A) \[2x+y=5\]
B) \[2x-y=5\]
C) \[2x+z=5\]
D) \[2x-z=5\]
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 \[2x+z=5\].
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