RD Sharma Solutions for Class 12 Maths Chapter 28 Straight Line in Space – PDF Download

RD Sharma Solutions for Class 12 Maths Chapter 28 – Straight Line in Space are available here to support students in mastering one of the key topics in 3D Geometry. These step-by-step RD Sharma solutions serve as a reliable reference while practicing the exercise questions from the RD Sharma Class 12 textbook.

Crafted by subject matter experts, these solutions aim to enhance conceptual understanding and boost problem-solving accuracy. They are particularly helpful for students preparing for both Class 12 CBSE board exams and competitive entrance exams like JEE Main and JEE Advanced.

Designed in line with the latest CBSE syllabus and marking scheme, RD Sharma Class 12 Chapter 28 solutions provide students with clear explanations and exam-oriented strategies, helping them secure high marks with confidence.

Download RD Sharma Solutions for Class 12 Maths Chapter 28 Straight Line in Space PDF

RD Sharma Class 12 Maths Solutions Chapter 28 Straight Line in Space cover all the questions from the textbook, crafted by expert Mathematics teachers at Infinity Learn. Download our free PDF of Chapter 28 Straight Line in Space RD Sharma Solutions for Class 12 to boost your performance in board exams and competitive exams.

Access answers to RD Sharma Solutions For Class 12 Chapter 28 Straight Line in Space

Q1. Find the vector equation of the line passing through the point (2, 3, -1) and parallel to the vector 3𝑖̂ + 4𝑗̂ + 5𝑘̂.

Solution:

Vector form of a line: 
r = a + λb

• a = 2𝑖̂ + 3𝑗̂ - 1𝑘̂
• b = 3𝑖̂ + 4𝑗̂ + 5𝑘̂

Therefore, the equation is:
r = (2𝑖̂ + 3𝑗̂ - 1𝑘̂) + λ(3𝑖̂ + 4𝑗̂ + 5𝑘̂)

Q2. Determine whether the lines are identical:

r₁ = 𝑖̂ + 2𝑗̂ + 3𝑘̂ + λ(2𝑖̂ + 3𝑗̂ + 4𝑘̂)

r₂ = 3𝑖̂ + 5𝑗̂ + 6𝑘̂ + μ(2𝑖̂ + 3𝑗̂ + 4𝑘̂)

Solution:

• Both lines have same direction vector: 2𝑖̂ + 3𝑗̂ + 4𝑘̂
• Check difference of position vectors: (3 - 1)𝑖̂ + (5 - 2)𝑗̂ + (6 - 3)𝑘̂ = 2𝑖̂ + 3𝑗̂ + 3𝑘̂
• This is not a multiple of the direction vector.

Conclusion: The lines are parallel but not identical.

Q3. Find the Cartesian equation of the line through (1, 2, 3) parallel to vector 4𝑖̂ + 5𝑗̂ + 6𝑘̂.

Solution:

Using symmetric form:
(x - 1)/4 = (y - 2)/5 = (z - 3)/6

Q4. Find the distance between two skew lines:

L₁: r₁ = 𝑖̂ + 2𝑗̂ + 3𝑘̂ + λ(𝑖̂ - 𝑘̂)

L₂: r₂ = 2𝑖̂ + 3𝑗̂ + 4𝑘̂ + μ(𝑗̂ + 𝑘̂)

Solution:

• Direction vectors:
  • b₁ = 𝑖̂ - 𝑘̂
  • b₂ = 𝑗̂ + 𝑘̂
• d = (2 - 1)𝑖̂ + (3 - 2)𝑗̂ + (4 - 3)𝑘̂ = 𝑖̂ + 𝑗̂ + 𝑘̂
• Cross product: b₁ × b₂ = 𝑖̂ - 𝑗̂ + 𝑘̂
• Distance = |(d • n)| / |n| = |1 - 1 + 1| / √3 = 1 / √3

Answer:1/√3 units