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  • Please solve RD Sharma class 12 chapter The Plane exercise 28.8 question 5 maths textbook solution

Please solve RD Sharma class 12 chapter The Plane exercise 28.8 question 5 maths textbook solution

Answers (1)

Answer:-  The answer of the given question is 28x-17y+9z=0.

Hint:- We know that, equation of a plane passing through the line of intersection of two planes a_{1} x+b_{1} y+c_{1} z+d_{1}=0 \text { and } a_{2} x+b_{2} y+c_{2} z+d_{2}=0 is given by

\left(a_{1} x+b_{1} y+c_{1} z+d_{1}\right)+k\left(a_{2} x+b_{2} y+c_{2} z+d_{2}\right)=0

Given:-2x-y=0 and 3z-y=0  and plane 4x+5y-3z=8

Solution:-  So equation of plane passing through the line of intersection of given two planes is

                        2x-y=0 and 3z-y=0 is

                        \begin{gathered} (2 x-y)+k(3 z-y)=0 \\\\ 2 x-y+3 k z-k y=0 \end{gathered}       

                    x(2)+y(-1-k)+z(3 k)=0           …(i)

We know that, two planes are perpendicular if

                        a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0                    …(ii)

 Given, plane (i) is perpendicular to plane

                        4x+5y-3z=8                                        …(iii)

Using (i) and (iii) in equation (ii)

                        \begin{gathered} (2)(4)+(-1-k)(5)+(3 k)(-3)=0 \\\\ 3-14 k=0 \end{gathered}

                                                           k=\frac{3}{14}

Putting value of k in equation (i)    

        

        \begin{gathered} x(2)+y(-1-k)+z(3 k)=0 \\\\ x(2)+y\left(-1-\frac{3}{14}\right)+z\left(3\left(\frac{3}{14}\right)\right)=0 \\\\ x(2)+y\left(-\frac{17}{14}\right)+z\left(\frac{9}{14}\right)=0 \end{gathered}

Multiplying with 14 we get

        28x-17y+9z=0

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