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  • Please Solve R.D.Sharma class 12 Chapter 28 The Plane Exercise 28.13 Question 5 Maths textbook Solution.

Please Solve R.D.Sharma class 12 Chapter 28 The Plane  Exercise 28.13 Question  5  Maths textbook Solution.

Answers (1)

Answer: equation of the plane 95x-17y+25z+53=0  point of intersection  is \left ( 2,4,-3 \right )

Hint: use simultaneous equation

Given:\frac{x+y}{3}=\frac{y+6}{5}=\frac{z-1}{-2} \text { and } 3 x-2 y+z+3=0=2 x+3 y+2-6

Solution: we have equation of the line is  \frac{x+y}{3}=\frac{y+6}{5}=\frac{z-1}{-2} =\lambda

                Point on the line is given by \left ( 3x-4,5x-6,-2x+1 \right )-------(1)

                Another equation of line is  3x-2y+z+5=0

                                                                    8z+2y+4=0

Let a, b, c be the direction ratio of the lines it will be perpendicular to normal of  3x-2y+z+5=0and 2x+3y+4z-4=0  so using a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}=0

                  \begin{aligned} &(3)(a)+(-2)(6)+(1)(c)=3 a-2 b+c=0 \\ \end{aligned}............(2)

Agian        \begin{aligned} &(2)(a)+(3)(b)+(4)(c)=0 \\ \end{aligned}

                  \begin{aligned} &2 a+3 b+4 c=0 \ldots-\cdots-(3) \\ \end{aligned}

Solving (2) and (3) by cross-multiplication

                  \begin{aligned} &\frac{a}{(-2)(4)-(3)(1)}=\frac{b}{(2)(1)-(2)(4)}=\frac{c}{(3)(3)-(-2) 2)} \\ \end{aligned}

                 \begin{aligned} &\frac{a}{-8-3}=\frac{b}{2-12}=\frac{c}{9+4} \\ \end{aligned}

                  \begin{aligned} &\frac{a}{-11}=\frac{b}{-10}=\frac{c}{13} \\ \end{aligned}

Directional ratios are proportional to

               \begin{aligned} &(-11,-10,13) \\ &\text { Let } z=0 \\ &3 x-2 y=-5 \ldots \cdots-(i) \\ &2 x+3 y=4 \ldots \ldots-(i i) \end{aligned}

Solving (i) and (ii) by elimination method

               \begin{aligned} 6 x-4 y=-10 \\ \pm 6 x \pm 3 y=\pm 12 \\ \hline-13 y=-22 \\ y=\frac{22}{13} \end{aligned}

Put y in equation (i)

               \begin{aligned} &3 x-2 y=-5 \\ &3 x-2 \frac{22}{13}=-5 \\ &3 x-\frac{44}{13}=-5 \\ &3 x=-5+\frac{44}{13} \\ &3 x=\frac{-21}{13} \\ &x=\frac{-7}{13} \end{aligned}

So, the equation of the line (2) is symmetrical form

\frac{x+\frac{7}{18}}{-11}=\frac{y-\frac{22}{13}}{-10}=\frac{2-0}{13}

Put the general point of a line from equation (i)

\begin{aligned} &\frac{3 x-4+\frac{7}{13}}{-11}=\frac{5 x-6-\frac{22}{13}}{-10}=\frac{-2 x+1}{13} \\ &\frac{39 x-52+7}{-11 x 13}=\frac{65 x-78-22}{-10 x 13}=\frac{-2 x+1}{13} \\ &\frac{39 x-45}{-11}=\frac{65 x-100}{-10}=\frac{-2 x+1}{1} \end{aligned}

The equation of the plane is 45x-17y+25z+53

There point of intersection is (2, 4, -3)

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