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  • Explain solution for RD Sharma maths class 12 chapter 28 The Plane exercise 28.6 question 10 maths textbook solution

Explain solution for RD Sharma maths class 12 chapter 28 The Plane exercise 28.6 question 10 maths textbook solution

Answers (1)

Answer:

              Therefore, required equation of the plane is 2x+2y+-3x+3=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 ( -1,1,1 \right )and\left ( 1,-1,1 \right ) plane x+2y+2z=5

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 is passing through \left ( -1,1,1 \right )

So, the equation of the plane is,

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

Plane (1) also passing through \left ( 1,-1,1 \right )

So \left ( 1,-1,1 \right )must satisfy the equation of plane

                                                      a\left ( 1+1 \right )+b\left ( -1-1 \right )+c\left ( 1-1 \right )=0

2a-2b=0                                                             (2)

Plane x+2y+2z=5is perpendicular to the required plane

We know that planes 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 right angle

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

Using (a) we have

                                                             a\left ( 1 \right )+b\left ( 2 \right )+c\left ( 2 \right )=0

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

Solving (2) and (3) we get

                                                      \begin{gathered} \frac{a}{(-2) \times 2-2 \times 0}=\frac{b}{1 \times 0-2 \times 2}=\frac{c}{2 \times 2-1 \times(-2)} \\ \frac{a}{-4-0}=\frac{b}{0-4}=\frac{c}{4+2} \\ \frac{a}{-4}=\frac{b}{-4}=\frac{c}{6}=\lambda \\ a=-4 \lambda, b=-4 \lambda, c=6 \lambda \end{gathered}

Putting the value of a, b and c in equation (1) we get

                                                                                      \begin{aligned} &(-4 \lambda)(x+1)+(-4 \lambda)(y-1)+6 \lambda(z-1)=0 \\ &-4 \lambda x-4 \lambda-4 \lambda y+4 \lambda+6 \lambda z-6 \lambda=0 \\ &-4 \lambda x-4 \lambda y+6 \lambda z-6 \lambda=0 \end{aligned}

Dividing by \left ( -2\lambda \right )we get

                                                                             2x+2y-3z+3=0

So the required plane is 2x+2y-3z+3=0

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