A) \[\mathbf{r}.(\sqrt{2}\mathbf{i}+\mathbf{j}+\mathbf{k})=4\]
B) \[\mathbf{r}.(\sqrt{2}\mathbf{i}+\mathbf{j}+\mathbf{k})=2\]
C) \[\mathbf{r}.(\mathbf{i}+\mathbf{j}+\mathbf{k})=4\]
D) None of these
Correct Answer: B
Solution :
Let \[\gamma \]be the angle made by \[\mathbf{n}\] with z-axis. Then direction cosines of \[\mathbf{n}\] are \[l=\cos {{45}^{o}}=\frac{1}{\sqrt{2}},\] \[m=\cos {{60}^{o}}=\frac{1}{2}\] and \[n=\cos \gamma \]. \ \[{{l}^{2}}+{{m}^{2}}+{{n}^{2}}=1\Rightarrow {{\left( \frac{1}{\sqrt{2}} \right)}^{2}}+{{\left( \frac{1}{2} \right)}^{2}}+{{n}^{2}}=1\] Þ \[{{n}^{2}}=\frac{1}{4}\Rightarrow n=\frac{1}{2}\], [\[\because \] \[\gamma \]is acute, \ \[n=\cos \gamma >0\]] We have \[|\mathbf{n}|=8\], \ \[\mathbf{n}=|\mathbf{n}|(l\mathbf{i}+m\mathbf{j}+n\mathbf{k})\] \[\Rightarrow \mathbf{n}=8\left( \frac{1}{\sqrt{2}}\mathbf{i}+\frac{1}{2}\mathbf{j}+\frac{1}{2}\mathbf{k} \right)\]\[=4\sqrt{2}\mathbf{i}+4\mathbf{j}+4\mathbf{k}\] The required plane passes through the point \[(\sqrt{2},-1,\,1)\] having position vector \[\mathbf{a}=\sqrt{2}\mathbf{i}-\mathbf{j}+\mathbf{k}\]. So, its vector equation is \[(\mathbf{r}-\mathbf{a}).\mathbf{n}=0\]or \[\mathbf{r}.\,\mathbf{n}=\mathbf{a}.\,\mathbf{n}\] Þ \[\mathbf{r}.(4\sqrt{2}\mathbf{i}+4\mathbf{j}+4\mathbf{k})=(\sqrt{2}\mathbf{i}-\mathbf{j}+\mathbf{k}).(4\sqrt{2}\mathbf{i}+4\mathbf{j}+4\mathbf{k})\] Þ \[\mathbf{r}.\,\,(\sqrt{2}\mathbf{i}+\mathbf{j}+\mathbf{k})=2\].
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