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# Class 12 RD Sharma Solutions – Chapter 28 The Straight Line in Space – Exercise 28.2 | Set 2

• Last Updated : 14 Jul, 2021

### Question 13. Find the equations of the line passing through the point (−1, 2, 1) and parallel to the line .

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, => ### Question 14. Find the equation of the line passing through the point (2, −1, 3) and parallel to the line .

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,

=> ### Question 15. Find the equations of the line passing through the point (2, 1, 3) and perpendicular to the lines and .

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

=> ### Question 16. Find the equation of the line passing through the point and perpendicular to the lines and .

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,

=> ### Question 17. Find the equation of the line passing through the point (1, −1, 1) and perpendicular to the lines joining the points (4, 3, 2), (1, −1, 0), and (1, 2, −1), (2, 1, 1).

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

=> ### Question 18. Determine the equations of the line passing through the point (1, 2, −4) and perpendicular to the two lines and .

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,

=> ### Question 19. Show that the lines and are perpendicular to each other.

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.

### Question 20. Find the vector equation of the line passing through the point (2, −1, −1) which is parallel to the line 6x − 2 = 3y + 1 = 2z − 2.

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,

=> ### Question 21. If the lines and are perpendicular, find the value of λ.

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.

### Question 22. If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.

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°.

### Question 23. Find the value of λ so that the following lines are perpendicular to each other. 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.

### Question 24. Find the direction cosines of the line . Also, find the vector equation of the line through the point A(−1, 2, 3) and parallel to the given line.

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,

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