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

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

Answers (1)

Answer:

              Therefore, required equation of the plane is 5x-4y-z=7

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 )             

              Plane 2x+3y-2z=5and x+2y-3z=8

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 passes through\left ( 1,-1,2 \right ).

So the equation plane is  a\left ( x-1\right )+b\left ( y+1 \right )+c\left ( z-2 \right )=0

ax+by+cz=a-b+2c                 (1)

Now the required plane is also perpendicular to the planes,

                                                   2x+3y-2z=5 & x+2y-3z=8

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}=0are the right angle

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

Using (a) we get

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

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

Solving (b) and (c) we get

                                                      \begin{aligned} &\frac{a}{3 \times(-3)-2 \times(-2)}=\frac{b}{1 \times(-2)-2 \times(-3)}=\frac{c}{2 \times 2-1 \times 3} \\ &\frac{a}{-9+4}=\frac{b}{-2+6}=\frac{c}{4-3} \\ &\frac{a}{-5}=\frac{b}{4}=\frac{c}{1}=\lambda \\ &a=-5 \lambda, b=4 \lambda, c=\lambda \end{aligned}

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

                                                     \begin{aligned} &\frac{a}{3 \times(-3)-2 \times(-2)}=\frac{b}{1 \times(-2)-2 \times(-3)}=\frac{c}{2 \times 2-1 \times 3} \\ &\frac{a}{-9+4}=\frac{b}{-2+6}=\frac{c}{4-3} \\ &\frac{a}{-5}=\frac{b}{4}=\frac{c}{1}=\lambda \\ &a=-5 \lambda, b=4 \lambda, c=\lambda \end{aligned}

Dividing both the sides by\left ( -\lambda \right ), we get

                                                 5x-4y-z=7

So, the required plane is 5x-4y-z=7

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