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  • Explain solution for RD Sharma maths class 12 chapter 28 The Plane exercise 28.6 question 5 maths textbook solution

Explain solution for RD Sharma maths class 12 chapter 28 The Plane exercise 28.6 question 5 maths textbook solution

Answers (1)

Answer:

              Therefore, required equation of the plane is 5x+9y+11z-8=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:

             \text { Point }(-1,-1,2) \text { and planes } 3 x+2 y-3 z=1 \text { and } 5 x-4 y+z=5

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,- 1,2\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

ax+by+cz=2c-a-b                 (1)

Now the required plane is also perpendicular to the planes. 3x+2y-3z=1And

                                                        5x-4y+z=5

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 at right angles

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

Using (a) we have,

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

5a-4b+c=0                                          (c)

Solving (b) and (c) we get

\begin{aligned} &\frac{a}{2 \times 1-(-3) \times(-4)}=\frac{b}{5 \times(-3)-3 \times(1)}=\frac{c}{3 \times(-4)-2 \times(5)} \\ &\frac{a}{2-12}=\frac{b}{-15-3}=\frac{c}{-12-10} \\ &\frac{a}{-10}=\frac{b}{-18}=\frac{c}{-22}=\lambda \\ &a=-10 \lambda, b=-18 \lambda, c=-22 \lambda \end{aligned}

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

\begin{aligned} (-10 \lambda) x+(-18 \lambda) y+(-22 \lambda) z &=2(-22 \lambda)-(-10 \lambda)-(-18 \lambda) \\ -10 \lambda x-18 \lambda y-22 \lambda z &=-44 \lambda+10 \lambda+18 \lambda \\ -10 \lambda x-18 \lambda y-22 \lambda z &=-16 \lambda \end{aligned}

Divide both sides by\left ( -2\lambda \right ), we get

                                                              5x+9y+11z=8

So, the required of the plane equation is 5x+9y+11z-8=0

 

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