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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

Answers (1)

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}

Posted by

infoexpert26

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