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  • Explain Solution R D Sharma Class 12 Chapter 28 The Plane Exercise 28 Point 13 Question 18 Sub Question 1 Maths Textbook Solution

Explain Solution R.D.Sharma Class 12 Chapter 28 The Plane  Exercise 28.13 Question 18 Sub Question 1 Maths Textbook Solution.

Answers (1)

Answer: y+z+1=0

Hint: use vector cross product

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

Solution: we know that the lines \frac{x-1}{2}=\frac{y+1}{1}=\frac{z}{2}  and  \frac{x+1}{5}=\frac{y+1}{2}=\frac{z}{k}  are coplanar if

              \left(\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{array}\right)=0

              here

              \begin{aligned} &x_{1}=1, x_{2}=-1, y_{1}=-1, y_{2}=-1, z_{1}=0, z_{2}=0\\ &I_{1}=2, l_{2}=5, m_{1}=k, m_{2}=2, n_{1}=2, n_{2}=k\\ &\left(\begin{array}{ccc} x-x_{1} & y-y_{1} & z-z_{1} \\ l_{1} & m_{1} & n_{1} \\ l_{2} & m_{2} & n_{2} \end{array}\right)=0\\ &\left(\begin{array}{ccc} -2 & 0 & 0 \\ 1 & k & 2 \\ 5 & 2 & k \end{array}\right)=0\\ &-2\left(k^{2}-4\right)=0\\ &k=\pm 2 \end{aligned}

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