A) \[2x-5y-z-15=0\]
B) \[2x-5y+z-15=0\]
C) \[2x-5y-z+15=0\]
D) \[2x+5y-z+15=0\]
E) \[2x+5y+z+15=0\]
Correct Answer: C
Solution :
Given equation can be rewritten as \[\overrightarrow{r}=3\hat{j}+(\hat{i}+2\hat{k})s+(-2\hat{i}-\hat{j}+\hat{k})t\] which is a plane passing through\[\overrightarrow{a}=3\hat{j}\]and parallel to the vectors\[\overrightarrow{b}=\hat{i}+2\hat{k}\]and\[\overrightarrow{c}=-2\hat{i}-\hat{j}+\hat{k}\]. Therefore, it is perpendicular to the vector \[\overrightarrow{n}=\overrightarrow{b}\times \overrightarrow{c}=2\hat{i}-5\hat{j}-\hat{k}\] Hence, its vector equation is \[(\overrightarrow{r}-\overrightarrow{a}).\overrightarrow{n}=0\] \[\Rightarrow \] \[\overrightarrow{r}.\overrightarrow{n}=\overrightarrow{a}.\overrightarrow{n}\] \[\Rightarrow \]\[(x\hat{i}+y\hat{j}+z\hat{k}).(2\hat{i}-5\hat{j}-\hat{k})\] \[=3\hat{j}.(2\hat{i}-5\hat{j}-\hat{k})\] \[\Rightarrow \] \[5x-5y-z+15=0\]
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