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  • Please Solve R.D.Sharma class 12 Chapter 28 The Plane Exercise 28.12 Question 1 Sub Question 2 Maths textbook Solution.

Please Solve R.D.Sharma class 12 Chapter 28 The Plane  Exercise 28.12 Question 1 Sub Question 2 Maths textbook Solution.

Answers (1)

Answer: Therefore, required coordinate is \left ( \frac{17}{3},0,\frac{23}{3} \right )

Hint: Use Formula \frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}

Given:\left ( 5,1,6 \right )and \left ( 3,4,1 \right )

Solution:

Equation of line through \left ( 5,1,6 \right )& \left ( 3,4,1 \right )is
\begin{aligned} &\frac{x-5}{3-5}=\frac{y-1}{4-1}=\frac{z-6}{1-6} \\ &\frac{x-5}{-2}=\frac{y-1}{3}=\frac{z-6}{-5}=\lambda \end{aligned}
∴Another point on the line
\left ( 5-2\lambda ,1+3\lambda ,6-5\lambda \right )
Equation of  ZX-plane is y=0
∴This point lies ony=0
1+3\lambda =0\Rightarrow \lambda =\frac{-1}{3}
The required point
\begin{aligned} &=\left(5+\frac{2}{3}, 1-\frac{3}{3}, 6+\frac{5}{3}\right) \\ &=\left(\frac{17}{3}, 0, \frac{23}{3}\right) \end{aligned}








 

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