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  • Need solution for RD Sharma maths class 12 chapter 28 The Plane exercise 28.6 question 8 maths textbook solution

Need solution for RD Sharma maths class 12 chapter 28 The Plane exercise 28.6 question 8 maths textbook solution

Answers (1)

Answer:

              Therefore, required equation of the plane is x+y-2z+4=0

Hint:

              Using the formula a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0

Given:

              Points \left ( 1,-1,2 \right )and\left ( 2,-2,2 \right ) plane 6x-2y+2z=9

Solution:

We know that solution of plane passing through \left ( x,y,z \right )is given as,

                                                       a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0

The require plane passes through \left ( 1,-1,2 \right ) 

So, the equation of plane is

a\left(x-1\right)+b\left(y+1\right)+c\left(z-2\right)=0                            (1)

Plane (1) is also passing through\left ( 2,-2,2 \right ). So, \left ( 2,-2,2 \right )must satisfy the equation of plane

So, we know

                               a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0       

a-b=0                                                          (2)

Plane 6x-2y+2z=9 is perpendicular to the required plane

We know that plane a_{1}x+b_{1}y+c_{1}z+d_{1}=0 and a_{2}x+b_{2}y+c_{2}z+d_{2}=0are at right angle

If  a_{1}a2+b_{1}b_{2}+c_{1}c_{2}=0                                                (a)

Using (a) we have

                                                            a\left ( 6 \right )+b\left ( -2 \right )+c\left ( 2 \right )=0

a\left ( 6 \right )+b\left ( -2 \right )+c\left ( 2 \right )=0                                                  (3)

Solving (2) and (3) we get

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

                                               \begin{aligned} &\frac{a}{-2}=\frac{b}{-2}=\frac{c}{4}=\lambda \\ &a=-2 \lambda, b=-2 \lambda, c=4 \lambda \end{aligned}

Putting values of a, b and c in equation (1)

We get

                                              \begin{aligned} &(-2 \lambda)(x-1)+(-2 \lambda)(y+1)+(4 \lambda)(z-2)=0 \\ &-2 \lambda x+2 \lambda-2 \lambda y-2 \lambda+4 \lambda z-8 \lambda=0 \\ &-2 \lambda x-2 \lambda y+4 \lambda z-8 \lambda=0 \end{aligned}
Divide by \left ( -2\lambda \right )we get

                                                  x+y-2z+4=0

So, the required plane is x+y-2z+4=0

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