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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.13 Question 8 Maths Textbook Solution.

Provide Solution For  R.D.Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.13 Question 8 Maths Textbook Solution.

Answers (1)

Answer: \vec{r}\left ( 5i+2j-3k \right )=17

Hint: first find the coordinates

Given: (3,4,2) and (7,0,6)

Solution: let the equation of the plane be \frac{x}{a}=\frac{y}{b}=\frac{z}{c}=1 --------(1)

Plane is passing through (3,4,2) and (7,0,6)

             \begin{aligned} &\frac{3}{a}=\frac{4}{b}=\frac{2}{c}=1 \\ &\frac{7}{a}=\frac{0}{b}=\frac{6}{c}=1 \end{aligned}

Required plane is perpendicular to

                    \begin{aligned} &2 x-5 y-15=0 \\ &\frac{2}{a}+\frac{-5}{b}+\frac{0}{c}=1 \\ &2 b=5 a \\ &b=2.5 a \\ &\frac{3}{a}+\frac{4}{2.5 a}+\frac{2}{c}=1 \\ &\frac{7}{a}+\frac{6}{2}=1 \end{aligned}

Solving the above equation

                \begin{aligned} &a=3.4=\frac{17}{5}, b=\frac{17}{2} \mathrm{and} \\ &c=\frac{-34}{6}=\frac{-17}{3} \end{aligned}

Substituting the values (1)

                   \begin{aligned} &\frac{x}{\frac{17}{5}}+\frac{y}{\frac{17}{2}}+\frac{z}{\frac{17}{-3}}=1 \\ &\frac{5 x}{17}+\frac{2 y}{17}-\frac{3 z}{17}=1 \\ &5 x+2 y-3 z=17 \end{aligned}

                \begin{aligned} &(x i+y j+2 k) \cdot(5 i+2 j-3 k)=17 \\ &\vec{r}(5 i+2 j-3 k)=17 \end{aligned}

Vector equation of the plane is \begin{aligned} &\vec{r}(5 i+2 j-3 k)=17 \end{aligned}

The line passes through B(1,3,-2) 5\left ( 1 \right )+2\left ( 3 \right )-3\left ( -2 \right )=17

The point B lies on the plane the line;  \vec{r}=i+3 j-2 k+\lambda(i+j+k) lies on the plane\begin{aligned} &\vec{r}(5 i+2 j-3 k)=17 \end{aligned}

 

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