MCQ
Let $L _1$ and $L _2$ denotes the lines

$\overrightarrow{ r }=\hat{ i }+\lambda(-\hat{ i }+2 \hat{ j }+2 \hat{ k }), \lambda \in R \text { and }$

$\overrightarrow{ r }=\mu(2 \hat{ i }-\hat{ j }+2 \hat{ k }), \mu \in R$

respectively. If $L _3$ is a line which is perpendicular to both $L _1$ and $L _2$ and cuts both of them, then which of the following options describe(s) $L _3$ ?

$(1)$ $\overrightarrow{ r }=\frac{1}{3}(2 \hat{ i }+\hat{ k })+ t (2 \hat{ i }+2 \hat{ j }-\hat{ k }), t \in R$

$(2)$ $\overrightarrow{ i }=\frac{2}{9}(2 \hat{ i }-\hat{ j }+2 \hat{ k })+ t (2 \hat{ i }+2 \hat{ j }-\hat{ k }), t \in R$

$(3)$ $\overrightarrow{ r }=t(2 \hat{ i }+2 \hat{ j }-\hat{ k }), t \in R$

$(4)$ $\overrightarrow{ r }=\frac{2}{9}(4 \hat{ i }+\hat{ j }+\hat{ k })+ t (2 \hat{ i }+2 \hat{ j }-\hat{ k }), t \in R$

  • $1,2,4$
  • B
    $1,2,3$
  • C
    $1,2$
  • D
    $1,3$

Answer

Correct option: A.
$1,2,4$
a
Points on $L _1$ and $L _2$ are respectively $A (1-\lambda, 2 \lambda, 2 \lambda)$ and $B (2 \mu,-\mu, 2 \mu)$

So, $\overline{ AB }=(2 \mu+\lambda-1) \hat{ i }+(-\mu-2 \lambda) \hat{ j }+(2 \mu-2 \lambda) \hat{ k }$

and vector along their shortest distance $=2 \hat{i}+2 \hat{j}-\hat{k}$.

Hence, $\frac{2 \mu+\lambda-1}{2}=\frac{-\mu-2 \lambda}{2}=\frac{2 \mu-2 \lambda}{-1}$

$\Rightarrow \lambda=\frac{1}{9} \& \mu=\frac{2}{9}$

Hence, $A \equiv\left(\frac{8}{9}, \frac{2}{9}, \frac{2}{9}\right)$ and $B \equiv\left(\frac{4}{9},-\frac{2}{9}, \frac{4}{9}\right)$

$\Rightarrow$ Mid point of $AB \equiv\left(\frac{2}{3}, 0, \frac{1}{3}\right)$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free