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  • Please Solve R.D. Sharma class 12 Chapter 28 The Plane Exercise 28.13 Question 4 Maths Textbook Solution.

Please Solve R.D. Sharma class 12 Chapter 28 The Plane  Exercise 28.13 Question  4 Maths Textbook Solution.

Answers (1)

Answer:11x-y-3z-35=0

Hint: use simultaneous equation to solve

Given:\frac{x-y}{1}=\frac{y-3}{-4}=\frac{z-2}{5} \text { and } \frac{x-3}{1}=\frac{y+2}{-4}=\frac{2}{5}

Solution: we know that equation of plane passing through \left (x _{1}+y_{1}+ 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....(i)

Since required plane contains lines  \frac{x-y}{1}=\frac{y-3}{-4}=\frac{z-2}{5} \text { and } \frac{x-3}{1}=\frac{y+2}{-4}=\frac{2}{5}  , so we required plane passes through (4, 3, 2) and (3, -8, 0) so equation of required plane is

a\left ( x-4_{1} \right )+b\left ( y-3\right )+c\left ( z-2 \right )=0....(2)

Plane (2) also passes through (3, -3, 0) , so

\begin{aligned} &a(3-4)+b(-2-3)+c(0-2) \\ &-a-5 b-2 z=0 \\ &a+5 b+2 c=0 \end{aligned}....(3)

Now plane (2) is also parallel to the line with direction ratio (1, -4, 5) so

\begin{aligned} &a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=(a)(1)+(6)(-4)+c(5)=0 \\ &a-4 c+5 c=0 \ldots-\ldots-(4) \\ \end{aligned}

Solving equation (3) and (4) by cross multiplication

                  \begin{aligned} &\frac{a}{(5)(5)-(-4)(2)}=\frac{b}{1(2)-(1)(5)}=\frac{c}{(z)(-4)-(1)(5)} \\ &\frac{a}{25+8}=\frac{b}{2-5}=\frac{c}{-5-5} \\ &\frac{a}{33}=\frac{b}{-3}=\frac{c}{-9} \\ \end{aligned}

  Multiplying by 3

                \begin{aligned} &\frac{a}{11}=\frac{b}{-1}=\frac{c}{-3}=x \\ &a=11 x, b=-x, c=3 x \\ \end{aligned}

Put a, b, c in equation (2)

            \begin{aligned} &a(x-4)+b(y-3)+c(z-2)=0 \\ &(11 x)(2-4)+(-2)(y-3)+(-3 x)(z-2)=0 \\ &11 x_{2}-44 x-\lambda y+3 x z+6 x=0 \\ &11 x-x y-3 x_{2}-35 x=0 \end{aligned}

Dividing by  x11x-y-3z-35=0

So equation of required plane is 11x-y-3z-35=0

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