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BITSAT 2023Paper 1
Exam
BITSAT
Year
2023
Paper
Paper 1
Subject
Mathematics
Chapter
Three Dimensional Geometry

BITSAT 2023 · Paper 1 · Mathematics · Three Dimensional Geometry

The lines (x-2)/(1)=(y-3)/(1)=(z-4)/(-k) and...

The lines x−21=y−31=z−4−k\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k} and x−1k=y−42=z−51\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1} are coplanar if

Options

  1. Ak=3\mathrm{k}=3 or -2
  2. Bk=0\mathrm{k}=0 or -1
  3. Ck=1\mathrm{k}=1 or -1
  4. Dk=0\mathrm{k}=0 or -3

Correct answer

D. k=0\mathrm{k}=0 or -3

Step-by-step solution

Two planes are coplanar if
\left|x2−x1y2−y1z2−z1l1m1n1l2m2n2\begin{array}{ccc}x_2-x_1 & y_2-y_1 & z_2-z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2\end{array}\right|=0
\left|1−1−111−kk21\begin{array}{ccc}1 & -1 & -1 \\ 1 & 1 & -k \\ k & 2 & 1\end{array}\right|=0<br/>Applying<br />Applying\mathrm{C}_2 \rightarrow \mathrm{C}_2+\mathrm{C}_1, \mathrm{C}_3 \rightarrow \mathrm{C}_3+\mathrm{C}_1<br/><br />\Rightarrow\left|100121−kkk+21+k\begin{array}{ccc}1 & 0 & 0 \\ 1 & 2 & 1-k \\ k & k+2 & 1+k\end{array}\right|=0<br/><br />\Rightarrow \quad 1[2+2 k-(k+2)(1-k)]=0<br/><br />\Rightarrow \quad 2+2 k-\left(-k^2-k+2\right)=0<br/><br />k^2+3 k=0 \Rightarrow k(k+3)=0
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