Let $L_{1}: \vec{r}=(\hat{i}-\hat{j}+2 \hat{k})+\lambda(\hat{i}-\hat{j}+2 \hat{k}), \lambda \in R$ $L_{2}: \vec{r}=(\hat{j}-\hat{k})+\mu(3 \hat{i}+\hat{j}+p \hat{k}), \mu \in R$ and $\mathrm{L}_{3}: \overrightarrow{\mathrm{r}}=\delta(\ell \hat{\mathrm{i}}+\mathrm{m} \hat{\mathrm{j}}+\mathrm{n} \hat{\mathrm{k}}) \delta \in \mathrm{R}$ Be three lines such that $\mathrm{L}_{1}$ is perpendicular to $\mathrm{L}_{2}$ and $\mathrm{L}_{3}$ is perpendicular to both $\mathrm{L}_{1}$ and $\mathrm{L}_{2}$. Then the point which lies on $\mathrm{L}_{3}$ is

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Three Dimensional Geometry

Let

L_{1} : r = (\hat{i} - \hat{j} + 2 \hat{k}) + λ (\hat{i} - \hat{j} + 2 \hat{k}), λ \in R

L_{2} : r = (\hat{j} - \hat{k}) + μ (3 \hat{i} + \hat{j} + p \hat{k}), μ \in R

and

L_{3} : r = δ (ℓ \hat{i} + m \hat{j} + n \hat{k}) δ \in R

Be three lines such that

L_{1}

is perpendicular to

L_{2}

and

L_{3}

is perpendicular to both

L_{1}

and

L_{2}

. Then the point which lies on

L_{3}

Answer: (A)

L_{1}

has direction

d_{1} = (1, - 1, 2)

;

L_{2}

has direction

d_{2} = (3, 1, p)

L_{1} ⊥ L_{2} \Rightarrow d_{1} \cdot d_{2} = 3 - 1 + 2 p = 0 \Rightarrow p = - 1

, so

d_{2} = (3, 1, - 1)

L_{3}

is perpendicular to both, so its direction is

d_{1} \times d_{2} = \hat{i} 13 \hat{j} - 1 1 \hat{k} 2 - 1 = \hat{i} (1 - 2) - \hat{j} (- 1 - 6) + \hat{k} (1 + 3) = (- 1, 7, 4)

L_{3}

passes through the origin along

(- 1, 7, 4)

; at parameter

δ = 1

this gives the point

(- 1, 7, 4)

JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Single correct option