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  • Please solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28 .9 Question 12 maths textbook solution.

Please solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28.9 Question 12 maths textbook solution.

Answers (1)

Answer : The answer of the given question is \frac{12}{\sqrt{29}} units

Hint :

       P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|

Given : Plane makes intercept -6,3,4 respectively on the coordinate axes

Solution :

The equation of the plane which makes intercept a, b and c with the x, y and z axes respectively is

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

Putting the values of a, b and c

Require equation of the plane

\begin{aligned} &\frac{x}{-6}+\frac{y}{3}+\frac{z}{4}=1 \\ &-2 x+4 y+3 z=12 \end{aligned}

We know, the distance of the point \left(x_{1}, y_{1}, z_{1}\right) from the plane

\pi: a x+b y+c z+d=0 Is given by

P=\left|\frac{a x_{1}+b y_{1}+c z_{1}+d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|

Distance from the origin i.e. (0,0,0):\left|\frac{d}{\sqrt{a^{2}+b^{2}+c^{2}}}\right|

Required distance

\begin{aligned} &=\left|\frac{-12}{\sqrt{(-2)^{2}+4^{2}+3^{2}}}\right| \\ &=\frac{12}{\sqrt{29}} \end{aligned}

The length of the perpendicular from the origin on the plane =\frac{12}{\sqrt{29}} \text { Units }

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