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  • Please Solve R.D.Sharma class 12 Chapter 27 Straight Line in Space Exercise 27.2 Question 4 Maths Textbook Solution.

Please Solve R.D.Sharma class 12 Chapter 27 Straight Line in Space Exercise 27.2 Question 4 Maths Textbook Solution.

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Answer : the Cartesian equation which is parallel to given lines will be \frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}

Given: find the Cartesian equation of the  line which passes through the points (-2,4,-5)and parallel to the given line by: \frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6}

Hint: let the Cartesian equation will be :- \frac{x+x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z+z_{1}}{c}

Solution:  \frac{x+x_{1}}{a}=\frac{y-y_{1}}{b}=\frac{z+z_{1}}{c}  [here a,b,c\; are 3,-5,6 and\; x,y,z \; is -2,4,-5]

now, \frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}

so the above line          \frac{x+3}{3}=\frac{y-4}{5}=\frac{z+8}{6} is parallel to \frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6} and passes through the given points.

 

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