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प्रश्न
Write the equation of a plane which is at a distance of \[5\sqrt{3}\] units from origin and the normal to which is equally inclined to coordinate axes.
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उत्तर
\[\text { Let } \alpha, \beta\text { and } \gamma\text { be the angles made by } \vec{n} withx, y andz-axes, respectively.\]
\[It is given that\]
\[\alpha = \beta = \gamma\]
\[ \Rightarrow \cos\alpha = \cos\beta = \cos\gamma\]
\[ \Rightarrow l = m = n, wherel, m, n\text { are direction cosines of } \vec{n} .\]
\[\text { But } l^2 + m^2 + n^2 = 1\]
\[ \Rightarrow l^2 + l^2 + l^2 = 1\]
\[ \Rightarrow 3 l^2 = 1\]
\[ \Rightarrow l^2 = \frac{1}{3}\]
\[ \Rightarrow l = \frac{1}{\sqrt{3}}\]
\[\text { So },l = m = n = \frac{1}{\sqrt{3}}\]
\[\text { It is given that the length of the perpendicular of the plane from the origin },p= 5\sqrt{3}\]
\[\text { The normal form of the plane is }lx + my + nz = p\]
\[ \Rightarrow \frac{1}{\sqrt{3}}x + \frac{1}{\sqrt{3}}y + \frac{1}{\sqrt{3}}z = 5\sqrt{3}\]
\[ \Rightarrow x + y + z = 5\sqrt{3} \left( \sqrt{3} \right) \]
\[ \Rightarrow x + y + z = 15\]
