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  • Please Solve R.D.Sharma class 12 Chapter 28 The Plane Exercise 28.13 Question 2 Maths textbook Solution.

Please Solve R.D.Sharma class 12 Chapter 28 The Plane  Exercise 28.13 Question  2 Maths textbook Solution.

Answers (1)

Answer: Proud L.H.S =R.H.S

Hint: use vector cross product

               Given:\frac{x+1}{-3}=\frac{y-3}{2}=\frac{z+2}{1} \text { and } \frac{x}{1}=\frac{y-7}{-3}=\frac{8+7}{2}

Solution :two lines

                  \begin{aligned} &\frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}} \text { and } \frac{x-x_{2}}{a_{2}} \\ &=\frac{y-y_{2}}{b_{2}}=\frac{z-z_{2}}{c_{2}} \end{aligned}are coplanar if\begin{aligned} \left(\begin{array}{ccc} x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right) \\ \end{aligned}

 Here,           \begin{aligned} &x_{1}=-1, y_{1}=-1, z_{1}=-2 \\ \end{aligned}

                  \begin{aligned} &{x_{2}}=0 \quad y_{2}=7 \quad z_{1}=-7 \\ &a_{1}=-3 \quad b_{1}=2 \quad c_{1}=1 \\ &a_{2}=1 \quad b_{2}=-3 \quad c_{2}=2 \\ \end{aligned}

              \begin{aligned} &\left(\begin{array}{ccc} 0-(-1) & 7-3 & -7-(-2) \\ -3 & 2 & 1 \\ 1 & -3 & 2 \end{array}\right) \\ &=\left(\begin{array}{ccc} 1 & 4 & -5 \\ -3 & 2 & 1 \\ 1 & -3 & 2 \end{array}\right) \\ &=1(7)-4(-7)-5(-7) \\ &=0 \end{aligned}

The given lines are coplarer . equation of the plane containing the given line is

                 \begin{aligned} &\left(\begin{array}{ccc} x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \end{array}\right)=0 \\ &\left(\begin{array}{ccc} x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ -3 & 2 & 1 \\ 1 & -3 & 2 \end{array}\right)=0 \\ &(x+1)(4+3)-(y-3)(-6+1)+(z+2)(9-0)=0 \\ &7 x+7 y+7 z=0 \\ &x+y+z=0 \end{aligned}

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