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  • Explain solution for RD Sharma maths class 12 chapter 28 The Plane exercise 28.6 question 6 maths textbook solution

Explain solution for RD Sharma maths class 12 chapter 28 The Plane exercise 28.6 question 6 maths textbook solution

Answers (1)

Answer:

              Therefore, required equation of the plane is 2x-4y+3z=8

Hint:

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

Given:

              Point \left ( 1,-3,-2 \right ) and planes x+2y+2z=5and 3x+3y+2z=8

Solution:

We know that equation of the 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 passes through \left ( 1,-3,-2 \right ) so the equation of the plane is

                                                      a\left ( x-1\right )+b\left ( y+3 \right )+c\left ( z+2 \right )=0

ax+by+cz=a-3b-2c               (1)

Now the required plane is also perpendicular to the planes

x+2y+2z=5 And 3x+3y+2z=8

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

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

Using (a) we have,

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

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

Solving (b) and (c) we get

                                             \begin{aligned} &\frac{a}{2 \times 2-3 \times 2}=\frac{b}{3 \times 2-1 \times 2}=\frac{c}{1 \times 3-2 \times 3} \\ &\frac{a}{4-6}=\frac{b}{6-2}=\frac{c}{3-6} \\ &\frac{a}{-2}=\frac{b}{4}=\frac{c}{-3}=\lambda \\ &a=-2 \lambda, b=4 \lambda, c=-3 \lambda \end{aligned}

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

                                           \begin{aligned} &(-2 \lambda) x+(4 \lambda) y+(-3 \lambda) z=(-2 \lambda)-3(4 \lambda)-2(-3 \lambda) \\ &-2 \lambda x+4 \lambda y-3 \lambda z=-2 \lambda-12 \lambda+6 \lambda \\ &-2 \lambda x+4 \lambda y-3 \lambda z=-8 \lambda \end{aligned}

Divide both sides by \left ( -\lambda \right ) we get

                                             2x-4y+3z=8

So, the equation of the required planes is 2x-4y+3z=8

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