Let $\vec{a}=\hat{i}+2 \hat{j}+\hat{k}$ and $\vec{b}=2 \hat{i}+7 \hat{j}+3 \hat{k}$. Let $L_{1}: \vec{r}=(-\hat{i}+2 \hat{j}+\hat{k})+\lambda \vec{a}, \lambda \in R$ and $L_{2}: \vec{r}=(\hat{j}+\hat{k})+\mu \vec{b}, \mu \in R$ be two lines. If the line $L_{3}$ passes through the point of intersection of $L_{1}$ and $L_{2}$, and is parallel to $\vec{a}+\vec{b}$, then $L_{3}$ passes through the point :

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Three Dimensional Geometry

Let

a = \hat{i} + 2 \hat{j} + \hat{k}

and

b = 2 \hat{i} + 7 \hat{j} + 3 \hat{k}

. Let

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

and

L_{2} : r = (\hat{j} + \hat{k}) + μ b, μ \in R

be two lines. If the line

L_{3}

passes through the point of intersection of

L_{1}

and

L_{2}

, and is parallel to

a + b

, then

L_{3}

passes through the point :

Answer: (A)

L_{1}

and

L_{2}

intersect where

(- 1 + λ, 2 + 2 λ, 1 + λ) = (2 μ, 1 + 7 μ, 1 + 3 μ) .

The third coordinate gives

λ = 3 μ

; substituting into the first gives

3 μ - 1 = 2 μ

, so

μ = 1

λ = 3

, and the point of intersection is

(2, 8, 4)

.

Since

a + b = 3 \hat{i} + 9 \hat{j} + 4 \hat{k}

, the line

L_{3}

r = (2 \hat{i} + 8 \hat{j} + 4 \hat{k}) + t (3 \hat{i} + 9 \hat{j} + 4 \hat{k}) .

Taking

t = 2

gives the point

(2 + 6, 8 + 18, 4 + 8) = (8, 26, 12)

, which lies on

L_{3}

JEE Main · 2025 · 29 Jan 2025 · Shift 1 · Single correct option