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RD Sharma Class 12 Chapter 28 Solutions (Straight Line In Space)

RD Sharma Class 12 Maths Solutions Chapter 28 ‘Straight Line in Space’ teaches you all the concepts about a straight line in space. In this chapter, you’ll learn about vector and cartesian equations of a line, associated theorems, reduction of the cartesian form of a line to vector form and vice-versa, vector to the cartesian, angle between two lines, the intersection of two lines, an algorithm for vector form, perpendicular distance of a line from a point, the shortest distance between two straight lines, skew lines, line of shortest distance, the shortest distance between two skew lines, the shortest distance between two parallel lines, etc.

Straight Line in Space Chapter 28 has a total of 5 exercises and 67 questions. You’ll answer questions like finding the vector equation of a line that satisfies the specific conditions and reducing them to a cartesian form, finding the line’s cartesian equation where the line will satisfy the specific condition and then reducing it to vector, checking collinearity of three points, or finding a point on a line, on finding the angle between two lines, equation of a line falling parallel to a specific line and passing through a given point, etc.

Instasolv provides best RD Sharma Class 12 Maths Solutions Chapter 28 ‘Straight Line in Space’. These solutions take every single concept that has been taught in the chapter. These solutions are available free of cost. Helping you to prepare well for your CBSE and all other exams like NEET and JEE, Instasolv is known for its years of experience and portfolio in helping students achieve their goals.

Topics Discussed in RD Sharma Class 12 Maths Solutions Chapter 28 – Straight Line in Space

Theorem on Vector and Cartesian Equations of a Line

  1. A straight line passes through a fixed point having position vector a and is found parallel to a given vector b. The vector equation will be vector r = vector a ∂vector b where ∂ is a scalar.
  2. A straight line passes through a specific point (x1, y1, z1). It has direction ratios found to be proportional with a, b, c. Then the cartesian equation of the given straight line will be 

             (x-x1)/a = (y-y1)/b = (z-z1)/c

  1. A line passes through points having position vectors a and b. The vector equation will be vector r = vector a + ∂(vector b – vector a)
  2. A line passes through the points(x1, y1, z1) and (x2, y2, z2). Then its cartesian equation will be 

            (x-x1)/(x2-x1) = (y-y1)/(y2-y1) = (z-z1)/(z2-z1)

The intersection of two lines

Here is an algorithm that you can use to check if the given lines meet or not and if they meet, a method to find the intersection point.

Algorithm

Consider the two lines as 

(x-x1)/a1 = (y-y1)/b1 = (z-z1)/c1    …(1)

(x-x2)/a2 = (y-y2)/b2 = (z-z2)/c2    …(2)

  • Mention the general points’ coordinates on equation 1 and 2

            Let’s call them ∂ and µ respectively.

      Which gives (a1∂ +x1, b1∂ + y1, c1∂+z1) and (a2µ +x2, b2µ + y2, c2µ + z2)

  • If the two lines 1 and 2 meet, they will have a point in common.

         Therefore, (a1∂ + x1) = (a2µ + x2)

                           (b1∂ + y1) = (b2µ + y2)

                           (c1∂ + z1) = (c2µ + z2)

  1. You will be required to solve two equations out of three in ∂ and µ provided in the above step. If the values of the two variables satisfy the third equation then the above lines i.e. 1 and 2 will meet or in other words intersect. However, if the value of ∂ and µ do not satisfy the third equation, then the lines 1 and 2 do not meet or in other words do not intersect.
  2. To find the coordinates of the intersection point, but the value of ∂ and/or µ in general point(s)’ coordinates that you calculated in the first step.

Discussion of Exercises in RD Sharma Class 12 Maths Solutions Chapter 28 – Straight Line in Space

  1. Exercise 28.1 discusses methods to find vector and cartesian equations of a given line, finding the vector equation of the line, finding the equation of the line in both the cartesian and vector form, finding the points of a given line, etc.
  2. Exercise 28.2 asks questions based on the angle between the two lines, the equation of a line that is parallel to a specific line and passes through a given point, or finding the equation of a line that passes through a specific point and is perpendicular to the two lines, etc.
  3. Exercise 28.3 talks about the intersection of two lines, the algorithm for vector form, and more.
  4. Exercise 28.4 discusses the perpendicular distance of a given line from a specific point, vector form, cartesian form, etc. 
  5. Exercise 28.5 discusses the shortest distance between two straight lines, two skew lines, two parallel lines, etc.

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Access RD Sharma Solutions for Class 12 Maths Solutions Chapter 28 ‘Straight Line in Space’ any time from anywhere and make your preparations easier. Our stepwise solutions will make you understand the complete methodology of getting the right answers in this chapter. Our experts are well-aware of the CBSE class 12 syllabus and have framed the solutions accordingly. These solutions are error-free and easy.