Question ID: #525
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:
- (1) (8, 26, 12)
- (2) (2, 8, 5)
- (3) (-1, -1, 1)
- (4) (5, 17, 4)
Solution:
First, find the intersection point of $L_1$ and $L_2$.
Line $L_1$: $\vec{r} = (-\hat{i}+2\hat{j}+\hat{k}) + \lambda(\hat{i}+2\hat{j}+\hat{k})$.
General point $P_1(\lambda-1, 2\lambda+2, \lambda+1)$.
Line $L_2$: $\vec{r} = (\hat{j}+\hat{k}) + \mu(2\hat{i}+7\hat{j}+3\hat{k})$.
General point $P_2(2\mu, 7\mu+1, 3\mu+1)$.
Equating coordinates:
(1) $\lambda – 1 = 2\mu$
(2) $2\lambda + 2 = 7\mu + 1$
(3) $\lambda + 1 = 3\mu + 1 \Rightarrow \lambda = 3\mu$.
Substitute $\lambda = 3\mu$ into (1):
$3\mu – 1 = 2\mu \Rightarrow \mu = 1$.
Then $\lambda = 3(1) = 3$.
Check consistency with (2): $2(3) + 2 = 8$ and $7(1) + 1 = 8$. (Consistent).
Intersection point $P$ is $(2(1), 7(1)+1, 3(1)+1) = (2, 8, 4)$.
Direction of $L_3$ is parallel to $\vec{a}+\vec{b}$.
$\vec{a}+\vec{b} = (\hat{i}+2\hat{j}+\hat{k}) + (2\hat{i}+7\hat{j}+3\hat{k}) = 3\hat{i} + 9\hat{j} + 4\hat{k}$.
Equation of line $L_3$ passing through $(2, 8, 4)$ with direction $(3, 9, 4)$:
$\vec{r} = (2\hat{i}+8\hat{j}+4\hat{k}) + k(3\hat{i}+9\hat{j}+4\hat{k})$.
Check options:
Option (1) $(8, 26, 12)$:
$2 + 3k = 8 \Rightarrow 3k = 6 \Rightarrow k = 2$.
$8 + 9k = 8 + 18 = 26$. (Matches).
$4 + 4k = 4 + 8 = 12$. (Matches).
Thus, the point $(8, 26, 12)$ lies on $L_3$.
Ans. (1)
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