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प्रश्न
\[\vec{n}\] is a vector of magnitude \[\sqrt{3}\] and is equally inclined to an acute angle with the coordinate axes. Find the vector and Cartesian forms of the equation of a plane which passes through (2, 1, −1) and is normal to \[\vec{n}\] .
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उत्तर
\[ \text{ Let } \alpha, \beta \text{ and } \gamma \text{ be the angles made by } \vec{n} \text{ with }x, y \text{ and z-axes respectively } .\]
\[ \text{ Given that } \]
\[\alpha = \beta = \gamma\]
\[ \Rightarrow \cos \alpha = \cos \beta = \cos \gamma\]
\[ \Rightarrow l = m = n, \text{ where l, m, n are direction cosines of } \vec{n} .\]
\[\text{ But } l^2 + m^2 + n^2 = 1\]
\[ \Rightarrow l^2 + l^2 + l^2 = 1\]
\[ \Rightarrow \text{ 3 }l^2 = 1\]
\[ \Rightarrow l^2 = \frac{1}{3}\]
\[ \Rightarrow l = \frac{1}{\sqrt{3}} (\text{ Since } \alpha \text{ is acute, l = cos } \alpha >0)\]
\[\text{ Thus }, \vec{n} = \sqrt{3} \left( \frac{1}{\sqrt{3}} \hat{i} + \frac{1}{\sqrt{3}} \hat{j} + \frac{1}{\sqrt{3}} \hat{k} \right) = \hat{i} + \hat{j} + \hat{k} (\text{ Using } \vec{r} =\left| \vec{r} \left( l \hat{i} + m \hat{j} + n \hat{k} \right) \right|)\]
\[\text{ We know that the vector equation of the plane passing through a point } \vec{a} \text{ and normal to } \vec{n} \text{ is } \]
\[ \vec{r} . \vec{n} = \vec{a} . \vec{n} \]
\[\text{ Substituting } \vec{a} = 2 \hat{i} + \hat{j} - \hat{k} \text{ and } \vec{n} = \hat{ i } + \hat{j} + \hat{k} , \text{ we get } \]
\[ \vec{r} . \left( \hat{i} + \hat{j} + \hat{k} \right) = \left( 2 \hat{i} + \hat{j} - \hat{k} \right) . \left( \hat{i} + \hat{j} + \hat{k} \right)\]
\[ \Rightarrow \vec{r} . \left( \hat{i} + \hat{j} + \hat{k} \right) = 2 + 1 - 1\]
\[ \Rightarrow \vec{r} . \left( \hat{i} + \hat{j} + \hat{k} \right) = 2\]
\[\text{ For the Cartesian form, we need to substitute } \vec{r} = x \hat{i} + y \hat{j} + z \hat{k} \text{ in the vector equation } .\]
\[ \text{ Then, we get } \]
\[\left( x \hat{i} + y \hat{j} + z \hat{k} \right) . \left( \hat{i} + \hat{j} + \hat{k} \right) = 2\]
\[ \Rightarrow x + y + z = 2\]
