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  • Explain solution for RD Sharma maths class 12 chapter 28 The Plane exercise 28.6 question 7 maths textbook solution

Explain solution for RD Sharma maths class 12 chapter 28 The Plane exercise 28.6 question 7 maths textbook solution

Answers (1)

Answer:

              Therefore, required equation of the plane is x+2y+5z=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:

              Passing through the origin is \left ( 0,0,0 \right )and planes x+2y-z=1and 3x-4y+z=5

Solution:

We know that solution 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 ( 0,0,0 \right )

So the equation of the plane is

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

ax+by+cz=0                      (1)

Now, the required plane is also perpendicular to the planes

x+2y-z=1 And 3x-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}=0 are at right angles

If a_{1}a2+b_{1}b_{2}+c_{1}c_{2}=0                      (a)

Using (a) we get

a+2b-c=0                                         (b)

3a-4b+c=0                        (c)

Solving (b) and (c) we get

                                                          \begin{aligned} &\frac{a}{2 \times 1-(-4)(-1)}=\frac{b}{3 \times(-1)-1 \times 1}=\frac{c}{1 \times(-4)-(2) \times 3} \\ &\frac{a}{2-4}=\frac{b}{-3-1}=\frac{c}{-4-6} \\ &\frac{a}{-2}=\frac{b}{-4}=\frac{c}{-10}=\lambda \\ &a=-2 \lambda, b=-4 \lambda, c=-10 \lambda \end{aligned}

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

                                                      \begin{aligned} &(-2 \lambda) x-4 \lambda y-10 \lambda z=0 \\ &-2 \lambda x-4 \lambda y-10 \lambda z=0 \end{aligned}

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

                                                        x+2y+5z=0

So the required equation of plane is

                                                        x+2y+5z=0

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