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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.13 Question 14 Sub Question 1 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 28 The Plane  Exercise 28.13 Question 14 Sub Question 1 Maths Textbook Solution.

Answers (1)

Answer: x-2y+z=0

Hint: use vector cross product

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

Solution: we know that lines

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

Are coplanar if

                \begin{aligned} &\left(\begin{array}{ccc} \lambda-\lambda_{1} & y-y_{1} & z-z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{array}\right)=0 \\ &\text { here }: \lambda_{1}=-3, \lambda_{2}=-1, y_{1}=1, y_{2}=2, z_{1}=5, z_{2}=5 \\ &l_{1}=-3, l_{2}=-1, m_{1}=1, m_{2}=2, n_{1}=5, \ln _{2}=5 \\ &\left(\begin{array}{ccc} \lambda-\lambda_{1} & y-y_{1} & z-z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{array}\right) \\ &\left(\begin{array}{lll} 2 & 1 & 0 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{array}\right) \\ &=2(5-10)-1(-15--5)+10(-6--1) \\ &=2(-5)-1(-10)=-10+10 \\ &=0 \end{aligned}

So the given lines are coplanar , the equation of plane contains line is

 

                 \begin{aligned} &\left(\begin{array}{ccc} \lambda+8 & y-1 & 3-5 \\ -3 & 1 & 5 \\ -1 & 2 & 5 \end{array}\right)=0 \\ &(\lambda+3)(5-10)-(y-1)(-15-(-5))+(2-5)(-6-(-1)=0 \\ &-5 \lambda-15+10 y-10-5 z+25=0 \\ &-5 \lambda+10 y-5 z=0 \end{aligned}

Divided by -5

                        x-2y+z=0
 

Posted by

infoexpert21

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