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  • Explain solution RD Sharma class 12 chapter The Plane exercise 28.8 question 16 maths

Explain solution RD Sharma class 12 chapter The Plane exercise 28.8 question 16 maths

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Answer:-  The answer of the given question is\vec{r} \cdot(\hat{\imath}-\hat{k})+2=0 \text {. }

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:- x+y+z=1    ,

            2x+3y+4z=5   

            x-y+z=0

Solution:-  Equation of the plane through the intersection of plane is 

            \begin{aligned} &(x+y+z-1)+\lambda(2 x+3 y+4 z-5)=0 \text { or } \\\\ &(1+2 \lambda) x+(1+3 \lambda) y+(1+4 \lambda) z-(1+5 \lambda)=0 \end{aligned}                            …(i)

This plane is perpendicular to x-y+z=0

We know that, two planes are perpendicular if  a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0


            \begin{array}{r} 1(1+2 \lambda)-1(1+3 \lambda)+1(1+4 \lambda)=0 \text { or } \\\\ \qquad \lambda=-\frac{1}{3} \end{array}

Equation of plane is

            \begin{gathered} (x+y+z-1)-\frac{1}{3}(2 x+3 y+4 z-5)=0 \\\\ \Rightarrow x-z+2=0 \end{gathered}

Vector form of plane is  \vec{r} \cdot(\hat{\imath}-\hat{k})+2=0

Yes, line lies on the plane on (-2, 3, 0) satisfies\vec{r} \cdot(\hat{\imath}-\hat{k})+2=0

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