Class 12 NCERT Solutions- Mathematics Part ii – Chapter 11 – Three Dimensional Geometry Exercise 11.2
In this article, we have covered solutions to exercise 11.2 of chapter chapter 11 – Three Dimensional Geometry Exercise for class 12 Mathematics. Now let’s learn the same.
Question 1: Show that the three lines with direction cosines
[Tex]\frac{12}{13},\frac{-3}{13},\frac{-4}{13};\frac{4}{13},\frac{12}{13},\frac{3}{13};\frac{3}{13},\frac{-4}{13},\frac{12}{13} \, are\,mutually\, perpendicular[/Tex]
Solution:
Two lines with direction cosines l1,m1,n1 and l2,m2,n2 are perpendicular to each other,
if l1l2 + m1m2 + n1n2 = 0
(i) For the lines with direction cosines,[Tex]\frac{12}{13},\frac{-3}{13},\frac{-4}{13} \, and \frac{4}{13},\frac{12}{13},\frac{3}{13} \, we \, obtain [/Tex]
l1l2 + m1m2 + n1n2 = [Tex]\frac{12}{13}*\frac{4}{13}+(\frac{-3}{13} )*\frac{12}{13}+(\frac{-4}{13})*\frac{3}{13}[/Tex]
[Tex]= \frac{48}{169} – \frac{36}{169} – \frac{12}{169} = 0[/Tex]
Hence, the lines are perpendicular.
(ii) For the lines with direction cosines [Tex] \frac{4}{13},\frac{12}{13},\frac{3}{13} \, and \, \frac{3}{13},\frac{-4}{13},\frac{12}{13} \, \, we \, obtain[/Tex]
l1l2 + m1m2 + n1n2 = [Tex] \frac{4}{13}*\frac{3}{13}+\frac{12}{13}*(\frac{-4}{13})+\frac{3}{13}*\frac{12}{13}[/Tex]
[Tex]= \frac{12}{169} – \frac{48}{169} + \frac{36}{169} = 0[/Tex]
Hence, the lines are perpendicular.
(iii) For the lines with direction cosines,[Tex] \frac{3}{13},\frac{-4}{13},\frac{12}{13} \, and \, \frac{12}{13},\frac{-3}{13},\frac{-4}{13} \, \, we \, obtain[/Tex]
l1l2 + m1m2 + n1n2 = [Tex] \frac{3}{13}*\frac{12}{13}+(\frac{-4}{13})*(\frac{-3}{13})+\frac{12}{13}*\frac{-4}{13}[/Tex]
[Tex]= \frac{36}{169} + \frac{12}{169} – \frac{48}{169} = 0[/Tex]
Hence, the lines are perpendicular.
So, all the lines are mutually perpendicular.
Question 2: Show that the line through the points (1, -1, 2) (3, 4, -2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).
Solution:
Let AB be the line joining the points, (1, -1, 2) and (3, 4, -2), and CD be the line through the
points (0, 3, 2) and (3, 5, 6).
a1 = (3 -1), b1 =(4 – (-1)), and c1 = (-2 -2) i.e., 2, 5, and -4.
a2 = (3 -0), b2 = (5 -3), and c2 = (6 -2) i.e., 3, 2, and 4.
AB ⊥ CD => a1a2 + b1b2 + c1c2 =0
a1a2 +b1b2 + c1c2 = 2⨯3 + 5⨯2 + (-4)⨯4 = 6 + 10 – 16 = 0
Hence, AB and CD are perpendicular to each other.
Question 3: Show that the line through the points (4, 7, 8) (2, 3, 4) is parallel to the line through the points (-1, -2, 1), (1, 2, 5).
Solution:
Let AB be the line through the points (4, 7, 8) and (2, 3, 4), CD be the line through the points, (-1, -2, 1) and (1, 2, 5).
a1 = (2 -4), b1 = (3 -7), and c1 = (4 -8) i.e., -2, -4, and -4.
a2 = (1 – (-1)), b2 = (2 – (-2)), and c2 = (5 -1) i.e., 2, 4, and 4.
AB||CD => a1 / a2 = b1 / b2 = c1 / c2
=> a1 / a2 = -2 / 2 = -1
=> b1 / b2 = -4 / 4 = -1
=> c1 / c2 = -4/4 = -1
So, a/a = b / b = c / c
Hence, AB is parallel to CD.
Question 4: Find the equation of the line which passes through point (1, 2, 3) and is parallel to the vector [Tex]3\hat{i}+2\hat{j}-2\hat{k}[/Tex]
Solution:
It is given that the line passes through the point A (1, 2, 3). Therefore, the position vector through
A is [Tex]\vec{a} = \hat{i}+2\hat{j}+3\hat{k}[/Tex]
[Tex] \vec{b} = 3\hat{i}+2\hat{j}-2\hat{k} [/Tex]
So, line passes through point A and parallel to [Tex]\vec{b}[/Tex] is given by
[Tex]\vec{r} = \vec{a}+\lambda\vec{b}[/Tex] , where [Tex]\lambda[/Tex] is a constant
=> [Tex]\vec{r} = \hat{i}+2\hat{j}+3\hat{k} + \lambda(3\hat{i}+2\hat{j}-2\hat{k})[/Tex]
This is required equation of the line.
Question 5: Find the equation of the line in vector and in Cartesian form that passes through the point with positive vector [Tex]2\hat{i}-\hat{j}+4\hat{k} [/Tex] and is in the direction [Tex]\hat{i}+2\hat{j}-\hat{k}[/Tex]
Solution:
It is given that the line passes through the point with positive vector
[Tex]\vec{a} = 2\hat{i}-\hat{j}+4\hat{k} …………..(i) [/Tex]
[Tex]\vec{b} = \hat{i}+2\hat{j}-\hat{k} ………..(ii)[/Tex]
So, line passes through point A and parallel to [Tex]\vec{b}[/Tex] is given by
=> [Tex]\vec{r} = \vec{a}+\lambda\vec{b}[/Tex] ,where [Tex]\lambda[/Tex] is a constant
[Tex]\vec{r} = 2\hat{i}-\hat{j}+4\hat{k} + \lambda(\hat{i}+2\hat{j}-\hat{k})[/Tex]
This is the required equation of the line in vector form.
[Tex]\vec{r} = x\hat{i}-y\hat{j}+z\hat{k} =>x\hat{i}-y\hat{j}+z\hat{k}=(\lambda+2)\hat{i}+(2\lambda-1)\hat{j}+(-\lambda+4)\hat{k}[/Tex]
Eliminating [Tex]\lambda[/Tex], we get the Cartesian form equation as
[Tex]\frac{x-2}{1} = \frac{y+1}{2} = \frac{z-4}{-1}[/Tex]
Question 6: Find the Cartesian equation of the line which passes through the point (-2, 4, -5) and parallel to the line given by [Tex]\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}[/Tex]
Solution:
It is given that the line passes through the point (-2, 4, -5) and is parallel to [Tex]\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}[/Tex]
Direction ratios of the line, [Tex]\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6} , are .3,5,6[/Tex]
Required line is parallel to [Tex]\frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}[/Tex]
Therefore, its direction ratios are 3k, 5k, and 6k, when k ≠0
It is known that the equation of the line through the point (x1 ,y1, z1 ) and with direction ratios,
a, b, c is given by [Tex]\frac{x-x1}{a} = \frac{y-y1}{b} = \frac{z-z1}{c}[/Tex]
Hence, equation of the required line is
[Tex]\frac{x+2k}{3k} = \frac{y-4}{5k} = \frac{z+5}{6k} => \frac{x+2}{3} = \frac{y-4}{5} = \frac{z+5}{6} = k[/Tex]
Question 7: Cartesian equation of a line is [Tex]\frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2}[/Tex] , Write its vector form.
Solution:
Cartesian equation of the line is given by
[Tex]\frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2} ……..(i)[/Tex]
The given line passes through the point (5, -4, 6). The position vector of this point is [Tex]\vec{a} = 5\hat{i}-4\hat{j}+6\hat{k} [/Tex]
Also, the direction ratios of the given line are 3, 7, and 2.
This means that the line is in the direction of the vector,[Tex]\vec{b} = 3\hat{i}-7\hat{j}+2\hat{k} [/Tex]
As we known that the line through positive vector [Tex]\vec{a} [/Tex] and in the direction of the vector [Tex]\vec{b}[/Tex] is given
by the equation, [Tex]\vec{r} = \vec{a} + \lambda\vec{b},\lambda\in{R}[/Tex]
[Tex]=> \vec{r} =(5\hat{i}-4\hat{j}+6\hat{k} ) + (3\hat{i}+7\hat{j}+2\hat{k} ) [/Tex]
This is the required equation of the given line in vector form.
Question8: Find the vector and the Cartesian equation of the lines that passes through the origin and (5, -2, 3).
Solution:
Required line passes through the origin. Therefore, its position vector is given by,
[Tex]\vec{a} = \vec{0} ……….(i)[/Tex]
The direction ratios of the line through origin and (5, -2, 3) are
(5 -0) = 5, (-2 -0) = -2, (3 – 0) = 3
The line is parallel to the vector given by the equation, [Tex]\vec{b} = 5\hat{i}-2\hat{j}+3\hat{k} [/Tex]
The equation of the line in vector form through a point with position vector [Tex]\vec{a}[/Tex]and parallel to [Tex]\vec{b}[/Tex] is,
[Tex]\vec{r} = \vec{a} + \lambda\vec{b},\lambda\in{R} [/Tex]
[Tex]=> \vec{r} = \vec{0} + \lambda(5\hat{i}-2\hat{j}+3\hat{k})[/Tex]
The equation of the line through the point ( x1,y1,z1) and direction ratios a, b, c is given by,
[Tex]\frac{x-x1}{a} = \frac{y-y1}{b} = \frac{z-z1}{c}[/Tex]
Hence, the equation of the required line in the Cartesian form is
[Tex]\frac{x-0}{5} = \frac{y-0}{-2} = \frac{z-0}{3}=> \frac{x}{5} = \frac{y}{2} = \frac{z}{3}[/Tex]
Question 9: Find the angle between the following pairs of lines:
[Tex] \vec{r} = 2\hat{i}-5\hat{j}+\hat{k} + \lambda(3\hat{i}-2\hat{j}+6\hat{k}) ,and ,\vec{r}= 7\hat{i}-6\hat{k}+\mu(\hat{i}+2\hat{j}+2\hat{k})[/Tex]
Solution:
Let θ be the angle between the given lines.
Angle between the given pairs of lines is given by,
[Tex]\cos(Q) = \frac{\vec{b}1 \cdot \vec{b}2}{\lVert \vec{b}1 \rVert \cdot \lVert \vec{b}2 \rVert} [/Tex]
Given lines are parallel to the vectors,[Tex]\vec{b}1 = 3\hat{i}+2\hat{j}+6\hat{k} , \vec{b}2 = \hat{i}+2\hat{j}+2\hat{k}[/Tex]
So, [Tex]{\left |\vec{b}1 \right|} = \sqrt{3^2+2^2+6^2} = 7 [/Tex]
[Tex]{\left |\vec{b}2 \right|} = \sqrt{(1)^2+(2)^2+(2)^2} = 3 [/Tex]
[Tex]\vec{b}1 \cdot \vec{b}2 = (3\hat{i}+2\hat{j}+6\hat{k}) \cdot (\hat{i}+2\hat{j}+2\hat{k}) = 3*1+2*2+6*2 = 19[/Tex]
[Tex]=> cosQ = \frac{19}{7*3} => Q = \cos^{-1}( \frac{19}{21})[/Tex]
(ii) [Tex] \vec{r} = 3\hat{i}+\hat{j}-2\hat{k} + \lambda(\hat{i}-\hat{j}-2\hat{k}) ,and ,\vec{r}= 2\hat{i}-\hat{j}-56\hat{k}+\mu(3\hat{i}-5\hat{j}-4\hat{k})[/Tex]
Solution:
Given line is parallel to the vectors, respectively [Tex]\vec{b}1 = \hat{i}-\hat{j}-2\hat{k} , \vec{b}2 = 3\hat{i}-5\hat{j}-4\hat{k},respectively.[/Tex]
[Tex]{\left |\vec{b}1 \right|} = \sqrt{(1)^2+(-1)^2+(-2)^2} = \sqrt{6}[/Tex]
[Tex]{\left |\vec{b}2 \right|} = \sqrt{(3)^2+(-5)^2+(-4)^2} = \sqrt{50}=5\sqrt{2}[/Tex]
[Tex]\vec{b}1 \cdot \vec{b}2 = (\hat{i}-\hat{j}-2\hat{k}) \cdot (3\hat{i}-5\hat{j}-4\hat{k})[/Tex]
[Tex]= 1\cdot3 – 1(-5)-2(-4) = 3+5+8 = 16[/Tex]
[Tex]\cos Q = \frac{\vec{b}1 \cdot \vec{b}2}{\lVert \vec{b}1 \rVert \cdot \lVert \vec{b}2 \rVert}[/Tex]
[Tex]=> cosQ = \frac{16}{\sqrt{6}.5\sqrt{2}} [/Tex]
[Tex]\frac{16}{10\sqrt{3}} => => Q = \cos^{-1}( \frac{18}{5\sqrt{3}})[/Tex]
Question 10: Find the angle between the following pair of lines:
[Tex](i \,) \,\frac{x-2}{2} = \frac{y-1}{5} = \frac{z+3}{-3} \,and\, \frac{x+2}{-1} = \frac{y-4}{8} =\frac{z-5}{4}[/Tex]
Solution:
[Tex]Let\, \vec{b}1 \, and\,\vec{b}2 \,be\, the\, vectors\, parallel\, to\, the \,pair \,of \,lines, [/Tex]
[Tex] \,\frac{x-2}{2} = \frac{y-1}{5} = \frac{z+3}{-3} \,and\, \frac{x+2}{-1} = \frac{y-4}{8} =\frac{z-5}{4} \,\,respectively[/Tex]
[Tex]So, \vec{b}1 = 2\hat{i}+5\hat{j}-3\hat{k} \, and \vec{b}2 = -\hat{i}+8\hat{j}+4\hat{k}[/Tex]
[Tex]{\left |\vec{b}1 \right|} = \sqrt{(2)^2+(5)^2+(-3)^2} = \sqrt{38}[/Tex]
[Tex]{\left |\vec{b}2 \right|} = \sqrt{(-1)^2+(8)^2+(4)^2} = \sqrt{81} = 9[/Tex]
[Tex]\vec{b}1 \cdot \vec{b}2 = (2\hat{i}+5\hat{j}-3\hat{k}) \cdot (-\hat{i}+8\hat{j}+4\hat{k})[/Tex]
[Tex]=2(-1)+5*8+(-3).4 = -2+40-12 = 26[/Tex]
Angle, ?, between the given pair of lines is given by the relation,
[Tex]\cos \theta = \frac{\vec{b}1 \cdot \vec{b}2}{\lVert \vec{b}1 \rVert \cdot \lVert \vec{b}2 \rVert}[/Tex]
[Tex]=> cos\theta = \frac{26}{9\sqrt{38}} [/Tex]
[Tex] => \theta = \cos^{-1}( \frac{26}{9\sqrt{38}})[/Tex]
[Tex](ii \,) \,\frac{x}{2} = \frac{y}{2} = \frac{z}{1} \,and\, \frac{x-5}{4} = \frac{y-2}{1} =\frac{z-3}{8}[/Tex]
Solution:
Let [Tex]\vec{b}1,\vec{b}2[/Tex] be the vectors parallel to the given pair of lines,
[Tex] \, \,\frac{x}{2} = \frac{y}{2} = \frac{z}{1} \,and\, \frac{x-5}{4} = \frac{y-2}{1} =\frac{z-3}{8},respectively.[/Tex]
[Tex]\vec{b}_1 = 2\hat{i}+2\hat{j}+\hat{k} \,[/Tex]
[Tex]\vec{b}_2 = 4\hat{i}+\hat{j}+8\hat{k}[/Tex]
[Tex]{\left |\vec{b}_1 \right|} = \sqrt{(2)^2+(2)^2+(1)^2} = \sqrt{9} = 3[/Tex]
[Tex]{\left |\vec{b}_2 \right|} = \sqrt{(4)^2+(1)^2+(8)^2} = \sqrt{81} = 9[/Tex]
[Tex]\vec{b}1 \cdot \vec{b}2 = (2\hat{i}+2\hat{j}+\hat{k}) \cdot (4\hat{i}+\hat{j}+8\hat{k})[/Tex]
[Tex]=> 2×4+2×1+1×8 = 8+2+8=18[/Tex]
Angle ?, between the given pair of lines, then
[Tex]\cos \theta = \frac{\vec{b}1 \cdot \vec{b}2}{\lVert \vec{b}1 \rVert \cdot \lVert \vec{b}2 \rVert}[/Tex]
[Tex]=> cos\theta = \frac{18}{{3*9}} = \frac{2}{{3}}[/Tex]
[Tex] => \theta = \cos^{-1}( \frac{2}{3})[/Tex]
Question 11: Find the values of p so that the lines [Tex] \, \,\frac{1-x}{3} = \frac{7y-14}{2p} = \frac{z-3}{2} \,and\, \frac{7-7x}{3p} = \frac{y-5}{1} =\frac{6-z}{5},are\, at\, right\,angle.[/Tex]
Solution:
Given equations can be written in the standard form as
[Tex] \, \,\frac{x-1}{-3} = \frac{y-2}{\frac{2p}7} = \frac{z-3}{2} \,and\, \frac{x-1}{\frac{-3p}7} = \frac{y-5}{1} =\frac{z-6}{-5}[/Tex]
The direction ratios of the lines are given by
Direction ratios of the lines are -3, 2p/7, 2 and -3p/7, 1, -5 respectively.
Two lines with direction ratios, a1, b1, c1 and a2, b2, c2 are perpendicular to each other if a1a2 + b1b2 + c1c2 = 0
So, (-3).(-3p/7) + (2p/7).(1) + 2.(-5) = 0
=> 9p/7 + 2p/7 = 10
=> 11p = 70
=> p = 70/11
Question 12: Show that the lines[Tex] \, \,\frac{x-5}{7} = \frac{y+2}{-5} = \frac{z}{1} \,and\, \frac{x}{1} = \frac{y}{2} =\frac{z}{3},are\, at\, perpendicular\,to \,each \,other.[/Tex]
Solution:
Equation of the given lines are [Tex]\frac{x-5}{7} = \frac{y+2}{-5} =\frac{z}{1} \,and\,\frac{x}{1} = \frac{y}{2} =\frac{z}{3}[/Tex]
Here a1 = 7, b1 = -5 c1 = 1 and a2 = 1 , b2 = 2 and c2 = 3
Two lines with direction ratios, a1,b1,c1 and a2,b2,c2 are perpendicular to each other, if a1a2+b1b2+c1c2 = 0
So, 7×1 + (-5)×2 + 1×3 = 7 – 10 + 3 = 0
Hence, given lines are perpendicular to each other.
Question 13: Find the shortest distance between the lines[Tex]\vec{r} = \hat{i}+2\hat{j}+\hat{k} + \lambda(\hat{i}-\hat{j}+\hat{k}) \,and\,[/Tex]
[Tex]\vec{r} = 2\hat{i}-\hat{j}-\hat{k} + \mu(2\hat{i}+\hat{j}+2\hat{k})[/Tex]
Solution:
Equation of the given lines are
[Tex]\vec{r} = \hat{i}+2\hat{j}+\hat{k} + \lambda(\hat{i}-\hat{j}+\hat{k}) \,[/Tex]
[Tex]\vec{r} = 2\hat{i}-\hat{j}-\hat{k} + \mu(2\hat{i}+\hat{j}+2\hat{k})[/Tex]
It is known that the shortest distance between the lines,
[Tex]\vec{r} = \vec{a}1+\lambda \vec{b}1\,and\,\ \vec{r} = \vec{a}2+\mu \vec{b}2,is[/Tex] given by,
[Tex] d = \left |\frac{(\vec{a_2} – \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right| –(i)[/Tex]
Comparing the given equations, we get
[Tex]\vec{a}_1 = \hat{i}+2\hat{j}+\hat{k}[/Tex]
[Tex]\vec{b}_1 = \hat{i}-\hat{j}+\hat{k}[/Tex]
[Tex]\vec{a}_2 = 2\hat{i}-\hat{j}-\hat{k}[/Tex]
[Tex]\vec{b}_2 = 2\hat{i}+\hat{j}+2\hat{k}[/Tex]
[Tex]\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}[/Tex]
[Tex]\vec{b}_1\times\vec{b}_2 = \begin{vmatrix} \hat{i} &\hat{j} & \hat{k} \\ 1 & -1 & 1\\ 2 & 1 & 2 \end{vmatrix}[/Tex]
[Tex]\vec{b}_1\times\vec{b}_2 = (-2-1)\hat{i}-(2-2)\hat{j}+(1+3)\hat{k}[/Tex]
[Tex]=> \left |\vec{b}1\times\vec{b}2 \right| = \sqrt{(-3)^2+(3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}[/Tex]
Substituting all the values in equation (1), we obtain
[Tex] d = \frac{|(-3\hat{i}+3\hat{k}) \cdot (\hat{i}-3\hat{j}-2\hat{k})}{|3\sqrt{2}|}[/Tex]
[Tex]=> d = \left |-3.1+3(-2) /3\sqrt{2} \right|[/Tex]
[Tex]=> d = \left |-9 / 3\sqrt{2} \right|[/Tex]
[Tex]=> d = \frac{3\sqrt{}2}{2}[/Tex]
Therefore, shortest distance between the two lines is [Tex] \frac{3\sqrt{}2}{2}[/Tex] units
Question 14: Find the shortest distance between the lines
[Tex] \, \,\frac{x+1}{7} = \frac{y+1}{-6} = \frac{z+1}{1} \,and\, \frac{x-3}{1} = \frac{y-5}{-2} =\frac{z-7}{1}[/Tex]
Solution:
Given lines are [Tex] \, \,\frac{x+1}{7} = \frac{y+1}{-6} = \frac{z+1}{1} \,and\, \frac{x-3}{1} = \frac{y-5}{-2} =\frac{z-7}{1}[/Tex]
It is known that the shortest distance between the two lines,
[Tex]\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1} \, and[/Tex]
[Tex]\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2} \,is\, given\, by,[/Tex]
[Tex]d = \frac{\begin{vmatrix} x_2-x_1 &y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1\\ a_2 & b_2 & c_2 \end{vmatrix}}{\sqrt{(b_1c_2-b_2c_1)^2+(c_1a_2-c_2a_1)^2+(a_1b_2-a_2b_1)^2}} ..(i)[/Tex]
x1 = -1, y1 = -1, z1 = -1
a1 = 7, b1 = -6, c1 = 1
x2 = 3, y2 = 5, z2 = 7
a2 = 1, b2 = -2 , c2 = 1
Then, [Tex]\begin{vmatrix} x_2-x_1 &y_2-y_1 & z_2-z_1 \\ a_1 & b_1 & c_1\\ a_2 & b_2 & c_2 \end{vmatrix}[/Tex]
[Tex]\begin{vmatrix} 4 & 6 & 8 \\ 7 & -6 & 1\\ 1 & -2 & 1 \end{vmatrix}[/Tex]
= 4(-6+2) – 6(7-1) + 8(-14+6)
= -16 – 36 – 64 = -116
[Tex]=> \sqrt{(b_1c_2-b_2c_1)^2+(c_1a_2-c_2a_1)^2+(a_1b_2-a_2b_1)^2}[/Tex]
[Tex]=> \sqrt{(-6+2)^2+(1+7)^2+(-14+6)^2}[/Tex]
[Tex]=> \sqrt{16+36+64} = \sqrt{116} = 2\sqrt{29}[/Tex]
[Tex]=> d = \frac{-116}{2\sqrt{29}} = \frac{-58}{\sqrt{29}} = -2\sqrt{29}[/Tex]
Thus, shortest distance between the lines whose vector equation are [Tex]2\sqrt{29}[/Tex] units
Question 15: [Tex]\vec{r} = \hat{i}+2\hat{j}+3\hat{k} + \lambda(\hat{i}-3\hat{j}+2\hat{k}) \,[/Tex]
[Tex]\vec{r} = 4\hat{i}+5\hat{j}+6\hat{k} + \mu(2\hat{i}+3\hat{j}+\hat{k})[/Tex]
Solution:
Given lines are [Tex]\vec{r} = \hat{i}+2\hat{j}+3\hat{k} + \lambda(\hat{i}-3\hat{j}+2\hat{k}) \,[/Tex] and
[Tex]\vec{r} = 4\hat{i}+5\hat{j}+6\hat{k} + \mu(2\hat{i}+3\hat{j}+\hat{k})[/Tex]
It is known that the shortest distance between the lines,
[Tex] d = \left |\frac{(\vec{a_2} – \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right| –(i)[/Tex]
Comparing the given equations, we get
[Tex]\vec{a}_1 = \hat{i}+2\hat{j}+3\hat{k}[/Tex]
[Tex]\vec{b}_1 = \hat{i}+-3\hat{j}+2\hat{k}[/Tex]
[Tex]\vec{a}_2 = 4\hat{i}+5\hat{j}+6\hat{k}[/Tex]
[Tex]\vec{b}_2 = 2\hat{i}+3\hat{j}+\hat{k}[/Tex]
[Tex]\vec{a}_2-\vec{a}_1 = (4\hat{i}+5\hat{j}+6\hat{k})-(\hat{i}+2\hat{j}+3\hat{k}) = 3\hat{i}+3\hat{j}+3\hat{k}[/Tex]
[Tex]\vec{b}_1\times\vec{b}_2 = \begin{vmatrix} \hat{i} &\hat{j} & \hat{k} \\ 1 & -3 & 2\\ 2 & 3 & 1 \end{vmatrix}[/Tex]
[Tex]\vec{b}_1\times\vec{b}_2 = (-3-6)\hat{i}-(1-4)\hat{j}+(3+6)\hat{k}[/Tex]
[Tex]-9\hat{i}+3\hat{j}+9\hat{k}[/Tex]
[Tex]=> \left |\vec{b}_1\times\vec{b}_2 \right| = \sqrt{(-9)^2+(3)^2+(9)^2} = \sqrt{81+9+81} = \sqrt{171} = 3\sqrt{19}[/Tex]
[Tex](\vec{b}_1\times\vec{b}_2)\cdot(\vec{a}_2-\vec{a}_1) = (-9\hat{i}+3\hat{j}+9\hat{k})\cdot (3\hat{i}+3\hat{j}+3\hat{k})[/Tex]
= -9×3 + 3×3 + 9×3 = 9
Substituting all the values in equation (1), we obtain
[Tex]=> d = \left| \frac{9}{3\sqrt{19}} \right|[/Tex]
[Tex]=> d = \left| \frac{3}{\sqrt{19}} \right|[/Tex]
Therefore, shortest distance between the two given line is [Tex] \left| \frac{3}{\sqrt{19}} \right| \,units[/Tex].
Question 16 Find the shortest distance between the lines whose vector equations are
[Tex]\vec{r} = (1-t)\hat{i}+(t-2)\hat{j}+(3-2t)\hat{k} \,and[/Tex]
[Tex]\vec{r} = (s+1)\hat{i}+(2s-1)\hat{j}-(2s+1)\hat{k} \,[/Tex]
Solution:
Given lines are
[Tex]\vec{r} = (1-t)\hat{i}+(t-2)\hat{j}+(3-2t)\hat{k} \,and[/Tex]
[Tex]=> \vec{r} = (\hat{i}-2\hat{j}+3\hat{k} )\,+t(-\hat{i}+\hat{j}-2\hat{k})[/Tex]
[Tex]\vec{r} = (s+1)\hat{i}+(2s-1)\hat{j}-(2s+1)\hat{k} \,[/Tex]
[Tex]=> \vec{r} = (\hat{i}-\hat{j}+\hat{k} )\,+s(\hat{i}+2\hat{j}-2\hat{k})[/Tex]
It is known that the shortest distance between the lines,
[Tex] d = \left |\frac{(\vec{a_2} – \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right| –(i)[/Tex]
For the given equations,
[Tex]\vec{a}_1 = \hat{i}-2\hat{j}+3\hat{k}[/Tex]
[Tex]\vec{b}_1 = -\hat{i}+\hat{j}-2\hat{k}[/Tex]
[Tex]\vec{a}_2 = \hat{i}-\hat{j}-\hat{k}[/Tex]
[Tex]\vec{b}_2 = \hat{i}+2\hat{j}-2\hat{k}[/Tex]
[Tex]\vec{a}_2-\vec{a}_1 = (\hat{i}-\hat{j}-\hat{k})-(\hat{i}-2\hat{j}+3\hat{k}) = \hat{j}-4\hat{k}[/Tex]
[Tex]\vec{b}_1\times\vec{b}_2 = \begin{vmatrix} \hat{i} &\hat{j} & \hat{k} \\ -1 & 1 & -2\\ 1 & 2 & -2 \end{vmatrix}[/Tex]
[Tex]=> ( 2+4)\hat{i}-(2+2)\hat{j}+(-2-1)\hat{k} = 2\hat{i}-4\hat{j}-3\hat{k}[/Tex]
[Tex]=> \left |\vec{b}_1\times\vec{b}_2 \right| = \sqrt{(2)^2+(-4)^2+(-3)^2} = \sqrt{4+16+9} = \sqrt{29} [/Tex]
[Tex](\vec{b})_1\times\vec{b}_2)\cdot(\vec{a}_2-\vec{a}_1) = (2\hat{i}-4\hat{j}-3\hat{k})\cdot (\hat{j}-4\hat{k}) = -4+12 = 8[/Tex]
Substituting all the values in equation (1), we obtain
[Tex]=> d = \left| \frac{8}{\sqrt{29}} \right|[/Tex]
Therefore, the shortest distance between the two given line is [Tex] \frac{8}{\sqrt{29}} [/Tex],units.
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