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  • Explain solution for RD Sharma maths class 12 chapter 28 The Plane exercise 28 point 6 question 15 maths textbook solution

Explain solution for RD Sharma maths class 12 chapter 28 The Plane exercise 28.6 question 15 maths textbook solution

Answers (1)

Answer:

              Therefore, required equation of the plane is 18x+17y+4z=49a\left ( x-x_{1} \right )+b\left ( y-y_{1} \right )+c\left ( z-z_{1} \right )=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 ( 2,1,-1 \right ) and \left ( -1,3,4 \right )               

              Plane x-2y+4z=10

Solution:

We know that solution of 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 require plane passes through \left ( 2,1,-1 \right )

So, the equation of plane is

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

Plane (1) is also passing through \left ( -1,3,4 \right ). So, \left ( -1,3,4 \right )must satisfy the equation of plane

So, we know

                                            a\left ( -1-2 \right )+b\left ( 3-1 \right )+c\left ( 4+1 \right )=0

-3a+2b+5c=0                                                             (2)

Planex-2y+4z=10 is perpendicular to the required plane

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

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

Using (a) we have

a-2b+4c=0          (3)

Solving (2) and (3) we get

\begin{aligned} &\frac{a}{4 \times 2+5 \times 2}=\frac{b}{5 \times 1+3 \times 4}=\frac{c}{3 \times 2-2 \times 1} \\ &\frac{a}{18}=\frac{b}{17}=\frac{c}{4} \\ &\frac{a}{18}=\frac{b}{17}=\frac{c}{4}=\lambda \\ &a=18 \lambda, b=17 \lambda, c=4 \lambda \end{aligned}

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

We get
\begin{aligned} &18 \lambda(x-2)+17 \lambda(y-1)+4 \lambda(z+1)=0 \\ &18 \lambda x-36 \lambda+17 \lambda y-17 \lambda+4 \lambda z+4 \lambda=0 \\ &18 \lambda x+17 \lambda y+4 \lambda z-49 \lambda=0 \end{aligned}

Divide by \left ( \lambda \right )we get

                                                    18x+17y+4z-49=0

So, the required plane is
                                                      18x+17y+4z-49=0

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