Vector Algebra Question 338
Question: The line through $ \mathbf{i}+3\mathbf{j}+2\mathbf{k} $ and perpendicular to the lines $ \mathbf{r}=(\mathbf{i}+2\mathbf{j}-\mathbf{k})+\lambda (2\mathbf{i}+\mathbf{j}+\mathbf{k}) $ and $ \mathbf{r}=(2\mathbf{i}+6\mathbf{j}+\mathbf{k})+\mu (\mathbf{i}+2\mathbf{j}+3\mathbf{k}) $ is
Options:
A) $ \mathbf{r}=(\mathbf{i}+2\mathbf{j}-\mathbf{k})+\lambda (-\mathbf{i}+5\mathbf{j}-3\mathbf{k}) $
B) $ \mathbf{r}=\mathbf{i}+3\mathbf{j}+2\mathbf{k}+\lambda (\mathbf{i}-5\mathbf{j}+3\mathbf{k}) $
C) $ \mathbf{r}=\mathbf{i}+3\mathbf{j}+2\mathbf{k}+\lambda (\mathbf{i}+5\mathbf{j}+3\mathbf{k}) $
D) $ \mathbf{r}=\mathbf{i}+3\mathbf{j}+2\mathbf{k}+\lambda (-\mathbf{i}+5\mathbf{j}-3\mathbf{k}) $
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Answer:
Correct Answer: D
Solution:
- The required line passes through the point $ \mathbf{i}+3\mathbf{j}+2\mathbf{k} $ and is perpendicular to the lines $ \mathbf{r}=(\mathbf{i}+2\mathbf{j}-\mathbf{k})+\lambda (2\mathbf{i}+\mathbf{j}+\mathbf{k}) $ and $ \mathbf{r}=(2\mathbf{i}+6\mathbf{j}+\mathbf{k})+\mu (\mathbf{i}+2\mathbf{j}+3\mathbf{k}) $ , therefore it is parallel to the vector $ \mathbf{b}=(2\mathbf{i}+\mathbf{j}+\mathbf{k})\times (\mathbf{i}+2\mathbf{j}+3\mathbf{k}) $ = $ (\mathbf{i}-5\mathbf{j}+3\mathbf{k}) $ Hence, the equation of the required line is $ \mathbf{r}=(\mathbf{i}+3\mathbf{j}+2\mathbf{k})+\lambda ‘(\mathbf{i}-5\mathbf{j}+3\mathbf{k}) $
Þ $ \mathbf{r}=(\mathbf{i}+3\mathbf{j}+2\mathbf{k})+\lambda (-\mathbf{i}+5\mathbf{j}-3\mathbf{k}) $ , where $ \lambda =-\lambda ’ $ .
BETA
