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  • Need Solution for R.D.Sharma Maths Class 12 Chapter 28 The Plane Exercise 28.13 Question 16 Sub Question 1 Maths Textbook Solution.

Need Solution for R.D.Sharma Maths Class 12 Chapter 28 The Plane  Exercise 28.13 Question 16 Sub Question 1 Maths Textbook Solution.

Answers (1)

Answer:  x=\pm \sqrt{2}

Hint: use vector cross product

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

Solution: we know that the lines

                 \begin{aligned} &\frac{x-x_{1}}{l_{1}}=\frac{y-y_{1}}{m_{1}}=\frac{z-z_{1}}{n_{1}}\\ &\text { and }\\ &\frac{x-x_{2}}{l_{2}}=\frac{y-y_{2}}{m_{2}}=\frac{z-z_{2}}{n_{2}} \text { are } \end{aligned}

                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 \cdot x_{2}=3 . y_{1}=2 \cdot y_{2} \\ &=2.21,=3.2=1 \\ &l_{1}=1 . l_{2}=m_{1}=2 m_{2} \\ &=\lambda^{2} n_{1}=\lambda^{2} n_{2}=2 \\ \end{aligned}

                 \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 \\ &\left(\begin{array}{ccc} 2 & 0 & -2 \\ 2 & 2 & \lambda^{2} \\ 1 & \lambda^{2} & 2 \end{array}\right)=0 \end{aligned}

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