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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.13 Question 13 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 28 The Plane  Exercise 28.13 Question 13 Maths Textbook Solution.

Answers (1)

Answer: x-y+z-1=0

Hint: first find the value of coordinates, a, b and c

Given: (3, 2, 0) and contain the line \frac{x-3}{1}=\frac{y-b}{5}=\frac{z-4}{7}

Solution: given that a plane is passes through the point (3, 2, 0) so equation will be;

              \begin{aligned} &a(x-3)+b(y-2)+c(z-0)=0 \\ \end{aligned}   -----(i)

                \begin{aligned} &a(x-3)+b(y-2)+c z=0 \end{aligned} --------------(ii)

Plane also contains the line

                \frac{x-3}{1}=\frac{y-b}{5}=\frac{z-4}{7}

So it passes through the point (3, 2, 10)

         \begin{aligned} &a(3-3)+b(6-2)+c(4-0)=0 \\ &b+4 c=0 \\ \end{aligned}        ------------------(2)

Also plane will be parallel to

           \begin{aligned} &a(1)+b(5)+c(4)=0 \\ &a+5 b+4 c=0------(3) \\ \end{aligned}

Solving (2) and (3) by cross multiplication,

                 \begin{aligned} &\frac{a}{16-20}=\frac{b}{4-0}=\frac{c}{0-4}=k \\ &\frac{a}{4}=\frac{b}{4}=-\frac{c}{4}=k \\ &a=-k, b=k, c=-k \\ \end{aligned}

Put a=-k,b=k,c=-kin equation (ii) we get

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

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