Open in App
Not now

# 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 = -2Î» + 4, y = 6Î», z = -3Î» + 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 = aÎ» + b

y = Î»

z = cÎ» + 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 + Î», y = -2 + 2Î», z = -3 – 2Î»

### 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 (3Î» – 2, 2Î» -1), 2Î» + 3)

Distance of points P from Q =

PQ =

(5)2 = (3Î» -3)2 + (2Î» – 4) + (2Î»)2

25 = 9Î»2 + 9 – 18Î» + 4Î»2 + 16 – 16Î» + 4Î»2

17Î»2 – 34Î» = 0

17Î» (Î» – 2) = 0

Î» = 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 , Î» = 3

3 – Î» = 0 , Î» = 3

3Î» = -1 , Î» =

Since, value of Î» 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 = Î¼ , b = 7Î¼, c =Î¼

So, equation of required line using equation (i) is,

Multiplying the denominators by 2

x = -2Î» + 1, y = 14Î» + 2, z = 3Î» + 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
Related Articles