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  • Explain solution for RD Sharma maths class 12 chapter 28 The Plane exercise 28.6 question 14 maths textbook solution

Explain solution for RD Sharma maths class 12 chapter 28 The Plane exercise 28.6 question 14 maths textbook solution

Answers (1)

Answer:

              Therefore, required equation of the plane is 7x-8y+3z+25=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 ( -1,3,2 \right )              

              Plane x+2y+3z=5and 3x+3y+z=0

Solution:

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

So, the equation of plane is

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

ax+by+cz=3b+2c-a                             (1)

Now, the required plane is also perpendicular to the planes,

                                             x+2y+3z=5  &   3x+3y+z=0

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

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

Using (a) we get

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

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

Solving (b) and (c) we get

                                                       \begin{aligned} &\frac{a}{2 \times 1-3 \times 3}=\frac{b}{3 \times 3-1 \times 1}=\frac{c}{1 \times 3-3 \times 2} \\ &\frac{a}{2-9}=\frac{b}{9-1}=\frac{c}{3-6} \\ &\frac{a}{-7}=\frac{b}{8}=\frac{c}{-3}=\lambda \\ &a=-7 \lambda, b=8 \lambda, c=-3 \lambda \end{aligned}
Putting values of a, b and c in equation (1) we get

                                                     \begin{aligned} &(-7 \lambda) x+(8 \lambda) y+(-3 \lambda) z=3(8 \lambda)+2(-3 \lambda)+7 \lambda \\ &-7 \lambda x+8 \lambda y-3 \lambda z=24 \lambda-6 \lambda+7 \lambda \\ &-7 \lambda x+8 \lambda y-3 \lambda z=25 \lambda \end{aligned}
Dividing both \left ( -\lambda \right )we get

                                                       7x-8y+3z+25=0

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