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Explain solution RD Sharma class 12 chapter The Plane exercise 28.15 question 8 maths

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Answer:   (-3,5,2)

Hint:

Formula ,

\begin{aligned} &\frac{x-x_{1}}{a}=\frac{y+y_{1}}{b}=\frac{z-z_{1}}{c} \\\\ &=\frac{-2\left(a x_{1}+b y_{1}+c z_{1}+d\right)}{\left(a^{2}+b^{2}+c^{2}\right)} \\\\ &\hat{k} \mathrm{P}(1,3,4) \end{aligned}                                     

 

                                    

                                2x-y+z+3=0

Given:

Find the image of the point with (1,3,4) in the plane 2x-y+z+3=0

Solution:

\begin{aligned} &\left(x_{1}, y_{1}, z_{1}\right) \\\\ &a x+b y+z+d=0 \\\\ &\frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{3-31}{c}=\frac{-2\left(a x_{1}+b y_{1}+c_{1}+d\right)}{\left(a^{2}+b^{2}+c^{2}\right)} \end{aligned}

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

point =(1,3,4)

\begin{aligned} &x=-3 \\ &y=5, \\ &z=2 \end{aligned}

Mirror image (-3,5,2)

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