Chapter 11 of the Class 12 NCERT Mathematics Part II textbook, titled "Three Dimensional Geometry," explores the concepts and techniques used to analyze and solve problems in three-dimensional space. Exercise 11.2 focuses on applying these concepts to specific problems involving three-dimensional coordinates and geometric calculations.
This section provides detailed solutions for Exercise 11.2 from Chapter 11 of the Class 12 NCERT Mathematics Part II textbook. The exercise includes problems related to three-dimensional geometry, such as finding distances, angles, and equations of planes. Solutions are presented step-by-step to help students understand and solve three-dimensional geometry problems effectively.
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.
Class 12 NCERT Mathematics Solutions– Exercise 11.2
Question 1: Show that the three lines with direction cosines
\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
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,\frac{12}{13},\frac{-3}{13},\frac{-4}{13} \, and \frac{4}{13},\frac{12}{13},\frac{3}{13} \, we \, obtain
l1l2 + m1m2 + n1n2 = \frac{12}{13}*\frac{4}{13}+(\frac{-3}{13} )*\frac{12}{13}+(\frac{-4}{13})*\frac{3}{13}
= \frac{48}{169} - \frac{36}{169} - \frac{12}{169} = 0
Hence, the lines are perpendicular.
(ii) For the lines with direction cosines \frac{4}{13},\frac{12}{13},\frac{3}{13} \, and \, \frac{3}{13},\frac{-4}{13},\frac{12}{13} \, \, we \, obtain
l1l2 + m1m2 + n1n2 = \frac{4}{13}*\frac{3}{13}+\frac{12}{13}*(\frac{-4}{13})+\frac{3}{13}*\frac{12}{13}
= \frac{12}{169} - \frac{48}{169} + \frac{36}{169} = 0
Hence, the lines are perpendicular.
(iii) For the lines with direction cosines, \frac{3}{13},\frac{-4}{13},\frac{12}{13} \, and \, \frac{12}{13},\frac{-3}{13},\frac{-4}{13} \, \, we \, obtain
l1l2 + m1m2 + n1n2 = \frac{3}{13}*\frac{12}{13}+(\frac{-4}{13})*(\frac{-3}{13})+\frac{12}{13}*\frac{-4}{13}
= \frac{36}{169} + \frac{12}{169} - \frac{48}{169} = 0
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 3\hat{i}+2\hat{j}-2\hat{k}
Solution:
It is given that the line passes through the point A (1, 2, 3). Therefore, the position vector through
A is \vec{a} = \hat{i}+2\hat{j}+3\hat{k}
\vec{b} = 3\hat{i}+2\hat{j}-2\hat{k}
So, line passes through point A and parallel to \vec{b} is given by
\vec{r} = \vec{a}+\lambda\vec{b} , where \lambda is a constant
=> \vec{r} = \hat{i}+2\hat{j}+3\hat{k} + \lambda(3\hat{i}+2\hat{j}-2\hat{k})
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 2\hat{i}-\hat{j}+4\hat{k} and is in the direction \hat{i}+2\hat{j}-\hat{k}
Solution:
It is given that the line passes through the point with positive vector
\vec{a} = 2\hat{i}-\hat{j}+4\hat{k} ..............(i)
\vec{b} = \hat{i}+2\hat{j}-\hat{k} ...........(ii)
So, line passes through point A and parallel to \vec{b} is given by
=> \vec{r} = \vec{a}+\lambda\vec{b} ,where \lambda is a constant
\vec{r} = 2\hat{i}-\hat{j}+4\hat{k} + \lambda(\hat{i}+2\hat{j}-\hat{k})
This is the required equation of the line in vector form.
\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}
Eliminating \lambda, we get the Cartesian form equation as
\frac{x-2}{1} = \frac{y+1}{2} = \frac{z-4}{-1}
Question 6: Find the Cartesian equation of the line which passes through the point (-2, 4, -5) and parallel to the line given by \frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}
Solution:
It is given that the line passes through the point (-2, 4, -5) and is parallel to \frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}
Direction ratios of the line, \frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6} , are .3,5,6
Required line is parallel to \frac{x+3}{3} = \frac{y-4}{5} = \frac{z+8}{6}
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 \frac{x-x1}{a} = \frac{y-y1}{b} = \frac{z-z1}{c}
Hence, equation of the required line is
\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
Question 7: Cartesian equation of a line is \frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2} , Write its vector form.
Solution:
Cartesian equation of the line is given by
\frac{x-5}{3} = \frac{y+4}{7} = \frac{z-6}{2} ........(i)
The given line passes through the point (5, -4, 6). The position vector of this point is \vec{a} = 5\hat{i}-4\hat{j}+6\hat{k}
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,\vec{b} = 3\hat{i}-7\hat{j}+2\hat{k}
As we known that the line through positive vector \vec{a} and in the direction of the vector \vec{b} is given
by the equation, \vec{r} = \vec{a} + \lambda\vec{b},\lambda\in{R}
=> \vec{r} =(5\hat{i}-4\hat{j}+6\hat{k} ) + (3\hat{i}+7\hat{j}+2\hat{k} )
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,
\vec{a} = \vec{0} ..........(i)
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, \vec{b} = 5\hat{i}-2\hat{j}+3\hat{k}
The equation of the line in vector form through a point with position vector \vec{a}and parallel to \vec{b} is,
\vec{r} = \vec{a} + \lambda\vec{b},\lambda\in{R}
=> \vec{r} = \vec{0} + \lambda(5\hat{i}-2\hat{j}+3\hat{k})
The equation of the line through the point ( x1,y1,z1) and direction ratios a, b, c is given by,
\frac{x-x1}{a} = \frac{y-y1}{b} = \frac{z-z1}{c}
Hence, the equation of the required line in the Cartesian form is
\frac{x-0}{5} = \frac{y-0}{-2} = \frac{z-0}{3}=> \frac{x}{5} = \frac{y}{2} = \frac{z}{3}
Question 9: Find the angle between the following pairs of lines:
\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})
Solution:
Let θ be the angle between the given lines.
Angle between the given pairs of lines is given by,
\cos(Q) = \frac{\vec{b}1 \cdot \vec{b}2}{\lVert \vec{b}1 \rVert \cdot \lVert \vec{b}2 \rVert}
Given lines are parallel to the vectors,\vec{b}1 = 3\hat{i}+2\hat{j}+6\hat{k} , \vec{b}2 = \hat{i}+2\hat{j}+2\hat{k}
So, {\left |\vec{b}1 \right|} = \sqrt{3^2+2^2+6^2} = 7
{\left |\vec{b}2 \right|} = \sqrt{(1)^2+(2)^2+(2)^2} = 3
\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
=> cosQ = \frac{19}{7*3} => Q = \cos^{-1}( \frac{19}{21})
(ii) \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})
Solution:
Given line is parallel to the vectors, respectively \vec{b}1 = \hat{i}-\hat{j}-2\hat{k} , \vec{b}2 = 3\hat{i}-5\hat{j}-4\hat{k},respectively.
{\left |\vec{b}1 \right|} = \sqrt{(1)^2+(-1)^2+(-2)^2} = \sqrt{6}
{\left |\vec{b}2 \right|} = \sqrt{(3)^2+(-5)^2+(-4)^2} = \sqrt{50}=5\sqrt{2}
\vec{b}1 \cdot \vec{b}2 = (\hat{i}-\hat{j}-2\hat{k}) \cdot (3\hat{i}-5\hat{j}-4\hat{k})
= 1\cdot3 - 1(-5)-2(-4) = 3+5+8 = 16
\cos Q = \frac{\vec{b}1 \cdot \vec{b}2}{\lVert \vec{b}1 \rVert \cdot \lVert \vec{b}2 \rVert}
=> cosQ = \frac{16}{\sqrt{6}.5\sqrt{2}}
\frac{16}{10\sqrt{3}} => => Q = \cos^{-1}( \frac{18}{5\sqrt{3}})
Question 10: Find the angle between the following pair of lines:
(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}
Solution:
Let\, \vec{b}1 \, and\,\vec{b}2 \,be\, the\, vectors\, parallel\, to\, the \,pair \,of \,lines,
\,\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
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}
{\left |\vec{b}1 \right|} = \sqrt{(2)^2+(5)^2+(-3)^2} = \sqrt{38}
{\left |\vec{b}2 \right|} = \sqrt{(-1)^2+(8)^2+(4)^2} = \sqrt{81} = 9
\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})
=2(-1)+5*8+(-3).4 = -2+40-12 = 26
Angle, ?, between the given pair of lines is given by the relation,
\cos \theta = \frac{\vec{b}1 \cdot \vec{b}2}{\lVert \vec{b}1 \rVert \cdot \lVert \vec{b}2 \rVert}
=> cos\theta = \frac{26}{9\sqrt{38}}
=> \theta = \cos^{-1}( \frac{26}{9\sqrt{38}})
(ii \,) \,\frac{x}{2} = \frac{y}{2} = \frac{z}{1} \,and\, \frac{x-5}{4} = \frac{y-2}{1} =\frac{z-3}{8}
Solution:
Let \vec{b}1,\vec{b}2 be the vectors parallel to the given pair of lines,
\, \,\frac{x}{2} = \frac{y}{2} = \frac{z}{1} \,and\, \frac{x-5}{4} = \frac{y-2}{1} =\frac{z-3}{8},respectively.
\vec{b}_1 = 2\hat{i}+2\hat{j}+\hat{k} \,
\vec{b}_2 = 4\hat{i}+\hat{j}+8\hat{k}
{\left |\vec{b}_1 \right|} = \sqrt{(2)^2+(2)^2+(1)^2} = \sqrt{9} = 3
{\left |\vec{b}_2 \right|} = \sqrt{(4)^2+(1)^2+(8)^2} = \sqrt{81} = 9
\vec{b}1 \cdot \vec{b}2 = (2\hat{i}+2\hat{j}+\hat{k}) \cdot (4\hat{i}+\hat{j}+8\hat{k})
=> 2×4+2×1+1×8 = 8+2+8=18
Angle ?, between the given pair of lines, then
\cos \theta = \frac{\vec{b}1 \cdot \vec{b}2}{\lVert \vec{b}1 \rVert \cdot \lVert \vec{b}2 \rVert}
=> cos\theta = \frac{18}{{3*9}} = \frac{2}{{3}}
=> \theta = \cos^{-1}( \frac{2}{3})
Question 11: Find the values of p so that the lines \, \,\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.
Solution:
Given equations can be written in the standard form as
\, \,\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}
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 \, \,\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.
Solution:
Equation of the given lines are \frac{x-5}{7} = \frac{y+2}{-5} =\frac{z}{1} \,and\,\frac{x}{1} = \frac{y}{2} =\frac{z}{3}
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\vec{r} = \hat{i}+2\hat{j}+\hat{k} + \lambda(\hat{i}-\hat{j}+\hat{k}) \,and\,
\vec{r} = 2\hat{i}-\hat{j}-\hat{k} + \mu(2\hat{i}+\hat{j}+2\hat{k})
Solution:
Equation of the given lines are
\vec{r} = \hat{i}+2\hat{j}+\hat{k} + \lambda(\hat{i}-\hat{j}+\hat{k}) \,
\vec{r} = 2\hat{i}-\hat{j}-\hat{k} + \mu(2\hat{i}+\hat{j}+2\hat{k})
It is known that the shortest distance between the lines,
\vec{r} = \vec{a}1+\lambda \vec{b}1\,and\,\ \vec{r} = \vec{a}2+\mu \vec{b}2,is given by,
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)
Comparing the given equations, we get
\vec{a}_1 = \hat{i}+2\hat{j}+\hat{k}
\vec{b}_1 = \hat{i}-\hat{j}+\hat{k}
\vec{a}_2 = 2\hat{i}-\hat{j}-\hat{k}
\vec{b}_2 = 2\hat{i}+\hat{j}+2\hat{k}
\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\times\vec{b}_2 = \begin{vmatrix} \hat{i} &\hat{j} & \hat{k} \\ 1 & -1 & 1\\ 2 & 1 & 2 \end{vmatrix}
\vec{b}_1\times\vec{b}_2 = (-2-1)\hat{i}-(2-2)\hat{j}+(1+3)\hat{k}
=> \left |\vec{b}1\times\vec{b}2 \right| = \sqrt{(-3)^2+(3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}
Substituting all the values in equation (1), we obtain
d = \frac{|(-3\hat{i}+3\hat{k}) \cdot (\hat{i}-3\hat{j}-2\hat{k})}{|3\sqrt{2}|}
=> d = \left |-3.1+3(-2) /3\sqrt{2} \right|
=> d = \left |-9 / 3\sqrt{2} \right|
=> d = \frac{3\sqrt{}2}{2}
Therefore, shortest distance between the two lines is \frac{3\sqrt{}2}{2} units
Question 14: Find the shortest distance between the lines
\, \,\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}
Solution:
Given lines are \, \,\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}
It is known that the shortest distance between the two lines,
\frac{x-x_1}{a_1} = \frac{y-y_1}{b_1} = \frac{z-z_1}{c_1} \, and
\frac{x-x_2}{a_2} = \frac{y-y_2}{b_2} = \frac{z-z_2}{c_2} \,is\, given\, by,
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)
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, \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}
\begin{vmatrix} 4 & 6 & 8 \\ 7 & -6 & 1\\ 1 & -2 & 1 \end{vmatrix}
= 4(-6+2) - 6(7-1) + 8(-14+6)
= -16 - 36 - 64 = -116
=> \sqrt{(b_1c_2-b_2c_1)^2+(c_1a_2-c_2a_1)^2+(a_1b_2-a_2b_1)^2}
=> \sqrt{(-6+2)^2+(1+7)^2+(-14+6)^2}
=> \sqrt{16+36+64} = \sqrt{116} = 2\sqrt{29}
=> d = \frac{-116}{2\sqrt{29}} = \frac{-58}{\sqrt{29}} = -2\sqrt{29}
Thus, shortest distance between the lines whose vector equation are 2\sqrt{29} units
Question 15: \vec{r} = \hat{i}+2\hat{j}+3\hat{k} + \lambda(\hat{i}-3\hat{j}+2\hat{k}) \,
\vec{r} = 4\hat{i}+5\hat{j}+6\hat{k} + \mu(2\hat{i}+3\hat{j}+\hat{k})
Solution:
Given lines are \vec{r} = \hat{i}+2\hat{j}+3\hat{k} + \lambda(\hat{i}-3\hat{j}+2\hat{k}) \, and
\vec{r} = 4\hat{i}+5\hat{j}+6\hat{k} + \mu(2\hat{i}+3\hat{j}+\hat{k})
It is known that the shortest distance between the lines,
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)
Comparing the given equations, we get
\vec{a}_1 = \hat{i}+2\hat{j}+3\hat{k}
\vec{b}_1 = \hat{i}+-3\hat{j}+2\hat{k}
\vec{a}_2 = 4\hat{i}+5\hat{j}+6\hat{k}
\vec{b}_2 = 2\hat{i}+3\hat{j}+\hat{k}
\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}
\vec{b}_1\times\vec{b}_2 = \begin{vmatrix} \hat{i} &\hat{j} & \hat{k} \\ 1 & -3 & 2\\ 2 & 3 & 1 \end{vmatrix}
\vec{b}_1\times\vec{b}_2 = (-3-6)\hat{i}-(1-4)\hat{j}+(3+6)\hat{k}
-9\hat{i}+3\hat{j}+9\hat{k}
=> \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}
(\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})
= -9×3 + 3×3 + 9×3 = 9
Substituting all the values in equation (1), we obtain
=> d = \left| \frac{9}{3\sqrt{19}} \right|
=> d = \left| \frac{3}{\sqrt{19}} \right|
Therefore, shortest distance between the two given line is \left| \frac{3}{\sqrt{19}} \right| \,units.
Question 16 Find the shortest distance between the lines whose vector equations are
\vec{r} = (1-t)\hat{i}+(t-2)\hat{j}+(3-2t)\hat{k} \,and
\vec{r} = (s+1)\hat{i}+(2s-1)\hat{j}-(2s+1)\hat{k} \,
Solution:
Given lines are
\vec{r} = (1-t)\hat{i}+(t-2)\hat{j}+(3-2t)\hat{k} \,and
=> \vec{r} = (\hat{i}-2\hat{j}+3\hat{k} )\,+t(-\hat{i}+\hat{j}-2\hat{k})
\vec{r} = (s+1)\hat{i}+(2s-1)\hat{j}-(2s+1)\hat{k} \,
=> \vec{r} = (\hat{i}-\hat{j}+\hat{k} )\,+s(\hat{i}+2\hat{j}-2\hat{k})
It is known that the shortest distance between the lines,
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)
For the given equations,
\vec{a}_1 = \hat{i}-2\hat{j}+3\hat{k}
\vec{b}_1 = -\hat{i}+\hat{j}-2\hat{k}
\vec{a}_2 = \hat{i}-\hat{j}-\hat{k}
\vec{b}_2 = \hat{i}+2\hat{j}-2\hat{k}
\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}
\vec{b}_1\times\vec{b}_2 = \begin{vmatrix} \hat{i} &\hat{j} & \hat{k} \\ -1 & 1 & -2\\ 1 & 2 & -2 \end{vmatrix}
=> ( 2+4)\hat{i}-(2+2)\hat{j}+(-2-1)\hat{k} = 2\hat{i}-4\hat{j}-3\hat{k}
=> \left |\vec{b}_1\times\vec{b}_2 \right| = \sqrt{(2)^2+(-4)^2+(-3)^2} = \sqrt{4+16+9} = \sqrt{29}
(\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
Substituting all the values in equation (1), we obtain
=> d = \left| \frac{8}{\sqrt{29}} \right|
Therefore, the shortest distance between the two given line is \frac{8}{\sqrt{29}} ,units.
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Chapter 11 of the Class 12 NCERT Mathematics Part II textbook, "Three Dimensional Geometry," introduces concepts such as distances, angles, and equations of planes in three-dimensional space. Exercise 11.2 focuses on solving problems related to these concepts, including finding distances from points to planes, the equations of lines and planes, and the volume of parallelepipeds. Detailed solutions are provided to help students understand and apply these geometric principles effectively.
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