${L_1}:\dfrac{{x - {x_1}}}{{{a_1}}} = \dfrac{{y - {y_1}}}{{{b_1}}} = \dfrac{{{z_{}} - {z_1}}}{{{c_1}}}$
with ${x_1} = 5,{y_1} = 0,{z_1} = 0,{a_1} = 1,{b_1} = 3 - \alpha ,{c_1} = - 2$
Also, line ${L_2}:x - \alpha = \dfrac{y}{{ - 1}} = \dfrac{z}{{3 - \alpha }}$
can be written as ${L_2}:\dfrac{{x - \alpha }}{1} = \dfrac{{y - 0}}{{ - 1}} = \dfrac{{z - 0}}{{2 - \alpha }}$
Now it is in the form ${L_2}:\dfrac{{x - {x_2}}}{{{a_2}}} = \dfrac{{y - {y_2}}}{{{b_2}}} = \dfrac{{{z_{}} - {z_2}}}{{{c_2}}}$
with ${x_2} = \alpha ,{y_2} = 0,{z_2} = 0,{a_2} = 1,{b_2} = - 1,{c_2} = 2 - \alpha$
${L_1}$ and ${L_2}$ are coplanar
\Rightarrow \det \left( {\begin{array}{*{20}{c}} {{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{array}} \right) = 0
Substituting the values
\det \left( {\begin{array}{*{20}{c}} {\alpha - 5}&0&0 \\\ 1&{3 - \alpha }&{ - 2} \\\ 1&{ - 1}&{2 - \alpha } \end{array}} \right) = 0
Calculating determinant using first row,
$(\alpha - 5)[(3 - \alpha )(2 - \alpha ) - ( - 2 \times - 1)] - 0[1(2 - \alpha ) - (1 \times - 2)] + 0[(1 \times - 1) - 1(3 - \alpha )] = 0$
$(\alpha - 5)[(3 - \alpha )(2 - \alpha ) - ( - 2 \times - 1)] = 0$ (since multiplying zero with something results zero)
Simplifying the terms in the bracket,
$\Rightarrow (\alpha - 5)[(6 - 3\alpha - 2\alpha + \alpha _{}^2) - 2] = 0$
$\Rightarrow (\alpha - 5)[6 - 5\alpha + \alpha _{}^2 - 2] = 0$
Rearranging the terms,
$(\alpha - 5)[\alpha _{}^2 - 5\alpha + 4] = 0$
Product of two terms equal to zero implies either one is zero.
$\alpha - 5 = 0$ or $\alpha _{}^2 - 5\alpha + 4 = 0$
$\Rightarrow \alpha = 5$ or $(\alpha - 1)(\alpha - 4) = 0$
Since $(\alpha - 1)(\alpha - 4) = \alpha _{}^2 - 4\alpha - \alpha + 4 = \alpha _{}^2 - 5\alpha + 4$
Product of two terms equal to zero implies either one is zero,
$\Rightarrow \alpha = 5$ or $\alpha - 1 = 0$ or $\alpha - 4 = 0$
$\Rightarrow \alpha = 5$ or $\alpha = 1$ or $\alpha = 4$
Therefore, for the lines ${L_1}$ and ${L_2}$ to be parallel, $\alpha$ can take values $1,4$ and $5$.

Note: If a directed line $L$ passes through the origin and makes angles $\alpha ,\beta ,\gamma$ with the $x,y,z$ axes, then $\cos \alpha ,\cos \beta ,\cos \gamma$ are called direction cosines of $L$. If $l,m,n$ are the direction cosines of the line equation of the line is $\dfrac{{x - {x_1}}}{l} = \dfrac{{y - {y_1}}}{m} = \dfrac{{z - {z_1}}}{n}$. We can use either Cartesian form or vector form to express lines.