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

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TextbookNCERT
ClassClass 12th
SubjectMaths
Chapter28
Exercise28.5
CategoryRD Sharma Solutions

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) (\vec{r}=3\hat{i}+8\hat{j}+3\hat{k}+λ(3\hat{i}-\hat{j}+\hat{k})  and \vec{r}=-3\hat{i}-7\hat{j}+6\hat{k}+μ(-3\hat{i}+2\hat{j}+4\hat{k})

Solution:

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})  and \vec{r}=\vec{a_2}+μ(\vec{b_2})  is:

D= |\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|

Now, \vec{a_2}-\vec{a_1}=-3\hat{i}-7\hat{j}+6\hat{k}-(3\hat{i}+8\hat{j}+3\hat{k})

-6\hat{i}-15\hat{j}+3\hat{k}

(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\3&-1&1\\-3&2&4\\\end{array}\right|

\hat{i}(-4-2)-\hat{j}(12+3)+\hat{k}(6-3)

-6\hat{i}-15\hat{j}+3\hat{k}

(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(-6\hat{i}-15\hat{j}+3\hat{k}).(-6\hat{i}-15\hat{j}+3\hat{k})

= 36 + 225 + 9

= 270

|\vec{b_1}×\vec{b_2}|=\sqrt{(-6)^2+(-15)^2+(3)^2}

\sqrt{36+225+9}

= √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) \vec{r}=3\hat{i}+5\hat{j}+7\hat{k}+λ(\hat{i}-2\hat{j}+7\hat{k})  and \vec{r}=-\hat{i}-\hat{j}-\hat{k}+μ(7\hat{i}-6\hat{j}+\hat{k}).

Solution:

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})  and \vec{r}=\vec{a_2}+μ(\vec{b_2})  is:

D= |\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|

Now, \vec{a_2}-\vec{a_1}=-\hat{i}-\hat{j}-\hat{k}-(3\hat{i}+5\hat{j}+7\hat{k})

-4\hat{i}-6\hat{j}-8\hat{k}

-2(2\hat{i}+3\hat{j}+4\hat{k})

(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\1&-2&7\\7&-6&1\\\end{array}\right|

8(5\hat{i}+6\hat{j}+\hat{k})

(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(-2(2\hat{i}+3\hat{j}+4\hat{k})).(8(5\hat{i}+6\hat{j}+\hat{k}))

= – 16 × 32

= – 512

|\vec{b_1}×\vec{b_2}|=8\sqrt{(5)^2+(6)^2+(1)^2}

8\sqrt{25+36+1}

8\sqrt{62}

On substituting the values in the formula, we have

SD = |\frac{-512}{8\sqrt{62}}|

Shortest distance between the given pair of lines is |\frac{512}{\sqrt{3968}}|  units.

(iii) \vec{r}=\hat{i}+2\hat{j}+3\hat{k}+λ(2\hat{i}+3\hat{j}+4\hat{k})  and \vec{r}=2\hat{i}+4\hat{j}+5\hat{k}+μ(3\hat{i}+4\hat{j}+5\hat{k})

Solution:

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})  and \vec{r}=\vec{a_2}+μ(\vec{b_2})   is:

D= |\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|

Now, \vec{a_2}-\vec{a_1}=2\hat{i}+4\hat{j}+5\hat{k}-(\hat{i}+2\hat{j}+3\hat{k})

\hat{i}+2\hat{j}+2\hat{k}

(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\2&3&4\\3&4&5\\\end{array}\right|

-\hat{i}+2\hat{j}-\hat{k}

(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(\hat{i}+2\hat{j}+2\hat{k}).(-\hat{i}+2\hat{j}-\hat{k})

= 1

|\vec{b_1}×\vec{b_2}|=\sqrt{(-1)^2+(2)^2+(-1)^2}

\sqrt{6}

On substituting the values in the formula, we have

SD = |\frac{1}{\sqrt6}|

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

(iv) \vec{r}=(1-t)\hat{i}+(t-2)\hat{j}+(3-t)\hat{k}  and \vec{r}=(s+1)\hat{i}+(2s-1)\hat{j}-(2s+1)\hat{k}

Solution:

Above equations can be re-written as:

\vec{r}=(\hat{i}-2\hat{j}+3\hat{k})+t(-\hat{i}+\hat{j}-\hat{k})

and, \vec{r}=(\hat{i}-\hat{j}-\hat{k})+s(\hat{i}+2\hat{j}-2\hat{k})

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})

and \vec{r}=\vec{a_2}+μ(\vec{b_2})  is:

D = |\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|

= 9/3√2

Shortest distance is 3/√2 units.

(v) \vec{r}=(λ-1)\hat{i}+(λ+1)\hat{j}-(1+λ)\hat{k}  and \vec{r}=(1-μ)\hat{i}+(2μ-1)\hat{j}+(μ+2)\hat{k}

Solution:

The given equations can be written as:

\\vec{r}=(-\hat{i}+\hat{j}-\hat{k})+λ(\hat{i}+\hat{j}-\hat{k})  and \vec{r}=\hat{i}-\hat{j}+2\hat{k}+µ(-\hat{i}+2\hat{j}+\hat{k})

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})  and \vec{r}=\vec{a_2}+μ(\vec{b_2})   is:

D= |\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|

Now, (\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(2\hat{i}-2\hat{j}+3\hat{k}).(3\hat{i}+3\hat{k})

= 15

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

= 3√2 

Thus, distance between the lines is |\frac{15}{3\sqrt2}| = \frac{5}{\sqrt2}          units.

(vi) \vec{r}=(2\hat{i}-\hat{j}-\hat{k})+λ(2\hat{i}-5\hat{j}+2\hat{k})  and \vec{r}=(\hat{i}+2\hat{j}+\hat{k})+μ(\hat{i}-\hat{j}+\hat{k})

Solution:

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})  and \vec{r}=\vec{a_2}+μ(\vec{b_2})  is:

D = |\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|

Now, (\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(-\hat{i}+3\hat{j}+2\hat{k}).(-3\hat{i}+3\hat{k})

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

= 3√2

Substituting the values in the formula, we have

The distance between the lines is |\frac{9}{3\sqrt2}| = \frac{3}{\sqrt2} units.

(vii) \vec{r}=\hat{i}+\hat{j}+λ(2\hat{i}-\hat{j}+\hat{k})  and \vec{r}=2\hat{i}+\hat{j}-\hat{k}+µ(3\hat{i}-5\hat{j}+2\hat{k})

Solution:

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})  and \vec{r}=\vec{a_2}+μ(\vec{b_2})  is:

D= |\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|

Now, (\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(-\hat{i}+3\hat{j}+2\hat{k}).(-3\hat{i}+3\hat{k})

= 10

Substituting the values in the formula, we have:

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

(viii) \vec{r}=(8+3λ)\hat{i}-(9+16λ)\hat{j}+(10+7λ)\hat{k}  and \vec{r}=15\hat{i}+29\hat{j}+5\hat{k}+µ(3\hat{i}+8\hat{j}-5\hat{k})

Solution:

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})  and \vec{r}=\vec{a_2}+μ(\vec{b_2})  is:

D= |\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|

Now, (\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(-\hat{i}+3\hat{j}+2\hat{k}).(-3\hat{i}+3\hat{k})

= 1176

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

= 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) \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-5}{5}  and \frac{x-2}{3}=\frac{y-3}{4}=\frac{z-5}{5}

Solution:

The given lines can be written as:

\vec{r}=\hat{i}+2\hat{j}+\hat{k}+λ(2\hat{i}+4\hat{j}+3\hat{k})  and \vec{r}=2\hat{i}+3\hat{j}+5\hat{k}+µ(3\hat{i}+4\hat{j} +5\hat{k})

\vec{a_2}-\vec{a_1}=2\hat{i}+3\hat{j}+5\hat{k}-(\hat{i}+2\hat{j}+3\hat{k})

\hat{i}+\hat{j}+2\hat{k}

(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\2&3&4\\3&4&5\\\end{array}\right|

-\hat{i}+2\hat{j}-\hat{k}

 (\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(\hat{i}+\hat{j}+2\hat{k}).(-\hat{i}+2\hat{j}-\hat{k})

= –1

|\vec{b_1}×\vec{b_2}|=\sqrt{(-1)^2+(2)^2+(-1)^2}

= √6 

On substituting the values in the formula, we have:

SD = 1/√6 units.

(ii) \frac{x-1}{2}=\frac{y+1}{3}=z  and \frac{x+1}{3}=\frac{y-2}{1};z=2

Solution:

The given equations can also be written as:

\vec{r}=\hat{i}-\hat{j}+λ(2\hat{i}+3\hat{j}+\hat{k})  and \\vec{r}=-\hat{i}+2\hat{j}+2\hat{k}+μ(3\hat{i}+\hat{j})

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})  and \vec{r}=\vec{a_2}+μ(\vec{b_2})  is:

D= |\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|

(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\2&3&1\\3&1&0\\\end{array}\right|

-\hat{i}+3\hat{j}-7\hat{k}

 (\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(\hat{i}+\hat{j}+2\hat{k}).(-\hat{i}+2\hat{j}-\hat{k})

=  3

SD = 3/√59 units.

(iii) \frac{x-1}{-1}=\frac{y+2}{1}=\frac{z-3}{-2}  and \frac{x-1}{1}=\frac{y+1}{2}=\frac{z+1}{-2}

Solution:

The given equations can be re-written as:

\vec{r}=\hat{i}-2\hat{j}+3\hat{k}+ λ(-\hat{i}+\hat{j}+2\hat{k})  and \vec{r}=\hat{i}-\hat{j}-\hat{k}+µ(\hat{i}+2\hat{j}-2\hat{k})

(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\-1&1&2\\1&2&-2\\\end{array}\right|

= √29 

 (\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(\hat{i}+\hat{j}+2\hat{k}).(-\hat{i}+2\hat{j}-\hat{k})

= 8

SD = 8/√29 units.

(iv) \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{-1}  and \vec{r}=\frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1}

Solution:

The given equations can be re-written as:

\vec{r}=3\hat{i}+5\hat{j}+7\hat{k}+λ(\hat{i}-2\hat{j}+\hat{k})  and \vec{r}=(-\hat{i}-\hat{j}-\hat{k})+µ(7\hat{i}-6\hat{j}+\hat{k})

(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\1&-2&1\\7&-6&1\\\end{array}\right|

4\hat{i}+6\hat{j}+8\hat{k}

SD = 58/√29 units.

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

(i) \vec{r}=(\hat{i}-\hat{j})+λ(2\hat{i}+\hat{k})  and \vec{r}=2\hat{i}-\hat{j}+µ(\hat{i}+\hat{j}-\hat{k})

Solution:

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})  and \vec{r}=\vec{a_2}+μ(\vec{b_2})  is:

D= |\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|

(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\2&0&1\\1&1&-1\\\end{array}\right|

-\hat{i}+3\hat{j}+2\hat{k}

(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(\hat{i}+\hat{j}+2\hat{k}).(-\hat{i}+2\hat{j}-\hat{k})

=  –1

|\vec{b_1}×\vec{b_2}|=\sqrt{(-1)^2+(3)^2+(2)^2}

= √14 

⇒ SD = 1/√14 units ≠ 0

Hence the given pair of lines does not intersect.

(ii) \vec{r}=\hat{i}+\hat{j}-\hat{k}+λ(3\hat{i}-\hat{j})  and \vec{r}=(4\hat{i}-\hat{k})+µ(2\hat{i}+3\hat{k})

Solution:

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})  and \vec{r}=\vec{a_2}+μ(\vec{b_2})  is:

D= |\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|

(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\3&-1&0\\2&0&3\\\end{array}\right|

-3\hat{i}-9\hat{j}+2\hat{k}\vec{r}=\hat{i}+\hat{j}-\hat{k}+λ(3\hat{i}-\hat{j})    \vec{r}=\hat{i}+\hat{j}-\hat{k}+λ(3\hat{i}-\hat{j})

(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(3\hat{i}-\hat{j}).(-3\hat{i}-9\hat{j}+2\hat{k})

= 0

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

= √94 

⇒ SD = 0/√94 units = 0

Hence the given pair of lines are intersecting.

(iii) \frac{x-1}{2}=\frac{y+1}{3}=z  and \frac{x+1}{5}=\frac{y-2}{1};z=2

Solution:

Given lines can be re-written as:

\vec{r}=(\hat{i}-\hat{j})+λ(2\hat{i}+3\hat{j}+\hat{k})  and \vec{r}=(-\hat{i}+2\hat{j}+2\hat{k})+μ(5\hat{i}+\hat{j})

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})  and \vec{r}=\vec{a_2}+μ(\vec{b_2})  is:

D= |\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|

(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\2&3&1\\5&1&0\\\end{array}\right|

-\hat{i}+5\hat{j}-13\hat{k}

(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(-2\hat{i}+3\hat{j}+2\hat{k}).(-\hat{i}+5\hat{j}-13\hat{k})

= −9

|\vec{b_1}×\vec{b_2}|=\sqrt{(-1)^2+(5)^2+(-13)^2}

= √195 

⇒ SD = 9/√195 units ≠ 0

Hence the given pair of lines does not intersect.

(iv) \frac{x-5}{4}=\frac{y-7}{-5}=\frac{z+3}{-5}  and \frac{x-8}{7}=\frac{y-7}{1}=\frac{z-5}{3}

Solution:

Given lines can be re-written as: 

\vec{r}=(5\hat{i}+7\hat{j}-3\hat{k})+λ(4\hat{i}-5\hat{j}-5\hat{k})   and \vec{r}=(8\hat{i}+7\hat{j}+5\hat{k})+μ(7\hat{i}+\hat{j}+3\hat{k})

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})  and \vec{r}=\vec{a_2}+μ(\vec{b_2})  is:

D= |\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|

(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\4&5&-5\\7&1&3\\\end{array}\right|

-10\hat{i}-47\hat{j}+39\hat{k}

(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(3\hat{i}+8\hat{k}).(-10\hat{i}-47\hat{j}+39\hat{k})

= 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) \vec{r}=(\hat{i}+2\hat{j}+3\hat{k})+λ(\hat{i}-\hat{j}+\hat{k})  and \vec{r}=(2\hat{i}-\hat{j}-\hat{k})+µ(-\hat{i}+\hat{j}-\hat{k})

Solution:

The second given line can be re-written as: \vec{r}=(2\hat{i}-\hat{j}-\hat{k})+µ(\hat{i}-\hat{j}+\hat{k})

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})  and \vec{r}=\vec{a_2}+μ(\vec{b_2})  is:

D= |\frac{(\vec{a_2}-\vec{a_1})×\vec{b}}{\vec{|b|}}|

(\vec{a_2}-\vec{a_1})×\vec{b}=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\1&-3&-4\\1&-1&1\\\end{array}\right|

-7\hat{i}-5\hat{j}+2\hat{k}

 |(\vec{a_2}-\vec{a_1})×\vec{b}|=\sqrt{(-7)^2+(-5)^2+2^2}

\sqrt{78}

\vec{|b|}=\sqrt3

⇒ SD = \frac{\sqrt{78}}{\sqrt{3}} = \sqrt{26}     units.

(ii) \vec{r}=(\hat{i}+\hat{j})+λ(2\hat{i}-\hat{j}+\hat{k})     and \vec{r}=(2\hat{i}-\hat{j}+\hat{k})+µ(4\hat{i}-2\hat{j}+2\hat{k})

Solution:

The second given line can be re-written as: \vec{r}=(2\hat{i}+\hat{j}-\hat{k})+µ'(2\hat{i}-\hat{j}+\hat{k})

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})  and \vec{r}=\vec{a_2}+μ(\vec{b_2})  is:

D= |\frac{(\vec{a_2}-\vec{a_1})×\vec{b}}{\vec{|b|}}|

(\vec{a_2}-\vec{a_1})×\vec{b}=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\1&0&-1\\2&-1&1\\\end{array}\right|

-\hat{i}-3\hat{j}-\hat{k}

⇒ |(\vec{a_2}-\vec{a_1})×\vec{b}|=\sqrt{(-1)^2+(-3)^2+(-1)^2}

= √11 

\vec{|b|}=\sqrt6

⇒ 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:

\vec{r}=(0\hat{i}+0\hat{j}+0\hat{k})+λ(\hat{i}+2\hat{k})

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

\vec{r}=(\hat{i}+3\hat{j}+0\hat{k})+µ(-\hat{i})

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})     and \vec{r}=\vec{a_2}+μ(\vec{b_2})    is:

D= |\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|

(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\1&0&2\\-1&0&0\\\end{array}\right|

-2\hat{j}

(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=(\hat{i}+3\hat{j}).(-2\hat{j})

= −6

|\vec{b_1}×\vec{b_2}|=\sqrt{(-2)^2}

= 2

⇒ SD = |-6/2| = 3 units.

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

\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+4}{6};\frac{x-3}{4}=\frac{y-3}{6}=\frac{z+5}{12}

Solution:

The given equations can be written as:

\vec{r}=(\hat{i}+2\hat{j}-4\hat{k})+λ(2\hat{i}+3\hat{j}+6\hat{k})    and \vec{r}=(3\hat{i}+3\hat{j}-5\hat{k})+µ2\hat{i}+3\hat{j}+6\hat{k})

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})  and \vec{r}=\vec{a_2}+μ(\vec{b_2})  is:

D= |\frac{(\vec{a_2}-\vec{a_1})×\vec{b}}{\vec{|b|}}|

(\vec{a_2}-\vec{a_1})×\vec{b}=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\2&1&-1\\2&3&6\\\end{array}\right|

9\hat{i}-14\hat{j}+4\hat{k}

⇒ |(\vec{a_2}-\vec{a_1})×\vec{b}|=\sqrt{(9)^2+(-14)^2+(4)^2}

\sqrt{293}

\vec{|b|}= 7

⇒ SD = √293/7 units.

Question 7. Find the shortest distance between the following:

(i) \vec{r}=\hat{i}+2\hat{j}+\hat{k}+λ(\hat{i}-\hat{j}+\hat{k})   and \vec{r}=(2\hat{i}-\hat{j}-\hat{k})+µ(2\hat{i}+\hat{j}+2\hat{k})

Solution:

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})  and \vec{r}=\vec{a_2}+μ(\vec{b_2})  is:

D= |\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|

Now, \vec{a_2}-\vec{a_1}=2\hat{i}-\hat{j}-\hat{k}-\hat{i}-2\hat{j}-\hat{k}

\hat{i}-3\hat{j}-2\hat{k}

(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\1&-1&1\\2&1&2\\\end{array}\right|

-3\hat{i}+3\hat{k}

(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=9
|\vec{b_1}×\vec{b_2}|=\sqrt{(-3)^2+(-3)^2}

= 3√2 

⇒ SD = 3/√2 units.

(ii) \frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1};\frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1}     

Solution:

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})  and \vec{r}=\vec{a_2}+μ(\vec{b_2})  is:

D= |\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|

Now, \vec{a_2}-\vec{a_1}=4\hat{i}+6\hat{j}+8\hat{k}

(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\7&-6&1\\1&-2&1\\\end{array}\right|

-4\hat{i}-6\hat{k}-8\hat{k}

(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=-116
|\vec{b_1}×\vec{b_2}|=\sqrt{(-3)^2+(-3)^2}

= √116 

⇒ SD = 2√29 units.

(iii) \vec{r}=\hat{i}+2\hat{j}+3\hat{k}+λ(\hat{i}-3\hat{j}+2\hat{k})   and \vec{r}=4\hat{i}+5\hat{j}+6\hat{k}+μ(2\hat{i}+3\hat{j}+\hat{k})

Solution:

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})  and \vec{r}=\vec{a_2}+μ(\vec{b_2})  is:

D= |\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|

Now, \vec{a_2}-\vec{a_1}=-9\hat{i}+3\hat{j}+9\hat{k}

(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\1&-3&2\\2&3&1\\\end{array}\right|

3\hat{i}+3\hat{k}+3\hat{k}

(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})=9
|\vec{b_1}×\vec{b_2}|=\sqrt{(-9)^2+(3)^2+(9)^2}

= √171 

⇒ SD = 3√19 units.

(iv) \vec{r}=(6\hat{i}+2\hat{j}+2\hat{k})+λ(\hat{i}-2\hat{j}+2\hat{k})   and \vec{r}=-4\hat{i}-\hat{k}+μ(3\hat{i}-2\hat{j}-2\hat{k})

Solution:

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})  and \vec{r}=\vec{a_2}+μ(\vec{b_2})  is:

D= |\frac{(\vec{a_2}-\vec{a_1}).(\vec{b_1}×\vec{b_2})}{|\vec{b_1}×\vec{b_2}|}|

Now, \vec{a_2}-\vec{a_1}=-10\hat{i}-2\hat{j}-3\hat{k}

(\vec{b_1}×\vec{b_2})=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\1&-2&2\\3&-2&-2\\\end{array}\right|

8\hat{i}+8\hat{k}+4\hat{k}

(\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: \vec{r}=(\hat{i}+2\hat{j}-4\hat{k})+λ(2\hat{i}+3\hat{j}+6\hat{k})   and \vec{r}=3\hat{i}+3\hat{j}-5\hat{k}+μ(2\hat{i}+3\hat{j}+6\hat{k})

Solution:

As we know that the shortest distance between the lines \vec{r}=\vec{a_1}+λ(\vec{b_1})  and \vec{r}=\vec{a_2}+μ(\vec{b_2})  is:

D= |\frac{(\vec{a_2}-\vec{a_1})×\vec{b}}{\vec{|b|}}|

(\vec{a_2}-\vec{a_1})×\vec{b}=\left|\begin{array}{cc}\hat{i}&\hat{j}&\hat{k}\\2&1&-1\\2&3&6\\\end{array}\right|

9\hat{i}-14\hat{j}+4\hat{k}

⇒  |(\vec{a_2}-\vec{a_1})×\vec{b}|=\sqrt{(9)^2+(-14)^2+(4)^2}

= √293  

\vec{|b|}=7

⇒ SD = √293/7 units.

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