Vector Algebra Question 304
Question: The vector equation of the plane containing the lines $ \mathbf{r}=(\mathbf{i}+\mathbf{j})+\lambda (\mathbf{i}+2\mathbf{j}-\mathbf{k}) $ and $ \mathbf{r}=(\mathbf{i}+\mathbf{j})+\mu (-\mathbf{i}+\mathbf{j}-2\mathbf{k}) $ is
Options:
A) $ \mathbf{r}.(\mathbf{i}+\mathbf{j}+\mathbf{k})=0 $
B) $ \mathbf{r}.(\mathbf{i}-\mathbf{j}-\mathbf{k})=0 $
C) $ \mathbf{r}.(\mathbf{i}+\mathbf{j}+\mathbf{k})=3 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
- Given two lines $ \mathbf{r}=(\mathbf{i}+\mathbf{j})+\lambda (\mathbf{i}+2\mathbf{j}-\mathbf{k}) $ and $ \mathbf{r}=(\mathbf{i}+\mathbf{j})+\mu (-\mathbf{i}+\mathbf{j}-2\mathbf{k}) $ pass through $ \mathbf{a}=\mathbf{i}+\mathbf{j} $ and are parallel to the vectors $ \mathbf{b}=\mathbf{i}+2\mathbf{j}-\mathbf{k} $ and $ \mathbf{c}=-\mathbf{i}+\mathbf{j}-2\mathbf{k} $ respectively. Therefore the plane containing them passes through $ \mathbf{a}=\mathbf{i}+\mathbf{j} $ and is perpendicular to $ \mathbf{n}=\mathbf{b}\times \mathbf{c}=(\mathbf{i}+2\mathbf{j}-\mathbf{k})\times (-\mathbf{i}+\mathbf{j}-2\mathbf{k})=-3\mathbf{i}+3\mathbf{j}+3\mathbf{k} $ . Hence, the equation of the plane is $ (\mathbf{r}-\mathbf{a}).\mathbf{n}=0\Rightarrow \mathbf{r}.\mathbf{n}=\mathbf{a}.\mathbf{n}\Rightarrow \mathbf{r}.(\mathbf{i}-\mathbf{j}-\mathbf{k})=0 $ .
BETA
