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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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Question

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

Sum

Solution

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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Chapter 29: The Plane - Exercise 29.09 [Page 49]

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RD Sharma Mathematics [English] Class 12
Chapter 29 The Plane
Exercise 29.09 | Q 12 | Page 49

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