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  • Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.8 question 18

Provide solution for RD Sharma maths class 12 chapter The Plane exercise 28.8 question 18

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Answer:-  The answer of the given question are

  1. 13 x+14 y+11 z=0, This plane doesn’t satisfies the given condition.
  2. The equation of the required plane is, 7 x+11 y+14 z=15.
  3. The equation of the required plane is, 7 x+11 y+4 z=33

Hint:-  By using intercept form of plane

            \frac{x}{a}+\frac{y}{b}+\frac{c}{z}=1

Given:-  The equation of the family of planes passing through the intersection of the planes

                x+2 y+3 z-4=0      and 2 x+y-z+5=0

Solution:- The equation of the family of planes passing through the intersection of the planes

            x+2 y+3 z-4=0 and 2 x+y-z+5=0

                (x+2 y+3 z-4)+k(2 x+y-z+5)=0 , where k is some constant.

                \begin{aligned} &(2 k+1) x+(k+2) y+(3-k) z=4-5 k \\\\ &\frac{x}{\left(\frac{4-5 k}{2 k+1}\right)}+\frac{y}{\left(\frac{y}{k+2}\right)}+\frac{z}{\left(\frac{4-5 k}{3-k}\right)}=1 \end{aligned}               

It is given that x-intercept of the required plane is twice its z intercept.

        \begin{aligned} &\left(\frac{4-5 k}{2 k+1}\right)=2\left(\frac{4-5 k}{3-k}\right) \\\\ &(4-5 k)(3-k)=(4 k+2)(4-5 k) \\\\ &(4-5 k)(3-k-4 k-2)=0 \end{aligned}

        (4-5 k)(1-5 k)=0       

Either

          \begin{aligned} &4-5 k=0 \quad \text { or } \quad 1-5 k=0\\\\ &k=4 / 5 \quad \text { or }\quad k=1 / 5 \end{aligned}

When k=4 / 5, the equation of the plane is
         \begin{aligned} &\left(2 \times \frac{4}{5}+1\right) x+\left(\frac{4}{5}+2\right) y+\left(3-\frac{4}{5}\right) z=4-5 \times \frac{4}{5} \\\\ &13 x+14 y+11 z=0 \end{aligned}

This plane does not satisfies the given condition, so this is rejected

When k=1 / 5 , the equation of the plane is
            \begin{aligned} &\left(2 \times \frac{1}{5}+1\right) x+\left(\frac{1}{5}+2\right) y+\left(3-\frac{1}{5}\right) z=4-5 \times \frac{1}{5} \\\\ &7 x+11 y+14 z=15 \end{aligned}

Thus, the equation of the required plane is 7 x+11 y+14 z=15.

Also, the equation of the plane passing through the point (2,3,-1) and parallel to the plane 7 x+11 y+14 z=15 is


            \begin{aligned} &7(x-2)+11(y-3)+14(z+1)=0 \\\\ &7 x+11 y+4 z=33 \end{aligned}

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