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  • Please solve RD Sharma class 12 chapter The Plane exercise 28.2 question 5 maths textbook solution

Please solve RD Sharma class 12 chapter The Plane exercise 28.2 question 5 maths textbook solution

Answers (1)

Answer:

Equation of plane is 6x-3y+2z=18

Hint:

The equation of plane with intercepts p,q \; \&\; r is xp+yq+zr=1

Given:

It is given that plane meets the co-ordinate axes at A, B \; \& \; C  with centroid of \Delta ABC  is (1,-2,3).

Explanation:

Centroid of triangle ABC is (1,-2,3).

We know,

Centroid of triangle =\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}, \frac{z_{1}+z_{2}+z_{3}}{3}\right)

                            \begin{aligned} &\Rightarrow(1,-2,3)=\left(\frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3}\right) \\ &\Rightarrow(1,-2,3)=\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right) \\ &\Rightarrow \frac{a}{3}=1, \quad \frac{b}{3}=-2, \quad \frac{c}{3}=3 \\ &\Rightarrow a=3, \quad b=-6, \quad c=9 \end{aligned}\begin{aligned} &\Rightarrow(1,-2,3)=\left(\frac{a+0+0}{3}, \frac{0+b+0}{3}, \frac{0+0+c}{3}\right) \\\\ &\Rightarrow(1,-2,3)=\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right) \\ \end{aligned}

                            \Rightarrow \frac{a}{3}=1, \quad \frac{b}{3}=-2, \quad \frac{c}{3}=3 \\\\

                            \Rightarrow a=3, \quad b=-6, \quad c=9

Put the values of a,b \; \& \: c in

                           \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1, to get the equation of plane

                            \begin{aligned} &\Rightarrow \frac{x}{3}+\frac{y}{-6}+\frac{z}{9}=1 \\\\ &\frac{6 x-3 y+2 z}{18}=1 \end{aligned}

Thus, equation of plane is 6x-3y+2z=18

 

Posted by

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