12th NCERT Three Dimensional Geometry Exercise 11.2 Questions 17
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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 vectors becomes perpendicular to each other when dot product of them is zero.
Here there are three vectors. If the product of every pair two vectors is equal to zero, then vectors are mutually perpendicular.
. let \[\overrightarrow a = \frac{{12}}{{13}},\frac{{ - 3}}{{13}},\frac{{ - 4}}{{13}},\overrightarrow b = \frac{4}{{13}},\frac{{12}}{{13}},\frac{3}{{13}},\;\overrightarrow c = \frac{3}{{13}},\frac{{ - 4}}{{13}},\frac{{12}}{{13}}\] \[\overrightarrow a \cdot \overrightarrow b = \left( {\frac{{12}}{{13}},\frac{{ - 3}}{{13}},\frac{{ - 4}}{{13}}} \right) \cdot \left( {\frac{4}{{13}},\frac{{12}}{{13}},\frac{3}{{13}}} \right)\] \[ = \frac{{48 - 36 - 12}}{{169}} = 0\] \[ \Rightarrow \overrightarrow a \bot \overrightarrow b \] Now \[\overrightarrow b \cdot \overrightarrow c = \left( {\frac{4}{{13}},\frac{{12}}{{13}},\frac{3}{{13}}} \right) \cdot \left( {\frac{3}{{13}},\frac{{ - 4}}{{13}},\frac{{12}}{{13}}} \right)\] \[ = \frac{{12 - 48 + 36}}{{169}} = 0\] \[ \Rightarrow \overrightarrow b \bot \overrightarrow c \] \[\overrightarrow a \cdot \overrightarrow c = \left( {\frac{{12}}{{13}},\frac{{ - 3}}{{13}},\frac{{ - 4}}{{13}}} \right) \cdot \left( {\frac{3}{{13}},\frac{{ - 4}}{{13}},\frac{{12}}{{13}}} \right)\] \[ = \frac{{36 + 12 - 48 + }}{{169}} = 0\] \[ \Rightarrow \overrightarrow a \bot \overrightarrow c \]
As every pair of vector is perpendicular , vectors are mutually perpendicular.

Question (2)

Show that the line throurh the points (1, -1, 2),(3, 4, -2) is perpendicular to the line through the points (0, 3, 2) and (3, 5, 6).

Solution

The two lines become perpendicular to each other, if dot product of the direction vectors of these lines is zero.
So first we will find the direction vectors of these lines.
\[let\;A{\rm{(1, - 1, 2),B(3, 4, - 2)}}\] \[\overrightarrow {AB} = \overline b - \overline a \] \[{\rm{ = (3, 4, - 2) - (1, - 1, 2) = ( 2,5, - 4)}}\] \[let\;P{\rm{(0, 3, 2) and Q(3, 5, 6)}}{\rm{.}}\] \[\overrightarrow {PQ} = \overline q - \overline p \] \[{\rm{ = (3, 5, 6) - (0, 3, 2) = }}\;{\rm{( 3,}}\,{\rm{2,}}\,{\rm{4)}}\] \[\overrightarrow {AB} \cdot \overrightarrow {PQ} \] \[{\rm{ = }}\;{\rm{( 3,}}\,{\rm{2,}}\,{\rm{4) }}{\rm{.( 2,}}\,{\rm{5, - 4) = 6 + 10 - 16 = 0}}\] \[ \Rightarrow \overrightarrow {AB} \bot \overrightarrow {PQ} \] So the vectors are perpendicular, so lines containing vectors are also perpendicular.

Question (3)

Show that the line through (4, 7, 8), (2, 3, 4) is parallel to the line through the points (-1, -2, 1),(1, 2, 5).

Solution

The two lines are parallel if the direction vectors of these lines are in propertion( one vector is multiple of the other). or cross product between them is zero.
So first we will find direction vectors of these lines which passes through the given points.
Let A(4, 7, 8), B(2, 3, 4),
\[\overrightarrow {AB} = \overline b - \overline a \] \[{\rm{ = (2, 3, 4) - (4, 7, 8)}}\] \[{\rm{ = ( - 2, - 4, - 4)}}\] Let P(-1, -2, 1), Q(1, 2, 5) \[\overrightarrow {PQ} = \overline q - \overline p \] \[{\rm{(1, 2, 5) - ( - 1, - 2, 1)}}\] \[{\rm{ = (2, 4, 4) }}\] \[{\rm{( - 2, - 4, - 4) = - 1 (2, 4, 4) }}\] \[\overrightarrow {AB} = - \overrightarrow {PQ} \] So vectors are parallel, so lines containing these vectors are parallel.

Question (4)

Find the equation of the line which passes through the points (1, 2, 3) and is parallel to the vector \[3\widehat i + 2\widehat j - 2\widehat k\]

Solution

The vector equation of the line is given by \[\overline r = \overline a + \lambda \overline b ,\lambda \in R\] where vector a is the passing point and vector b is the direction vector.
here \[\overrightarrow a = \widehat i + 2\widehat j + 3\widehat k\;and\,\overrightarrow b = \widehat i + 2\widehat j - 2\widehat k\] So the vector equation of the line is given by \[\overrightarrow r = \widehat i + 2\widehat j + 3\widehat k + \lambda \left( {\widehat i + 2\widehat j - 2\widehat k} \right),\lambda \in R\]

Question (5)

Find the equation of the line in vectore and cartesian form that passes through the point with positions vector \[2\widehat i - \widehat j + 4\widehat k\] and is in the direction \[\widehat i + 2\widehat j - \widehat k\]

Solution

The vector equation of the line is given by \[\overline r = \overline a + \lambda \overline b ,\lambda \in R\] where vector a is the passing point and vector b is the direction vector.
Here \[\overrightarrow a = 2\widehat i - \widehat j + 4\widehat k\;and\,\overrightarrow b = \widehat i + 2\widehat j - \widehat k\] So vector equation of the line is given by, \[\overrightarrow r = 2\widehat i - \widehat j + 4\widehat k + \lambda \left( {\widehat i + 2\widehat j - \widehat k} \right),\lambda \in R\]
The cartesian equation of the line is given by \[\frac{{x - {x_1}}}{l} = \frac{{y - {y_1}}}{m} = \frac{{z - {z_1}}}{n}\] where (x1, y1, z1) is the passing point and (l, m, n) is the direction of the line.
So (x1, y1, z1) = (2, -1, 4) and (l, m, n) = ( 1, 2, -1) So the cartesian equation of the line is given by \[\frac{{x - 2}}{1} = \frac{{y + 1}}{4} = \frac{{z - 4}}{{ - 1}}\]

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

The cartesian equation of the line is given by \[\frac{{x - {x_1}}}{l} = \frac{{y - {y_1}}}{m} = \frac{{z - {z_1}}}{n}\] where (x1, y1, z1) is the passing point and (l, m, n) is the direction of the line.
The required line is parallel to given line , so direction of the required line is same of the given line
The direction of the given line is ( 3, 5, 6). The line is passing through ( -2, 4, -5),
So (x1, y1, z1) = (-2, 4, -5) and (l, m, n) = ( 3, 5, 6) So the cartesian equation of the line is given by \[\frac{{x + 2}}{3} = \frac{{y -4}}{5} = \frac{{z +5 }}{{ 6}}\]

Question (7)

The cartesian equation of the line is \[\frac{{x - 5}}{3} = \frac{{y + 4}}{7} = \frac{{z - 6}}{2}\]. Write its vector form.

Solution

The vector equation of the line is given by \[\overline r = \overline a + \lambda \overline b ,\lambda \in R\] where vector a is the passing point and vector b is the direction vector.
The given line is passing through (x1, y1, z1) = (5,-4, 6) and direction vector is (l, m, n) = ( 3, 7, 2)
So the vector equation of the line is \[\overrightarrow r = \left( {5\widehat i - 4\widehat j + 6\widehat k} \right) + \lambda \left( {\widehat {3i} + 7\widehat j + 2\widehat k} \right),\lambda \in R\]

Question (8)

Find the vector and cartesian equations of the lines that passes through the origin and (5, -2, 3).

Solution

The vector equation of the line is given by \[\overline r = \overline a + \lambda \overline b ,\lambda \in R\] where vector a is the passing point and vector b is the direction vector.
The given line is passing through origin so (x1, y1, z1) = (0, 0, 0).
The line is passes through the other point ( 5, -2, 3), so the direction vector = ( 5, -2, 3)- (0, 0, 0) = ( 5, -2, 3)
So the vector equation of the line is \[\overrightarrow r = \lambda \left( {5\widehat i - 2\widehat j + 3\widehat k} \right),\lambda \in R\]
The cartesian equation of the line passing through two points (x1, y1, z1) and (x2, y2, z2) is given by \[\frac{{x - {x_1}}}{{{x_2} - {x_1}}} = \frac{{y - {y_1}}}{{{y_2} - {y_1}}} = \frac{{z - {z_1}}}{{{z_2} - {z_1}}}\]
Here (x1, y1, z1) = (0, 0, 0) and (x2, y2, z2) = (5, -2, 3).
So the cartesian equation of the line passing through the given points is \[\frac{{x - 0}}{{5 - 0}} = \frac{{y - 0}}{{ - 2 - 0}} = \frac{{z - 0}}{{3 - 0}}\] \[\frac{x}{5} = \frac{y}{{ - 2}} = \frac{z}{3}\].

Question (9)

Find the vector and the cartesian equations of the line that passes through the points (3, -2, -5),(3, -2, 6).

Solution

The vector equation of the line is given by \[\overline r = \overline a + \lambda \overline b ,\lambda \in R\] where vector a is the passing point and vector b is the direction vector.
The given line is passing through (3, -2, -5),so (x1, y1, z1) =(3, -2, -5).
The line is passes through the other point (3, -2, 6), so the direction vector = (3, -2, 6) - (3, -2, -5) = ( 0, 0,11)
So the vector equation of the line is \[\overrightarrow r = \left( {3\widehat i - 2\widehat j - 5\widehat k} \right) + \lambda \left( { 11\widehat k} \right),\lambda \in R\]
The cartesian equation of the line passing through two points (x1, y1, z1) and (x2, y2, z2) is given by \[\frac{{x - {x_1}}}{{{x_2} - {x_1}}} = \frac{{y - {y_1}}}{{{y_2} - {y_1}}} = \frac{{z - {z_1}}}{{{z_2} - {z_1}}}\]
Here (x1, y1, z1) = (3, -2, -5) and (x2, y2, z2) = (3, -2, 6).
So the cartesian equation of the line passing through the given points is \[\frac{{x - 3}}{{3 - 3}} = \frac{{y + 2}}{{ - 2 + 0}} = \frac{{z + 5}}{{-5 - 6}}\] \[\;\frac{{x - 3}}{0} = \frac{{y + 2}}{{ - 0}} = \frac{{z + 5}}{{11}}\]

Question (10)

Find the angles between the following pairs of lines: (i) \[\;\overrightarrow r = 2\widehat i - 5\widehat j + \widehat k + \lambda (3\widehat i + 2\widehat j + 6\widehat k)\] and \[\overrightarrow r = 7\widehat i - 6\widehat k + \mu (\widehat i + 2\widehat j + 2\widehat k)\]

Solution

If the angle between two vectors is θ,then \[\cos \theta = \frac{{\overline a \cdot \overline b }}{{\left| {\overline a } \right|\left| {\overline b } \right|}}\]
The angle between two lines is the same as angle between the direction vectors of the lines.
Here vector b1 is the direction vector of line l1, so\[\overrightarrow {{b_1}} = (3\hat i + 2\hat j + 6\hat k)\] Here vector b2 is the direction vector of line l2, so \[\overrightarrow {{b_2}} = (1\hat i + 2\hat j + 2\hat k)\] \[\left| {\overline {{b_1}} } \right| = \sqrt {{3^2} + {2^2} + {6^2}} = \sqrt {9 + 4 + 36} = \sqrt {49} = 7\] \[\left| {\overline {{b_2}} } \right| = \sqrt {{1^2} + {2^2} + {2^2}} = \sqrt {1 + 4 + 4} = \sqrt 9 = 3\] \[\overrightarrow {{b_1}} \cdot \overrightarrow {{b_2}} {\rm{ = }}\;{\rm{( 3,}}\,{\rm{2,}}\,6{\rm{) }}{\rm{.( 1,}}\,2{\rm{, 2) = 3 + 4 + 12 = 19}}\] \[\cos \theta = \frac{{\overline {{b_1}} \cdot \overline {{b_2}} }}{{\left| {\overline {{b_1}} } \right|\left| {\overline {{b_2}} } \right|}}\] \[\cos \theta = \frac{{19}}{{21}} \Rightarrow \theta = {\cos ^{ - 1}}\left( {\frac{{19}}{{21}}} \right)\] (ii) \[\overrightarrow r = 3\widehat i + \widehat j - 2\widehat k + \lambda (\widehat i - \widehat j - 2\widehat k)\] and \[\overrightarrow r = 2\widehat i - \widehat j - 56\widehat k + \mu (\widehat {3i} - 5\widehat j - 4\widehat k)\]

Solution

If the angle between two vectors is θ,then \[\cos \theta = \frac{{\overline a \cdot \overline b }}{{\left| {\overline a } \right|\left| {\overline b } \right|}}\]
The angle between two lines is the same as angle between the direction vectors of the lines.
Here vector b1 is the direction vector of line l1, so\[\overrightarrow {{b_1}} = (1\hat i - 1\hat j - 2\hat k)\] Here vector b2 is the direction vector of line l2, so \[\overrightarrow {{b_2}} = (3\hat i - 5\hat j - 4\hat k)\] \[\left| {\overline {{b_1}} } \right| = \sqrt {{1^2} + {-1^2} + {-2^2}} = \sqrt {1 + 1 + 4} = \sqrt {6} \] \[\left| {\overline {{b_2}} } \right| = \sqrt {{3^2} + {-5^2} + {-4^2}} = \sqrt {9 + 25 + 16} = \sqrt{50} =5\sqrt{2}\] \[\overrightarrow {{b_1}} \cdot \overrightarrow {{b_2}} {\rm{ = }}\;{\rm{( 1,}}\, - 1{\rm{,}}\, - 2{\rm{) }}{\rm{.( 3,}}\, - 5{\rm{, - 4) = 3 + 5 + 8 = 16}}\] \[\cos \theta = \frac{{\overline {{b_1}} \cdot \overline {{b_2}} }}{{\left| {\overline {{b_1}} } \right|\left| {\overline {{b_2}} } \right|}}\] \[\cos \theta = \frac{{16}}{{5\sqrt 12 }} \Rightarrow \theta = {\cos ^{ - 1}}\left( {\frac{{8}}{{5\sqrt 3 }}} \right)\]

Question (11)

Find the angle between the following pais 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

If the angle between two vectors is θ,then \[\cos \theta = \frac{{\overline a \cdot \overline b }}{{\left| {\overline a } \right|\left| {\overline b } \right|}}\]
The angle between two lines is the same as angle between the direction vectors of the lines.
Here vector b1 is the direction vector of line l1, so\[\overrightarrow {{b_1}} = (2\hat i + 5\hat j - 3\hat k)\] Here vector b2 is the direction vector of line l2, so \[\overrightarrow {{b_2}} = (-1\hat i + 8\hat j + 4\hat k)\] \[\left| {\overline {{b_1}} } \right| = \sqrt {{2^2} + {5^2} + - {3^2}} = \sqrt {4 + 25 + 9} = \sqrt {38} \] \[\left| {\overline {{b_2}} } \right| = \sqrt { - {1^2} + {8^2} + {4^2}} = \sqrt {1 + 64 + 16} = \sqrt {81} = 9\] \[\overrightarrow {{b_1}} \cdot \overrightarrow {{b_2}} = \left( {2,5, - 3} \right) \cdot \left( { - 1,8,4} \right) = - 2 + 40 - 12 = 26\] \[\cos \theta = \frac{{\overrightarrow {{b_1}} \cdot \overrightarrow {{b_2}} }}{{\left| {\overline {{b_1}} } \right|\left| {\overline {{b_2}} } \right|}} = \frac{{26}}{{9\sqrt {38} }}\] \[ \Rightarrow \theta = {\cos ^{ - 1}}\left( {\frac{{26}}{{9\sqrt {38} }}} \right)\] (ii) \[\frac{x}{2} = \frac{y}{2} = \frac{z}{1}\] and \[\frac{{x - 5}}{4} = \frac{{y - 2}}{1} = \frac{{z - 3}}{8}\]

Solution

If the angle between two vectors is θ,then \[\cos \theta = \frac{{\overline a \cdot \overline b }}{{\left| {\overline a } \right|\left| {\overline b } \right|}}\]
The angle between two lines is the same as angle between the direction vectors of the lines.
Here vector b1 is the direction vector of line l1, so\[\overrightarrow {{b_1}} = (2\hat i + 2\hat j + 1\hat k)\] Here vector b2 is the direction vector of line l2, so \[\overrightarrow {{b_2}} = (4\hat i + 1\hat j + 8\hat k)\] \[\left| {\overline {{b_1}} } \right| = \sqrt {{2^2} + {2^2} + {1^2}} = \sqrt {4 + 4 + 1} = \sqrt {9} = 3 \] \[\left| {\overline {{b_2}} } \right| = \sqrt { {4^2} + {1^2} + {8^2}} = \sqrt {16 + 1 + 64} = \sqrt {81} = 9\] \[\overrightarrow {{b_1}} \cdot \overrightarrow {{b_2}} = \left( {2,2, 1} \right) \cdot \left( { 4,1,8} \right) = - 8 + 2 + 8 = 18\] \[\cos \theta = \frac{{\overrightarrow {{b_1}} \cdot \overrightarrow {{b_2}} }}{{\left| {\overline {{b_1}} } \right|\left| {\overline {{b_2}} } \right|}} = \frac{{18}}{{9(3)}} = \frac{2}{3}\] \[ \Rightarrow \theta = {\cos ^{ - 1}}\left( {\frac{2}{3}} \right)\]

Question (12)

Find the value 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 angles.

Solution

First we will represent the given lines in standard cartesian form \[\frac{{x - {x_1}}}{l} = \frac{{y - {y_1}}}{m} = \frac{{z - {z_1}}}{n}\] and find out the direction vectors.
\[\frac{{1 - x}}{3} = \frac{{7y - 14}}{{2p}} = \frac{{z - 3}}{2}\] \[ \Rightarrow \frac{{x - 1}}{{ - 3}} = \frac{{y - 2}}{{2p/7}} = \frac{{z - 3}}{2}\] \[\overrightarrow {{b_1}} = \left( { - 3,\frac{{2p}}{7},2} \right)\] \[\frac{{7 - 7x}}{{3p}} = \frac{{y - 5}}{1} = \frac{{6 - z}}{5}\] \[ \Rightarrow \frac{{x - 1}}{{ - 3p/7}} = \frac{{y - 5}}{1} = \frac{{z - 6}}{{ - 5}}\] \[\overrightarrow {{b_2}} = \left( {\frac{{ - 3p}}{7},1, - 5} \right)\] The lines are perpendicular, so dot product between the direction vector is zero.
\[\overrightarrow {{b_1}} \cdot \overrightarrow {{b_2}} = 0\] \[\left( { - 3,\frac{{2p}}{7},2} \right) \cdot \left( {\frac{{ - 3p}}{7},1, - 5} \right) = 0\] \[\frac{{9p}}{7} + \frac{{2p}}{7} - 10 = 0\] \[\frac{{11}}{7}p = 10 \Rightarrow p = \frac{{70}}{{11}}\]

Question (13)

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 perpendicular to each other.

Solution

\[\overrightarrow {{b_1}} = \left( {7, - 5,1} \right),\quad \overrightarrow {{b_2}} = \left( {1,2,3} \right)\] \[\overrightarrow {{b_1}} \cdot \overrightarrow {{b_2}} = \left( {7, - 5,1} \right) \cdot \left( {1,2,3} \right) = 7 - 10 + 3 = 0\] \[ \Rightarrow \overrightarrow {{b_1}} \bot \overrightarrow {{b_2}} \] So the lines are perpendicular to each other.

Question (14)

Find the shortest distance between the lines \[\overrightarrow r = \widehat i + \widehat {2j} + \widehat k + \lambda (\widehat i - \widehat j + \widehat k)\] and \[\overrightarrow r = 2\widehat i - \widehat j - \widehat k + \mu (2\widehat i + \widehat j + 2\widehat k)\]

Solution

\[{l_1}:\quad \vec r = \hat i + \widehat {2j} + \hat k + \lambda (\hat i - \hat j + \hat k)\] \[so,\;\overrightarrow {{a_1}} = (\widehat i + 2\widehat j + \widehat k)\;and\quad \overrightarrow {{b_1}} = \widehat i - \widehat j + \widehat k\] \[{l_2}:\;\;\vec r = 2\hat i - \hat j - \hat k + \mu (2\hat i + \hat j + 2\hat k)\] \[so,\;\;\,\overrightarrow {{a_2}} = (2\widehat i - \widehat j - \widehat k)\;and\quad \overrightarrow {{b_2}} = 2\widehat i + \widehat j + 2\widehat k\] Calculate b1 x b2 as \[\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} = \left| {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\1&{ - 1}&1\\2&1&2\end{array}} \right| = - 3\widehat i + 3\widehat k \ne \overline 0 \] Since b1 x b2 ≠ 0 , lines are either intersecting or skew. \[\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right| = \sqrt {9 + 9} = \sqrt {18} = 3\sqrt 2 \] \[\overrightarrow {{a_2}} - \overrightarrow {{a_1}} = (2\widehat i - \widehat j - \widehat k) - (\widehat i + 2\widehat j + \widehat k) = \widehat i - 3\widehat j - 2\widehat k\] \[\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \cdot \left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) = \left( {\widehat i - 3\widehat j - 2\widehat k} \right) \cdot \left( { - 3\widehat i + 3\widehat k} \right) = - 3 - 6 = - 9 \ne 0\] The lines are skew. The shortest perpendicular distance be 'p' , then \[p = \frac{{\left| {\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \cdot \left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right)} \right|}}{{\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}} = \frac{{| - 9|}}{{3\sqrt 2 }} = \frac{3}{{\sqrt 2 }} = \frac{{3\sqrt 2 }}{2}\]

Question (15)

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

\[{l_1}:\quad \frac{{x + 1}}{7} = \frac{{y + 1}}{{ - 6}} = \frac{{z + 1}}{1}\] \[so,\;\overrightarrow {{a_1}} = ( - \widehat i - \widehat j - \widehat k)\;and\quad \overrightarrow {{b_1}} = 7\widehat i - 6\widehat j + \widehat k\] \[{l_2}:\;\;\frac{{x - 3}}{1} = \frac{{y - 5}}{{ - 2}} = \frac{{z - 7}}{1}\] \[so,\;\;\,\overrightarrow {{a_2}} = (3\widehat i + 5\widehat j + 7\widehat k)\;and\quad \overrightarrow {{b_2}} = \widehat i - 2\widehat j + \widehat k\] Calculate b1 x b2 as \[\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} = \left| {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\7&{ - 6}&1\\1&{ - 2}&1\end{array}} \right| = - 4\widehat i - 6\widehat j - 8\widehat k \ne \overline 0 \] Since b1 x b2 ≠ 0 , lines are either intersecting or skew. \[\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right| = \sqrt {16 + 36 + 64} = \sqrt {116} = 2\sqrt {29} \] \[\overrightarrow {{a_2}} - \overrightarrow {{a_1}} = (3\widehat i + 5\widehat j + 7\widehat k) - ( - \widehat i - \widehat j - \widehat k) = 4\widehat i + 6\widehat j + 8\widehat k\] \[\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \cdot \left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) = \left( {4\widehat i + 6\widehat j + 8\widehat k} \right) \cdot \left( { - 4\widehat i - 6\widehat j - 8\widehat k} \right) = - 16 - 36 - 64 = - 116 \ne 0\] The lines are skew. The shortest perpendicular distance be 'p' , then \[p = \frac{{\left| {\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \cdot \left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right)} \right|}}{{\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}} = \frac{{| - 116|}}{{\sqrt {116} }} = \sqrt {116} = 2\sqrt {29} \]

Question (16)

Find the shortest distance between the lines whose vector equations are \[\;\overrightarrow r = \widehat i + \widehat {2j} + 3\widehat k + \lambda (\widehat i - 3\widehat j + 2\widehat k)\] and \[\overrightarrow r = 4\widehat i + 5\widehat j + 6\widehat k + \mu (2\widehat i + 3\widehat j + \widehat k)\]

Solution

\[{l_1}:\quad \;\vec r = \hat i + 2\widehat j + 3\hat k + \lambda (\hat i - 3\hat j + 2\hat k)\] \[so,\;\overrightarrow {{a_1}} = (\widehat i + 2\widehat j + 3\widehat k)\;and\quad \overrightarrow {{b_1}} = \widehat i - 3\widehat j + 2\widehat k\] \[{l_2}:\;\;\vec r = 4\hat i + 5\hat j + 6\hat k + \mu (2\hat i + 3\hat j + \hat k)\] \[so,\;\;\,\overrightarrow {{a_2}} = (4\widehat i + 5\widehat j + 6\widehat k)\;and\quad \overrightarrow {{b_2}} = 2\widehat i + 3\widehat j + \widehat k\] Calculate b1 x b2 as \[\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} = \left| {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\1&{ - 3}&2\\2&3&1\end{array}} \right| = - 9\widehat i + 3\widehat j + 9\widehat k \ne \overline 0 \] Since b1 x b2 ≠ 0 , lines are either intersecting or skew. \[\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right| = \sqrt {81 + 9 + 81} = \sqrt {171} = 3\sqrt {19} \] \[\overrightarrow {{a_2}} - \overrightarrow {{a_1}} = (4\widehat i + 5\widehat j + 6\widehat k) - (\widehat i + 2\widehat j + 3\widehat k) = 3\widehat i + 3\widehat j + 3\widehat k\] \[\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \cdot \left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) = \left( {3\widehat i + 3\widehat j + 3\widehat k} \right) \cdot \left( { - 9\widehat i + 3\widehat j + 9\widehat k} \right) = - 27 + 9 + 27 = 9 \ne 0\] The lines are skew. The shortest perpendicular distance be 'p' , then \[p = \frac{{\left| {\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \cdot \left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right)} \right|}}{{\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}} = \frac{{|9|}}{{3\sqrt {19} }} = \frac{3}{{\sqrt {19} }}\]

Question (17)

Find the shortest distance between the lines whose vector equations are \[\;\overrightarrow r = (1 - t)\widehat i + (t - 2)\widehat j + (3 - 2t)\widehat k\] and \[\;\overrightarrow r = (s + 1)\widehat i + (2s - 1)\widehat j - (2s + 1)\widehat k\]

Solution

\[\vec r = (1 - t)\hat i + (t - 2)\hat j + (3 - 2t)\hat k\] \[\vec r\quad = \hat i - t\hat i + t\hat j - 2\hat j + 3\hat k - 2t\hat k\] \[\vec r = \hat i - 2\hat j + 3\hat k + t( - \hat i + \hat j - 2\hat k)\] \[so,\;\overrightarrow {{a_1}} = (\widehat i - 2\widehat j + 3\widehat k)\;and\quad \overrightarrow {{b_1}} = - \widehat i + \widehat j - 2\widehat k\] \[{l_2}:\;\;\;\vec r = (s + 1)\hat i + (2s - 1)\hat j - (2s + 1)\hat k\] \[\vec r = s\hat i + \hat i + 2s\hat j - \hat j - 2s\hat k - \hat k\] \[\vec r = \hat i - \hat j - \hat k + s(\hat i + 2\hat j - 2\hat k)\] \[so,\;\;\,\overrightarrow {{a_2}} = (\widehat i - \widehat j - \hat k)\;and\quad \overrightarrow {{b_2}} = \widehat i + 2\widehat j - 2\widehat k\] Calculate b1 x b2 as \[\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} = \left| {\begin{array}{*{20}{c}}{\widehat i}&{\widehat j}&{\widehat k}\\{ - 1}&1&{ - 2}\\1&2&{ - 2}\end{array}} \right| = 2\widehat i - 4\widehat j - 3\hat k \ne \overline 0 \] Since b1 x b2 ≠ 0 , lines are either intersecting or skew. \[\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right| = \sqrt {4 + 16 + 9 + } = \sqrt {29} \] \[\overrightarrow {{a_2}} - \overrightarrow {{a_1}} = (\widehat i - \widehat j - \widehat k) - (\widehat i - 2\widehat j + 3\widehat k) = \widehat j - 4\widehat k\] \[\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \cdot \left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right) = \left( {\widehat j - 4\widehat k} \right) \cdot \left( {2\widehat i - 4\widehat j - 3\widehat k} \right) = 0 - 4 + 12 = 8 \ne 0\] The lines are skew. The shortest perpendicular distance be 'p' , then \[p = \frac{{\left| {\left( {\overrightarrow {{a_2}} - \overrightarrow {{a_1}} } \right) \cdot \left( {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right)} \right|}}{{\left| {\overrightarrow {{b_1}} \times \overrightarrow {{b_2}} } \right|}} = \frac{{|8|}}{{\sqrt {29} }} = \frac{8}{{\sqrt {29} }}\]
Exercise 11.1 ⇐
⇒ Exercise 11.3