In geometry, the study of lines in three-dimensional space extends our understanding of the linear relationships from the plane to a more complex environment. Chapter 28 of RD Sharma's Class 12 textbook focuses on the "The Straight Line in Space" providing an in-depth exploration of how straight lines can be represented and analyzed within the 3D coordinate system. Exercise 28.5 delves into specific problems and applications helping the students grasp the concepts of the line equations and spatial relationships in three dimensions.
The Straight Line in Space
In three-dimensional space a straight line can be described using the variety of the methods including the vector equations, parametric equations and symmetric equations. The general form of the line in space is often given by a vector equation:
r=a+λb
where r is the position vector of any point on the line a is a point on the line b is a direction vector and λ is a scalar parameter. By solving different equations and problems related to the lines in space students learn how to manipulate and interpret these representations to the solve real-world problems and geometric challenges.
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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Conclusion
Understanding the straight line in space is crucial for the advanced studies in geometry and physics where spatial relationships play a significant role. Exercise 28.5 in RD Sharma's Class 12 textbook equips students with necessary skills to analyze and solve problems involving the lines in a 3D coordinate system. Mastery of these concepts lays the groundwork for the more complex topics and practical applications in the various scientific fields.
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