Vector Algebra Question 307
Question: The cartesian equation of the plane $ \mathbf{r}=(1+\lambda -\mu )\mathbf{i}+(2-\lambda )\mathbf{j}+(3-2\lambda +2\mu )\mathbf{k} $ is
Options:
A) $ 2x+y=5 $
B) $ 2x-y=5 $
C) $ 2x+z=5 $
D) $ 2x-z=5 $
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Answer:
Correct Answer: C
Solution:
- We have $ \mathbf{r}=(1+\lambda -\mu )\mathbf{i}+(2-\lambda )\mathbf{j}+(3-2\lambda +2\mu )\mathbf{k} $
Þ $ \mathbf{r}=(\mathbf{i}+2\mathbf{j}+3\mathbf{k})+\lambda (\mathbf{i}-\mathbf{j}-2\mathbf{k})+\mu (-\mathbf{i}+2\mathbf{k}) $ , which is a plane passing through $ \mathbf{a}=\mathbf{i}+2\mathbf{j}+3\mathbf{k} $ and parallel to the vectors $ b=\mathbf{i}-\mathbf{j}-2\mathbf{k} $ and $ \mathbf{c}=-\mathbf{i}+2\mathbf{k} $ Therefore, it is perpendicular to the vector $ \mathbf{n}=\mathbf{b}\times \mathbf{c}=-2\mathbf{i}-\mathbf{k} $ Hence, its vector equation is $ (\mathbf{r}-\mathbf{a}).\mathbf{n}=0 $
Þ $ \mathbf{r}.\mathbf{n}=\mathbf{a}.\mathbf{n}\Rightarrow \mathbf{r}.(-2\mathbf{i}-\mathbf{k})=-2-3 $
$ \Rightarrow \mathbf{r}.(2\mathbf{i}+\mathbf{k})=5 $ So, the cartesian equation is $ (x\mathbf{i}+y\mathbf{j}+z\mathbf{k}).(2\mathbf{i}+\mathbf{k}) $ =5 or $ 2x+z=5 $ .
BETA
