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

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

Answers (1)

Answer: Therefore, required coordinate is \left ( 0,\frac{17}{3},\frac{-13}{2} \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: (5,1,6) and (3,4,1)

Solution: Equation of line through (5,1,6)& (3,4,1) 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  YZ-plane is x=0
∴This point lies on x=0
5-2\lambda =0\Rightarrow \lambda =\frac{5}{2}
The required point
\begin{aligned} &=\left(5-\frac{10}{2}, 1+\frac{15}{2}, 6-\frac{25}{2}\right) \\ &=\left(0, \frac{17}{2}, \frac{-13}{2}\right) \end{aligned}

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