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

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 vectorand parallel to the line joining the points with the position vectorand. Also, find the Cartesian equivalent of this equation.

Solution:

We know that, equation of a line passing throughand parallel to vectoris

……. (i)

Here,

and, \vec{b} = line joiningand

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 lineat 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 areandare collinear.

Solution:

Let the given points are A,B.C with position vectorsrespectively.

We know that, equation of a line passing throughandare

If A, B, C are collinear thenmust 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,