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

Here we provideRD Sharma Class 12 Ex 28.5 Solutions Chapter 28 The Straight Line in Space for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 12 Ex 28.5 Solutions Chapter 28 The Straight Line in Space book pdf download. Now you will get step-by-step solutions to each question.

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

### (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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