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  • Please solve RD Sharma class 12 chapter The Plane exercise 28.15 question 1 maths textbook solution

Please solve RD Sharma class 12 chapter The Plane exercise 28.15 question 1 maths textbook solution

Answers (1)

Answer:   \left(-\frac{6}{61}, \frac{-8}{61}, \frac{12}{61}\right)

Hint:

Suppose point P(0,0,0), LetP M \perp \text { Plane }. Direction of P M \div 3,4,-6

Equation of line=\frac{x-x_{1}}{\alpha}=\frac{y-y_{1}}{\beta}=\frac{z-z_{1}}{\gamma}

P(0,0,0)

 

Given:

 

Find the image of the point (0,0,b) in the plane 3x+4y-6z+1=0

Solution:

\begin{aligned} &3 x+4 y-6 z+1=0 \\\\ &\text { Let } x_{1}=0=y_{1}=31 \end{aligned}

Equation of line =\frac{x-0}{3}=\frac{y-0}{4}=\frac{z-0}{-6}

\begin{aligned} &\text { Say } \lambda, \\ &x=3 \lambda \Rightarrow x=3 \times\left(\frac{-1}{61}\right)=\frac{-3}{61} \\ &y=4 \lambda \Rightarrow y=4 \times\left(\frac{-1}{61}\right)=\frac{-4}{61} \\ &z=-6 \lambda \Rightarrow y=-6 \times\left(\frac{-1}{61}\right)=\frac{6}{61} \end{aligned}

\begin{aligned} &\therefore x=\frac{-3}{61}, y=\frac{-4}{61}, z=\frac{6}{61} \\\\ &\frac{A}{z}=\frac{-3}{61}, \frac{B}{z}=\frac{-4}{61}, \frac{C}{z}=\frac{6}{61} \end{aligned}

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