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  • Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.8 question 6

Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.8 question 6

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Answer:-  The answer of the given question is 33x+45y+50z-41=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:-x+2y+3z-4=0 and

             2x+y-z+5=0

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

        x+2y+3z-4=0 and2x+y-z+5=0 is given by

        \begin{aligned} &(x+2 y+3 z-4)+k(2 x+y-z+5)=0 \\\\ &x+2 y+3 z-4+2 k x+k y-k z+5 k=0 \end{aligned}

        x(1+2 k)+y(2+k)+z(3-k)-4+5 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,

        5x+3y-6z+8=0                                                                  … (iii)

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

                                            \begin{gathered} 5(1+2 k)+3(2+k)+(-6)(3-k)=0 \\\\ 5+10 k+6+3 k-18+6 k=0 \\\\ k=\frac{7}{19} \end{gathered}

Putting the value of k in equation (iii)

        \begin{gathered} x\left(1+\frac{14}{19}\right)+y\left(2+\frac{7}{19}\right)+z\left(3-\frac{7}{19}\right)-4+\frac{35}{19}=0 \\\\ x \frac{(19+14)}{19}+y\left(\frac{38+7}{19}\right)+z\left(\frac{57-7}{19}\right)+\frac{-76+35}{19}=0 \end{gathered}

        x\left(\frac{33}{19}\right)+y\left(\frac{45}{19}\right)+z\left(\frac{50}{19}\right)-\frac{41}{19}=0

Multiplying with 19 we get

        33x+45y+50z-41=0

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