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Question
Find the distance of the plane 2x − 3y + 4z − 6 = 0 from the origin.
Solution
\[\text{The given equation of the plane is } \]
\[2x - 3y + 4z = 6 . . . \left( 1 \right)\]
\[\text{ Now } ,\sqrt{2^2 + \left( - 3 \right)^2 + 4^2}=\sqrt{4 + 9 + 16}=\sqrt{29}\]
\[ \text{ Dividing (1) by } \sqrt{29}, \text{ we get }\]
\[\frac{2}{\sqrt{29}}x - \frac{3}{\sqrt{29}}y + \frac{4}{\sqrt{29}}z = \frac{6}{\sqrt{29}}, \text{ which is the normal form of plane (1) } .\]
\[\text{ So, the length of the perpendicular from the origin to the plane} = \frac{6}{\sqrt{29}}\]
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