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N is a Vector of Magnitude √ 3 and is Equally Inclined to an Acute Angle with the Coordinate Axes. Find Vector and Cartesian Forms of Equation of a Plane Which Passes Through (2, 1, −1) . . - Mathematics

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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 \[\vec{n}\] .

 

योग
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उत्तर

\[ \text{ Let }  \alpha, \beta \text{ and }  \gamma \text{ be the angles made by } \vec{n} \text{ with }x, y \text{ and z-axes respectively } .\]

\[ \text{ Given that } \]

\[\alpha = \beta = \gamma\]

\[ \Rightarrow \cos \alpha = \cos \beta = \cos \gamma\]

\[ \Rightarrow l = m = n, \text{ where l, m, n are direction cosines of }  \vec{n} .\]

\[\text{ But } l^2 + m^2 + n^2 = 1\]

\[ \Rightarrow l^2 + l^2 + l^2 = 1\]

\[ \Rightarrow \text{ 3  }l^2 = 1\]

\[ \Rightarrow l^2 = \frac{1}{3}\]

\[ \Rightarrow l = \frac{1}{\sqrt{3}} (\text{ Since } \alpha \text{ is acute, l = cos }  \alpha >0)\]

\[\text{ Thus }, \vec{n} = \sqrt{3} \left( \frac{1}{\sqrt{3}} \hat{i}  + \frac{1}{\sqrt{3}} \hat{j} + \frac{1}{\sqrt{3}} \hat{k} \right) = \hat{i} + \hat{j} + \hat{k} (\text{ Using } \vec{r} =\left| \vec{r} \left( l \hat{i}  + m \hat{j} + n \hat{k} \right) \right|)\]

\[\text{ We know that the vector equation of the plane passing through a point }  \vec{a} \text{ and normal to } \vec{n} \text{ is } \]

\[ \vec{r} . \vec{n} = \vec{a} . \vec{n} \]

\[\text{ Substituting  } \vec{a} = 2 \hat{i} + \hat{j}  - \hat{k}  \text{ and  } \vec{n} = \hat{ i } + \hat{j} + \hat{k} , \text{ we get } \]

\[ \vec{r} . \left( \hat{i}  + \hat{j}  + \hat{k}  \right) = \left( 2 \hat{i} + \hat{j} - \hat{k} \right) . \left( \hat{i} + \hat{j}  + \hat{k}  \right)\]

\[ \Rightarrow \vec{r} . \left( \hat{i} + \hat{j} + \hat{k}  \right) = 2 + 1 - 1\]

\[ \Rightarrow \vec{r} . \left( \hat{i}  + \hat{j}  + \hat{k}  \right) = 2\]

\[\text{ For the Cartesian form, we need to substitute } \vec{r} = x \hat{i}  + y \hat{j}  + z \hat{k}  \text{ in the vector equation } .\]

\[ \text{ Then, we get } \]

\[\left( x \hat{i} + y \hat{j} + z \hat{k}  \right) . \left( \hat{i}  + \hat{j}  + \hat{k}  \right) = 2\]

\[ \Rightarrow x + y + z = 2\]

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अध्याय 29: The Plane - Exercise 29.03 [पृष्ठ १३]

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आरडी शर्मा Mathematics [English] Class 12
अध्याय 29 The Plane
Exercise 29.03 | Q 6 | पृष्ठ १३

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