Question

Which of the following lines lie on the plane $x+2y-z=0$?

• A. $x-1=y-1=1$
• B. $x-y+z=2x+y-z=0$
• C. $\overrightarrow{r}=2\hat{i}-\hat{j}+4\hat{k}+\lambda (3\hat{i}+\hat{j}+5\hat{k})$
• D. None of these.

Answer 1

The correct option is D None of these.

consider $P:x+2y-z=0$
direction ratio of normal of $P$ is $(1,2,-1)$
a) for $L_1:\frac{x-1}{1}=\frac{y}{-1}=\frac{z-5}{-1}$
Since, $(1,0,5)$ does not lie in $P$
Therefore, $L_1$ does not lie in $P$
b) for $L_2:x-y+z=2x+y-z=0$
direction ratio of $L_2$ is $(i-j+k)\times (2i+j-k)=3(j+k)$
since, $(i+2j-k).(3j+3k)=6-3=3\ne =0$
Therefore, $L_2$ does not lie on $P$
c) for $L_3:\overrightarrow{r}=2\hat{i}-\hat{j}+4\hat{k}+\lambda (3\hat{i}+\hat{j}+5\hat{k})$
Since, $(2,-1,4)$ does not lie in $P$
Therefore, $L_3$ does not lie in $P$
Ans: D