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  • Need solution for RD Sharma maths class 12 chapter 28 The Plane exercise 28.6 question 9 maths textbook solution

Need solution for RD Sharma maths class 12 chapter 28 The Plane exercise 28.6 question 9 maths textbook solution

Answers (1)

Answer:

              Therefore, required equation of the plane is 3x+4y-5z=9

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 ( 2,2,1\right )and\left ( 9,3,6\right ) plane 3x+4y-5z=9

Solution:

We know that solution of a plane passing through \left ( x_{1} ,y_{1},z_{1}\right )is given as,

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

The required plane passing through \left ( 2,2,1 \right )

So the equation of the plane is

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

Plane (1) is also passing through \left ( 9,3,6 \right )

So \left ( 9,3,6 \right ) must satisfy the equation of plane

                                                               a\left ( 9-2 \right )+b\left ( 3-2 \right )+c\left ( 6-1 \right )=0

7a+b+5c=0                                                     (2)

Plane 2x+6y+6z=1is perpendicular to required plane

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

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

Using (a) we have

2a+6b+6c=0                                                  (3)

Solving (2) and (3) we get

                                                               \begin{aligned} &\frac{a}{1 \times 6-5 \times 6}=\frac{b}{2 \times 5-7 \times 6}=\frac{c}{7 \times 6-2 \times 1} \\ &\frac{a}{6-30}=\frac{b}{10-42}=\frac{c}{42-2} \\ &\frac{a}{-24}=\frac{b}{-32}=\frac{c}{40}=\lambda \\ &a=-24 \lambda, b=-32 \lambda, c=40 \lambda \end{aligned}

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

                                                               \begin{aligned} &(-24 \lambda)(x-2)+(-32 \lambda)(y-2)+(40 \lambda)(z-1)=0 \\ &-24 \lambda x+48 \lambda-32 \lambda y+64 \lambda+40 \lambda z-40 \lambda=0 \\ &-24 \lambda x-32 \lambda y+40 \lambda z+72 \lambda=0 \end{aligned}
Divide by \left ( -8\lambda \right )we get

                                                             3x+4y-5z-9=0

So, required plane is 3x+4y-5z-9=0

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