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  • Need Solution for R.D. Sharma Maths Class 12 Chapter 27 Straight Line in Space Exercise 27.2 Question 17 Maths Textbook Solution.

Need Solution for R.D. Sharma Maths Class 12 Chapter 27 Straight Line in Space  Exercise 27.2 Question 17  Maths Textbook Solution.

Answers (1)

Answer : The equation will be  \frac{x-1}{10}=\frac{y+1}{y}=\frac{3}{7}

Given: Find the equation of the line passing through the point(1, -1, 1) and perpendicular to the lines joining the points (4,3,2)(1,-1,0) and (1,2,-1), (2,1,1

Hint: equation will be in the form of

          \frac{x-x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z-z_{1}}{c}

Solution:

                Direction ratio of a_{1}b_{1}c_{1}=3,4,2

                Direction ratio of  a_{2}b_{2}c_{2}=\left ( -1,1,-2 \right )

Now

               \begin{aligned} &3 a+4 b+2 c=0 \\ &-a+b-2 c=0 \\ &\frac{a}{10}=\frac{b}{y}=\frac{c}{7} \\ &\frac{x-1}{10}=\frac{y+1}{4}=\frac{3-0}{7} \end{aligned}

The equation will be\frac{x-1}{10}=\frac{y+1}{y}=\frac{3}{7}
 

 

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