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Mathematics · 2022 · 3 marks
CBSE 2022 · Region 3 · Set 2 · Q9
Show that the lines: (1-x)/2=(y-3)/4=z/-1 and (x-4)/3=(2 y-2)/-4=z-1 are coplanar.
Show that the lines : \[\frac{1-x}{2}=\frac{y-3}{4}=\frac{z}{-1} \text { and } \frac{x-4}{3}=\frac{2 y-2}{-4}=z-1 \text { are coplanar. } \]
Marking-scheme solution
Equations of given lines in standard form are
\[\begin{aligned}
& \frac{x-1}{-2}=\frac{y-3}{4}=\frac{z}{-1} \text { and } \\
& \frac{x-4}{3}=\frac{y-1}{-2}=\frac{z-1}{1}
\end{aligned}
\]
Lines are coplanar if
\[\left|\begin{array}{ccc}
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2}
\end{array}\right|=0
\]
\(\displaystyle \mathrm{LHS}=\left|\begin{array}{rrr} 3 & -2 & 1 \\ -2 & 4 & -1 \\ 3 & -2 & 1 \end{array}\right|\)
\[\begin{aligned}
& =3(4-2)+2(-2+3)+1(4-12) \\
& =6+2-8 \\
& =0
\end{aligned}
\]
⇒ Coplanar.
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