Three Dimensional Geometry · Class 12 · CBSE · Science Mathematics
The vector equation of the line passing through the points A(1, -2, 3) and B(3, 1, -1) is:
Practice question from Three Dimensional Geometry (Class 12 · CBSE · Science Mathematics), with the correct answer and a step-by-step solution.
easy · MCQ · Q-C8539B
The vector equation of the line passing through the points A(1, -2, 3) and B(3, 1, -1) is:
- A.r = (i - 2j + 3k) + λ(2i + 3j - 4k)Correct answer
- B.r = (i - 2j + 3k) + λ(3i + 2j - 4k)
- C.r = (i + 2j + 3k) + λ(2i + 3j + 4k)
- D.r = (3i + j - k) + λ(i - 2j + 3k)
Solution
The direction vector of the line is . Hence, the required equation is .
More from Three Dimensional Geometry
- The direction cosines of a line whose direction ratios are 2, -1 and 2 are:
- The shortest distance between the skew lines r = (1, 0, 1) + λ(1, 1, 0) and r = (0, 1, 0) + μ(1, 0, 1) is:
- Find the equation of the plane passing through the point (1, 2, -1) and perpendicular to the line whose direction ratios are 2, -1 and 2.
- The acute angle between the planes 2x - y + 2z = 5 and x + 2y + 2z = 7 has cosine equal to:
Practise more Mathematics
Sign up free for adaptive practice on this chapter, with instant feedback and hints on every question.
Start practising freeFree to start · No credit card needed