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

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.