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# Class 12 RD Sharma Solutions – Chapter 28 The Straight Line in Space – Exercise 28.1 | Set 2

• Last Updated : 04 May, 2021

### Question 11. Find the direction cosines of the line . Also, reduce it to vector form

Solution:

Given:  x = -2λ + 4, y = 6λ, z = -3λ + 1

So, Direction ratios of the line are = -2, 6, -3

Direction cosines of the lines are, ### Question 12. The Cartesian equations of a line are x = ay + b, z = cy + d. Find its direction ratios and reduce it to vector form.

Solution:

x = ay + b

z = cy + d So, DR’s of line are (a, 1, c)

From above equation, we can write

x = aλ + b

y = λ

z = cλ + d

So vector equation of line is ### Question 13. Find the vector equation of a line passing through the point with position vector and parallel to the line joining the points with the position vector and . Also, find the Cartesian equivalent of this equation.

Solution:

We know that, equation of a line passing through and parallel to vector is ……. (i)

Here, and, \vec{b} = line joining and  Equation of the line is For Cartesian form of equation put  Equating coefficients of x = 1 + λ, y = -2 + 2λ, z = -3 – 2λ ### Question 14. Find the points on the line at a distance of 5 units from the points P(1, 3, 3).

Solution:

Given, line is General points Q on line is (3λ – 2, 2λ -1), 2λ + 3)

Distance of points P from Q = PQ = (5)2 = (3λ -3)2 + (2λ – 4) + (2λ)2

25 = 9λ2 + 9 – 18λ + 4λ2 + 16 – 16λ + 4λ2

17λ2 – 34λ = 0

17λ (λ – 2) = 0

λ = 0 or 2

So, points on the line are (3(0) – 2, 2(0) – 1, 2(0) + 3)

(3(2) – 2, 2(2) – 1, 2(2) + 3)

= (-2, -1, 3), (4, 3, 7)

### Question 15. Show that the points whose position vectors are and are collinear.

Solution:

Let the given points are A,B.C with position vectors respectively. We know that, equation of a line passing through and are If A, B, C are collinear then must satisfy equation (i) Equation the coefficients of -2 + 3 = 7 , λ = 3

3 – λ = 0 , λ = 3

3λ = -1 , λ = Since, value of λ are not equal, so,

Given points are collinear.

### Question 16. Find the Cartesian and vector equations of a line which passes through the points (1, 2, 3) and is parallel to the line Solution:

We know that, equation of a line passing through a point (x1, y1, z1) and having direction ratios proportional to a, b, c is Here,

(x1, y1, z1) = (1, 2, 3) and

Given line  Its parallel to the required line, so

a = μ , b = 7μ, c = μ

So, equation of required line using equation (i) is, Multiplying the denominators by 2 x = -2λ + 1, y = 14λ + 2, z = 3λ + 3

So, vector form of the equation of required line, ### Question 17. The Cartesian equations of a line are 3x + 1 = 6y – 2 = 1 – z. Find the fixed point through which it passes, its direction ratios, and also its vector equation.

Solution:

Given equation of line is,

3x + 1 = 6y -2 = 1 – z

Dividing all by 6 Comparing it with equation of line equation of line passing through (x,1 y1, z1) and the direction ratios a, b, c, a = 2, b = 1, -6

So, direction ratios of the line are -2, 1, -6

From equation (i) So, vector equation of the given line is, My Personal Notes arrow_drop_up
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