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.

Textbook | NCERT |

Class | Class 12th |

Subject | Maths |

Chapter | 28 |

Exercise | 28.5 |

Category | RD Sharma Solutions |

Table of Contents

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

**Question 1: Ten cards numbered 1 through 10 are placed in a box, mixed **

### Question 1. Find the shortest distance between the pair of lines whose vector equation is:

### (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.**

### Question 2. Find the shortest distance between the pair of lines whose cartesian equation is:

### (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.**

### Question 3. By computing the shortest distance determine whether the pairs of lines intersect or not:

### (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.**

### Question 4. Find the shortest distance between the following:

### (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.**

### Question 5. Find the equations of the lines joining the following pairs of vertices and then find the shortest distance between the lines:

### (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.**

### Question 7. Find the shortest distance between the following:

### (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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