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

### (i)  and

Solution:

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

D=

Now,

= 36 + 225 + 9

= 270

= âˆš270

On substituting the values in the formula, we have

SD = 270/âˆš270

= âˆš270

Shortest distance between the given pair of lines is 3âˆš30 units.

### (ii) and

Solution:

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

D=

Now,

= â€“ 16 Ã— 32

= â€“ 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/âˆš6 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 =

= 9/3âˆš2

Shortest distance is 3/âˆš2 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

= 3âˆš2

Thus, distance between the lines is  units.

### (vi)  and

Solution:

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

D =

Now,

= 3âˆš2

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/âˆš59 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

= â€“1

= âˆš6

On substituting the values in the formula, we have:

SD = 1/âˆš6 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/âˆš59 units.

### (iii)  and

Solution:

The given equations can be re-written as:

and

= âˆš29

= 8

SD = 8/âˆš29 units.

### (iv)  and

Solution:

The given equations can be re-written as:

and

SD = 58/âˆš29 units.

### (i)  and

Solution:

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

D=

=  â€“1

= âˆš14

â‡’ SD = 1/âˆš14 units â‰  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

= âˆš94

â‡’ SD = 0/âˆš94 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=

= âˆ’9

= âˆš195

â‡’ SD = 9/âˆš195 units â‰  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

â‡’ SD = 282/âˆš3 units â‰  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=

â‡’ 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=

â‡’

= âˆš11

â‡’ SD = âˆš11/âˆš6 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=

= âˆ’6

= 2

â‡’ 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=

â‡’

\vec{|b|}= 7

â‡’ SD = âˆš293/7 units.

### (i) and

Solution:

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

D=

Now,

= 3âˆš2

â‡’ SD = 3/âˆš2 units.

### (ii)

Solution:

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

D=

Now,

= âˆš116

â‡’ SD = 2âˆš29 units.

### (iii) and

Solution:

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

D=

Now,

= âˆš171

â‡’ SD = 3âˆš19 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}Ã—\vec{b_2})=108

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

= 12

â‡’ 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=

â‡’

= âˆš293

â‡’ SD = âˆš293/7 units.

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