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VOOZH | about |
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VOOZH | about |
In Class 12 Mathematics, Chapter 28 focuses on the "Straight Line in Space". This chapter delves into the geometric properties of the lines in three-dimensional space and their equations. Understanding these concepts is crucial for solving complex problems in coordinate geometry and for applications in the various fields of science and engineering.
A straight line in space can be defined in several ways such as through parametric equations or vector equations. In three-dimensional space, a line can be represented using the point through which it passes and a direction vector. The parametric form of the line equation in space is often used to describe its position and orientation precisely. It provides a way to represent lines when given a point and a direction vector which is useful for solving problems involving intersections and distances between the lines.
Solution:
The equation of line can be re-written as,
The direction ratios of the line parallel to line are proportional to 2, 2/3, -3.
Equation of the required line passing through the point ( -1, 2, 1) having direction ratios proportional to (2, 2/3, -3) is,
=>
Solution:
The given line is parallel to the vector
And the required line is also parallel to the given line.
So, the required line is parallel to the vector
Hence, the equation of the required line passing through the point (2,-1, 3) and parallel to the vector is,
=>
Solution:
Let,
Since the required line is perpendicular to the lines parallel to the vectors and , it is also parallel to the vector
Now,
=
=
=
Thus, the direction ratios of the required line are proportional to 2, -7, 4.
The equation of the required line passing through the point (2, 1, 3) and having direction ratios proportional to 2, -7, 4 is
=>
Solution:
The required line is perpendicular to the lines parallel to the vectors and .
So, the required line is parallel to the vector,
=
=
Equation of the required line passing through the point and parallel to is,
=>
Solution:
The direction ratios of the line joining the points (4, 3, 2), (1, -1, 0) and (1, 2, -1), (2, 1, 1) are -3, -4, -2 and 1, -1, 2 respectively.
Let,
Since the required line is perpendicular to the lines parallel to the vectors and , it is parallel to the vector
Now,
=
=
So, the direction ratios of the required line are proportional to -10, 4, 7.
The equation of the required line passing through the point (1,-1, 1) and having direction ratios proportional to -10, 4, 7 is
=>
Solution:
We have,
Let,
Since the required line is perpendicular to the lines parallel to the vectors and , it is parallel to the vector
Now,
=
=
The direction ratios of the required line are proportional to 24, 61, 112.
The equation of the required line passing through the point (1, 2,-4) and having direction ratios proportional to 24, 61, 112 is,
=>
Solution:
The direction ratios of the line are proportional to 7, -5, 1 respectively.
And the direction ratios of the line are proportional to 1, 2, 3 respectively.
Let,
Now,
= 7 - 10 + 3
= 0
So,
Therefore, the given lines are perpendicular to each other.
Hence proved.
Solution:
The equation of the line 6x − 2 = 3y + 1 = 2z − 2 can be re-written as
=>
Since the required line is parallel to the given line, the direction ratios of the required line are proportional to 1,2,3.
The vector equation of the required line passing through the point (2,-1,-1) and having direction ratios proportional to 1,2,3 is,
=>
Solution:
The equations of the given lines are,
Since the given lines are perpendicular to each other, we have
=> -3 (3λ) + 2λ (1) + 2 (-5) = 0
=> -9λ + 2λ - 10 = 0
=> -7λ = 10
=> λ = -10/7
Therefore, the value of λ is -10/7.
Solution:
The direction ratios of AB and CD are proportional to 3, 3, 4 and 6, 6, 8, respectively.
Let θ be the angle between AB and CD. Then,
=
= 1
Now cos θ = 1
=> θ = 0°
Therefore, the angle between AB and CD is 0°.
Solution:
The equation of the given line can be re-written as,
The equation of the given line can be re-written as,
Since the given lines are perpendicular to each other, we have
=> (5λ + 2) (1) - 5 (2λ) + 1 (3) = 0
=> 5λ + 2 - 10λ + 3 = 0
=> -5λ = -5
=> λ = 1
Therefore, the value of λ is 1.
Solution:
The equation of the given line is,
The given equation can be re-written as
This line passes through the point (-2, 7/2, 5) and has direction ratios proportional to 2, 3, −6.
So, its direction cosines are
Or,
The required line passes through the point having position vector .
And also it is parallel to the vector .
So, its vector equation is,
=>
Read More:
Understanding the straight line in space is fundamental for the mastering three-dimensional geometry. It helps in visualizing and solving problems related to the lines and their interactions in space. Mastery of this topic provides the strong foundation for the more advanced topics in the geometry and spatial analysis.
