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  • Please Solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28.11 Question 24 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 28 The Plane Exercise 28.11 Question 24 Maths Textbook Solution.

Answers (1)

Answer:  -2

Hint: Use formula  \frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}

Given:  \frac{x-2}{6}=\frac{y-1}{\lambda}=\frac{z+5}{-4} \text { and } 3 x-y-2 z=7

Solution: Here given midline

            \frac{x-2}{6}=\frac{y-1}{\lambda}=\frac{z+5}{4}  is perpendicular to plane  3 x-y-2 z=7

We know that line

            \frac{x-x_{1}}{a_{1}}=\frac{y-y_{1}}{b_{1}}=\frac{z-z_{1}}{c_{1}}  is perpendicular to plane  a_{2} x+b_{2} y+c_{2} z+d_{2}=0  if

            \frac{a_{1}}{a_{2}}+\frac{b_{1}}{b_{2}}+\frac{c_{1}}{c_{2}}=0

So, normal vector of plane is parallel to line.

So, direction ratios of normal to plane are proportional to the direction of line.

            Here,  \frac{6}{3}=\frac{\lambda}{-1}=\frac{-4}{-2}

By cross multiplying the last two we have

            -2 \lambda=4 \Rightarrow \lambda=-2

 

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