# Class 12 RD Sharma Solutions – Chapter 28 The Straight Line in Space – Exercise 28.5

• Last Updated : 16 Jun, 2021

### (i)  and

Solution:

As we know that the shortest distance between the lines and is:

D=

Now,

= 36 + 225 + 9

= 270

= √270

On substituting the values in the formula, we have

SD = 270/√270

= √270

Shortest distance between the given pair of lines is 3√30 units.

### (ii) and

Solution:

As we know that the shortest distance between the lines  and  is:

D=

Now,

= – 16 × 32

= – 512

On substituting the values in the formula, we have

SD =

Shortest distance between the given pair of lines is units.

### (iii)  and

Solution:

As we know that the shortest distance between the lines and  is:

D=

Now,

= 1

On substituting the values in the formula, we have

SD =

Shortest distance between the given pair of lines is 1/√6 units.

### (iv)  and

Solution:

Above equations can be re-written as:

and,

As we know that the shortest distance between the lines

and  is:

D =

= 9/3√2

Shortest distance is 3/√2 units.

### (v)  and

Solution:

The given equations can be written as:

\and

As we know that the shortest distance between the lines and  is:

D=

Now,

= 15

= 3√2

Thus, distance between the lines is  units.

### (vi)  and

Solution:

As we know that the shortest distance between the lines and is:

D =

Now,

= 3√2

Substituting the values in the formula, we have

The distance between the lines is units.

### (vii)  and

Solution:

As we know that the shortest distance between the lines and is:

D=

Now,

= 10

Substituting the values in the formula, we have:

The distance between the lines is 10/√59 units.

### (viii)  and

Solution:

As we know that the shortest distance between the lines and is:

D=

Now,

= 1176

= 84

Substituting the values in the formula, we have:

The distance between the lines is 1176/84 = 14 units.

### (i) and

Solution:

The given lines can be written as:

and

= –1

= √6

On substituting the values in the formula, we have:

SD = 1/√6 units.

### (ii)  and

Solution:

The given equations can also be written as:

and \

As we know that the shortest distance between the lines and is:

D=

=  3

SD = 3/√59 units.

### (iii)  and

Solution:

The given equations can be re-written as:

and

= √29

= 8

SD = 8/√29 units.

### (iv)  and

Solution:

The given equations can be re-written as:

and

SD = 58/√29 units.

### (i)  and

Solution:

As we know that the shortest distance between the lines and is:

D=

=  –1

= √14

⇒ SD = 1/√14 units ≠ 0

Hence the given pair of lines does not intersect.

### (ii)  and

Solution:

As we know that the shortest distance between the lines and is:

D=

= 0

= √94

⇒ SD = 0/√94 units = 0

Hence the given pair of lines are intersecting.

### (iii)  and

Solution:

Given lines can be re-written as:

and

As we know that the shortest distance between the lines and is:

D=

= −9

= √195

⇒ SD = 9/√195 units ≠ 0

Hence the given pair of lines does not intersect.

### (iv)  and

Solution:

Given lines can be re-written as:

and

As we know that the shortest distance between the lines and is:

D=

= 282

⇒ SD = 282/√3 units ≠ 0

Hence the given pair of lines does not intersect.

### (i) and

Solution:

The second given line can be re-written as:

As we know that the shortest distance between the lines and is:

D=

⇒ SD =  units.

### (ii)  and

Solution:

The second given line can be re-written as:

As we know that the shortest distance between the lines and is:

D=

⇒

= √11

⇒ SD = √11/√6 units.

### (i) (0, 0, 0) and (1, 0, 2)       (ii) (1, 3, 0) and (0, 3, 0)

Solution:

Equation of the line passing through the vertices (0, 0, 0) and (1, 0, 2) is given by:

Similarly, the equation of the line passing through the vertices (1, 3, 0) and (0, 3, 0):

As we know that the shortest distance between the lines   and  is:

D=

= −6

= 2

⇒ SD = |-6/2| = 3 units.

### Question 6. Write the vector equations of the following lines and hence find the shortest distance between them:

Solution:

The given equations can be written as:

and

As we know that the shortest distance between the lines and is:

D=

⇒

\vec{|b|}= 7

⇒ SD = √293/7 units.

### (i) and

Solution:

As we know that the shortest distance between the lines and is:

D=

Now,

= 3√2

⇒ SD = 3/√2 units.

### (ii)

Solution:

As we know that the shortest distance between the lines and is:

D=

Now,

= √116

⇒ SD = 2√29 units.

### (iii) and

Solution:

As we know that the shortest distance between the lines and is:

D=

Now,

= √171

⇒ SD = 3√19 units.

### (iv) and

Solution:

As we know that the shortest distance between the lines and is:

D=

Now,

(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=108

|\vec{b_1}×\vec{b_2}|=\sqrt{(-9)^2+(3)^2+(9)^2}

= 12

⇒ SD = 9 units.

### Question 8. Find the distance between the lines: and

Solution:

As we know that the shortest distance between the lines and is:

D=

⇒

= √293

⇒ SD = √293/7 units.

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