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

Leave a Comment / Class 12 Maths / By

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.

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.

I think you got complete solutions for this chapter. If You have any queries regarding this chapter, please comment in the below section our subject teacher will answer you. We tried our best to give complete solutions so you got good marks in your exam.

If these solutions have helped you, you can also share rdsharmasolutions.in to your friends.

Related Posts

Leave a Comment

Your email address will not be published. Required fields are marked *