Question

Consider the following three lines in space $L_1:\overline{r}=3\hat{i}-\hat{j}+2\hat{k}+\lambda (2\hat{i}+4\hat{j}-\hat{k});$ $L_2:\overline{r}=\hat{i}+\hat{j}-3\hat{k}+\mu (4\hat{i}+2\hat{j}+4\hat{k});$ $L_3:\overline{r}=3\hat{i}-2\hat{j}-2\hat{k}+t(2\hat{i}+4\hat{j}-2\hat{k})$ Which one of the following pair(s) are in the same plane

• A. Only $L_1{L}_2$
• B. Only $L_2{L}_3$
• C. Only $L_3{L}_1$
• D. $L_1{L}_2$ and $L_2{L}_3$

Answer 1

The correct option is A Only $L_1{L}_2$

We need to verify whether the determinant formed by three vectors, two directions and one by subtracting the given points on line is zero or not.

Thus, for $L_1{L}_2,\left|\begin{array}{ccc}2& 4& -1\\ 4& 2& 4\\ 2& -2& 5\end{array}\right|$

$=2(10+8)-4(20-8)-1(-8-4)=36-48+12=0,\Rightarrow L_1{L}_2$ are in the same plane.
Similarly, for $L_1{L}_3,\left|\begin{array}{ccc}2& 4& -1\\ 2& 4& -2\\ 0& 1& 4\end{array}\right|$ = $2(16+2)-4\left(8\right)-1\left(2\right)=36-32-2\ne 0$
For $L_2{L}_3,\left|\begin{array}{ccc}2& 4& -2\\ 4& 2& 4\\ 2& -3& 1\end{array}\right|$ = $2(2+12)-4(4-8)-2(-12-4)=28+16+32\ne 0$

Hence option $A$