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  • Please solve RD Sharma class 12 chapter The Plane exercise 28.8 question 13 maths textbook solution

Please solve RD Sharma class 12 chapter The Plane exercise 28.8 question 13 maths textbook solution

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Answer:-  The answer of the given question is \vec{r} \cdot(20 \hat{\imath}+23 \hat{\jmath}+26 \hat{k})=69

Hints:- 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:  \vec{r} \cdot(\hat{\imath}+\hat{\jmath}+\hat{k})=6        and

            \vec{r} \cdot\left(2 \hat{\imath}+\widehat{3}_{j}+4 \hat{k}\right)=-5  and Point (1,1,1)

Solution:-  The Cartesian equation of the given planes are x+y+z-6=0 and

                2x+3y+4z+5=0

The family of planes isx+y+z-6+\lambda(2 x+3 y+4 z+5)=0       …(i)

Since it pass through (1,1,1)
                \begin{gathered} 1+1+1-6+\lambda(2+3+4+5)=0 \\\\ -3+\lambda(14)=0 \\ \lambda=\frac{3}{14} \end{gathered}
            \begin{gathered} x+y+z-6+\frac{3}{14}(2 x+3 y+4 z+5)=0 \\\\ 14 x+14 y+14 z-84+6 x+9 y+12 z+15=0 \\\\ 20 x+23 y+26 z-69=0 \end{gathered}

Vector equation is  \vec{r} \cdot(20 \hat{\imath}+23 \hat{\jmath}+26 \hat{k})=69

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