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  • Need solution for RD Sharma Maths Class 12 Chapter 28 The Plane Excercise 28.11 Question 25

Need solution for RD Sharma Maths Class 12 Chapter 28 The Plane Excercise 28.11 Question 25

Answers (1)

Answer:  x+2 y+3 z-3=0

Hint: Use formula  \frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}

Given: Points  (-1,2,0) \&(2,2,-1)   and line  \frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}

Solution: We know that, the equation of plane passing through  \left(\mathrm{x}_{1}, \mathrm{y}_{1}, \mathrm{z}_{1}\right) Is given by

            a\left(x-x_{1}\right)+b\left(y-y_{1}\right)+c\left(z-z_{1}\right)=0                 …………. (1)

So, equation of plane passing through  \left ( -1,2,0 \right )  is

            a(x+1)+b(y-2)+c(z-0)=0                     …………. (2)

It also passes through \left ( 2,2,-1 \right )

So, equation (2) must satisfy the point \left ( 2,2,-1 \right )

            \begin{aligned} &a(2+1)+b(2-2)+c(-1-0)=0 \\ & \end{aligned}

            \Rightarrow 3 a-c=0                                                    ……….. (3)

We know that line

           \frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}  is parallel to plane  a_{2} x+b_{2} y+c_{2} z+d_{2}=0

If  a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0                                              ………… (4)

Here the plane is parallel to line

            \frac{x-1}{1}=\frac{2 y+1}{2}=\frac{z+1}{-1}

So, a \times 1+b \times 1+c \times-1=0 \\

            \Rightarrow a+b-c=0                                                               ………… (5)

Solving equation (3) and (5) by cross multiplication we have,

            \begin{aligned} &\frac{a}{0-(-1) \times 1}=\frac{b}{1 \times(-1)+3 \times 1}=\frac{c}{3 \times 1-0} \\ & \end{aligned}

            \Rightarrow \frac{a}{1}=\frac{b}{2}=\frac{c}{3}=k(\text { say })

            a=k, b=2 k, c=3 k

Putting the value in equation (2) we get

             \begin{aligned} &k(x+1)+2 k(y-2)+3 k(z-0)=0 \\ & \end{aligned}

             k x+k+2 k y-4 k+3 k z=0

Dividing by k we have,

            x+2 y+3 z-3=0

The required equation is  x+2 y+3 z-3=0

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