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A Plane Makes Intercepts −6, 3, 4 Respectively on the Coordinate Axes. Find the Length of the Perpendicular from the Origin on It. - Mathematics

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प्रश्न

A plane makes intercepts −6, 3, 4 respectively on the coordinate axes. Find the length of the perpendicular from the origin on it.

बेरीज

उत्तर

We know that the equation of the plane which makes intercepts ab and c on the coordinate axes is 

\[\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\]

So, the equation of the plane which makes intercepts −6, 3, 4 on the x-axis, y-axis and z-axis, respectively is \[\frac{x}{- 6} + \frac{y}{3} + \frac{z}{4} = 1\]
\[ \Rightarrow - 2x + 4y + 3z = 12\]
\[ \Rightarrow 2x - 4y - 3z + 12 = 0\]

∴ Length of the perpendicular from (0, 0, 0) to the plane

\[2x - 4y - 3z + 12 = 0\]
\[= \left| \frac{2 \times 0 - 4 \times 0 - 3 \times 0 + 12}{\sqrt{2^2 + \left( - 4 \right)^2 + \left( - 3 \right)^2}} \right|\]
\[ = \left| \frac{12}{\sqrt{4 + 16 + 9}} \right|\]
\[ = \frac{12}{\sqrt{29}} \text{ units } \]
Thus, the length of the perpendicular from the origin to the plane is \[\frac{12}{\sqrt{29}}\]  units . 
 
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पाठ 29: The Plane - Exercise 29.09 [पृष्ठ ४९]

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आरडी शर्मा Mathematics [English] Class 12
पाठ 29 The Plane
Exercise 29.09 | Q 12 | पृष्ठ ४९

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