Vector Algebra Question 136
Question: The line of intersection of the planes $ \mathbf{r}.(\mathbf{i}-3\mathbf{j}+\mathbf{k})=1 $ and $ \mathbf{r}.(2\mathbf{i}+5\mathbf{j}-3\mathbf{k})=2 $ is parallel to the vector
Options:
A) $ -4\mathbf{i}+5\mathbf{j}+11\mathbf{k} $
B) $ 4\mathbf{i}+5\mathbf{j}+11\mathbf{k} $
C) $ 4\mathbf{i}-5\mathbf{j}+11\mathbf{k} $
D) $ 4\mathbf{i}-5\mathbf{j}-11\mathbf{k} $
Show Answer
Answer:
Correct Answer: B
Solution:
- The line of intersection of the planes $ \mathbf{r}.(\mathbf{i}-3\mathbf{j}+\mathbf{k})=1 $ and $ \mathbf{r}.(2\mathbf{i}+5\mathbf{j}-3\mathbf{k})=2 $ is perpendicular to each of the normal vectors $ {{\mathbf{n}}_1}=\mathbf{i}-3\mathbf{j}+\mathbf{k} $ and $ {{\mathbf{n}}_2}=2\mathbf{i}+5\mathbf{j}-3\mathbf{k} $
$ \therefore $ It is parallel to the vector $ {{\mathbf{n}}_1}\times {{\mathbf{n}}_2}=(\mathbf{i}-3\mathbf{j}+\mathbf{k})\times (2\mathbf{i}+5\mathbf{j}-3\mathbf{k}) $ $ =. \begin{vmatrix} ,\mathbf{i}, \\ 1 \\ 2 \\ \end{matrix}\begin{matrix} ,\mathbf{j} \\ -3 \\ 5 \\ \end{matrix}\begin{matrix} \mathbf{k} \\ ,1 \\ -3, \\ \end{matrix} . | $ = $ 4\mathbf{i}+5\mathbf{j}+11\mathbf{k} $ .
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