Let $L_{1}$ and $L_{2}$ denote the lines $\vec{r} = \hat{i} + \lambda(-\hat{i} + 2\hat{j} + 2\hat{k}), \lambda \in R$ and $\vec{r} = \mu(2\hat{i} - \hat{j} + 2\hat{k}), \mu \in R$ 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) $L_{3}$ ?

Let L₁ and L₂ denote the lines r⃗ = î + λ(-î + 2ĵ + 2k̂), λ ∈ R and r⃗ = μ(2î - ĵ + 2k̂), μ ∈ R respectively. If L₃ is a line which is perpendicular to both L₁ and L₂ and cuts both of them, then which of the following options describe(s) L₃ ?

$\text{Given: } L_1:\ \vec{r} = \hat{i} + \lambda(-\hat{i} + 2\hat{j} + 2\hat{k})$

$L_2:\ \vec{r} = \mu(2\hat{i} - \hat{j} + 2\hat{k})$

$\text{To find: line } L_3 \text{ with } L_3 \perp L_1,\ L_3 \perp L_2,\ \text{cutting both}$

$L_1:\ \vec{r} = \hat{i} + \lambda(-\hat{i}+2\hat{j}+2\hat{k})$ $A(1,0,0),\quad \vec{a} = (-1,2,2)$ $L_2:\ \vec{r} = \mu(2\hat{i}-\hat{j}+2\hat{k})$ $B(0,0,0),\quad \vec{b} = (2,-1,2)$ $\text{Plan: } 1.\ \vec{d} = \vec{a}\times\vec{b}$ $2.\ \text{feet } P\in L_1,\ Q\in L_2 \text{ with } PQ \perp \vec{a},\ \vec{b}$ $3.\ L_3:\ \vec{r} = P + t\vec{d};\ \text{then match each option}$ $\vec{a}\times\vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 2 \\ 2 & -1 & 2 \end{vmatrix}$ $= \hat{i}(4+2) - \hat{j}(-2-4) + \hat{k}(1-4) = 6\hat{i}+6\hat{j}-3\hat{k}$ $\vec{d} = 2\hat{i}+2\hat{j}-\hat{k}$ $P = A + \lambda\vec{a} = (1-\lambda,\ 2\lambda,\ 2\lambda)$ $Q = B + \mu\vec{b} = (2\mu,\ -\mu,\ 2\mu)$ $Q-P = (2\mu-1+\lambda,\ -\mu-2\lambda,\ 2\mu-2\lambda)$ $(Q-P)\cdot\vec{a} = 0, \qquad (Q-P)\cdot\vec{b} = 0$ $(Q-P)\cdot\vec{a} = -(2\mu-1+\lambda) + 2(-\mu-2\lambda) + 2(2\mu-2\lambda)$ $= 1 - 9\lambda = 0 \Rightarrow \lambda = \dfrac{1}{9}$ $(Q-P)\cdot\vec{b} = 2(2\mu-1+\lambda) - (-\mu-2\lambda) + 2(2\mu-2\lambda)$ $= 9\mu - 2 = 0 \Rightarrow \mu = \dfrac{2}{9}$ $P = \left(\dfrac{8}{9},\ \dfrac{2}{9},\ \dfrac{2}{9}\right),\qquad Q = \left(\dfrac{4}{9},\ -\dfrac{2}{9},\ \dfrac{4}{9}\right)$ $L_3:\ \vec{r} = \left(\dfrac{8}{9},\ \dfrac{2}{9},\ \dfrac{2}{9}\right) + t(2\hat{i}+2\hat{j}-\hat{k})$ $\text{(1) base } \dfrac{2}{9}(4\hat{i}+\hat{j}+\hat{k}) = \left(\dfrac{8}{9},\dfrac{2}{9},\dfrac{2}{9}\right) = P\ \checkmark$ $\text{(2) base } \dfrac{2}{9}(2\hat{i}-\hat{j}+2\hat{k}) = \left(\dfrac{4}{9},-\dfrac{2}{9},\dfrac{4}{9}\right) = Q\ \checkmark$ $\text{(3) base } (0,0,0):\ \text{origin} \notin L_3\ \text{✗}$ $P + t\vec{d} = 0 \Rightarrow t = -\dfrac{4}{9},\ \text{but } y = -\dfrac{2}{3} \neq 0$ $\text{(4) base } \dfrac{1}{3}(2\hat{i}+\hat{k}) = \left(\dfrac{2}{3},\ 0,\ \dfrac{1}{3}\right) = \text{midpoint of } PQ\ \checkmark$ $\text{Answer: } (1),\ (2)\ \text{and}\ (4)$