RD Sharma Solutions for Class 12 Maths Chapter 28 Straight Line in Space Exercise 28.1 is provided here. The BYJU’S team has prepared solutions which help students attain good marks in Maths. From the exam point of view, the solutions are solved in a simple and structured manner, which will help students to secure excellent scores in the board exams. RD Sharma Solutions PDF can be downloaded from the links given below.
RD Sharma Solution for Class 12 Maths Chapter 28 Exercise 1
Access Answers to RD Sharma Solution Class 12 Maths Chapter 28 Exercise 1
Q1.
Solution:
Q2.
Solution:
It is given that,
The direction ratios of the line are: (3+1, 4-0, 6-2) = (4, 4, 4)
Since the given line passes through (-1, 0, 2)
We know that the vector equation of a line is given as
Q3.
Solution:
Let us consider,
The vector equation of the line passing through a fixed point vector a and parallel to vector b is given as
Q4.
Solution:
Let us consider,
The vector equation of the line passing through a fixed point vector a and parallel to vector b is given as
Hence,
The Cartesian form of the equation of the line is
Q5.
Solution:
It is given that ABCD is a parallelogram.
Let us consider AC and BD bisects each other at point O.
So,
Let us consider position vector of point O and B are represented by
So,
The equation of the line BD is the line passing through O and B is given by
Now, let us compare the coefficients of vector i, j, R
x = 2 – λ, y = -3 – 13λ, z = 4 – 17λ
By equating to λ, we get,
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