Q71 / 160 · TS EAMCET 2024 (10 May Shift 1) · Mathematics
The position vector of the point of intersection of the line joining the points $2 \hat{i}-\hat{j}+6 \hat{k}, 3 \hat{i}-\hat{j}-7 \hat{k}$ and the line joining the points $2 \hat{i}+\hat{j}-6 \hat{k}, 3 \hat{i}-\hat{j}-7 \hat{k}$ is
- A$\hat{i}-3 \hat{j}+4 \hat{k}$
- B$4 \hat{i}-3 \hat{j}-8 \hat{k}$
- Correct answerC$\hat{i}+3 \hat{j}-5 \hat{k}$
- D$\hat{i}+\hat{j}-2 \hat{k}$
Solution
Line joining $\hat{i}-\hat{j}+\hat{k}$ and $\hat{i}+\hat{j}-2 \hat{k}$ is
$\vec{r}=(\hat{i}-\hat{j}+\hat{k})+\lambda(2 \hat{j}-3 \hat{k})$...(i)
Line joining $2 \hat{i}+\hat{j}-6 \hat{k}$ and $3 \hat{i}-\hat{j}-7 \hat{k}$ is
$\vec{r}=2 \hat{i}+\hat{j}-6 \hat{k}+\mu(\hat{i}-2 \hat{j}-\hat{k})$...(ii)
From (i) and (ii) $\hat{i}+(2 \lambda-1) \hat{j}+(1-3 \lambda) \hat{k}=(2+\mu) \hat{i}+(1-2 \mu) \hat{j}-(6+\mu) \hat{k}$
On comparing we get, $2+\mu=1 \Rightarrow \mu=-1$ Putting in (ii) we get Position vector of intersection point, $\hat{i}+3 \hat{j}-5 \hat{k}$.
From (i) and (ii) $\hat{i}+(2 \lambda-1) \hat{j}+(1-3 \lambda) \hat{k}=(2+\mu) \hat{i}+(1-2 \mu) \hat{j}-(6+\mu) \hat{k}$
On comparing we get, $2+\mu=1 \Rightarrow \mu=-1$ Putting in (ii) we get Position vector of intersection point, $\hat{i}+3 \hat{j}-5 \hat{k}$.
Hence, the correct answer is option: 'C'.