Miscellaneous Exercise Three Dimensional Geometry Solution - NCERT Class 12

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Chapter 11 Ex.11.ME Question 1

Show that the line joining the origin to the point \(\left( {2,1,1} \right)\) is perpendicular to the line determined by the points \(\left( {3,5, - 1} \right),\left( {4,3, - 1} \right)\).

Solution

Let OA be the line joining the origin \({\rm{O}}\left( {0,0,0} \right)\) and the point \(A\left( {2,1,1} \right)\).

Also, let BC be the line joining the points, \(B\left( {3,5, - 1} \right)\) and \(C\left( {4,3, - 1} \right)\).

The direction ratios of OA are \(2, 1\) and \(1\) and of \(BC\) are \(\left( {4 - 3} \right) = 1,\left( {3 - 5} \right) = - 2\) and \(\left( { - 1 + 1} \right) = 0\)

If, \(OA \bot BC \Rightarrow {a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0\)

Here,

\[\begin{align}{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} &= 2 \times 1 + 1\left( { - 2} \right) + 1 \times 0\\ &= 2 - 2\\& = 0\end{align}\]

Thus, \(OA \bot BC\) proved.

Chapter 11 Ex.11.ME Question 2

If \({l_1},{m_1},{n_1}\) and \({l_2},{m_2},{n_2}\) are the direction cosines of two mutually perpendicular lines. Show that the direction cosines of the perpendicular to both of these are \({m_1}{n_2} - {m_2}{n_1},\;\;{n_1}{l_2} - {n_2}{l_1},\;\;{l_1}{m_2} - {l_2}{m_1}\).

Solution

\[\begin{align}&{l_1}{l_2} + {m_1}{m_2} + {n_1}{n_2} = 0\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\&{l_1}^2 + {m_1}^2 + {n_1}^2 = 1\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\\&{l_2}^2 + {m_2}^2 + {n_2}^2 = 1\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}\]

Let \(l,m,n\) be the direction cosines of the line which is perpendicular to the line with direction cosines \({l_1},{m_1},{n_1}\) and \({l_2},{m_2},{n_2}\).

Therefore,

\[\begin{align}&l{l_1} + m{m_1} + n{n_1} = 0\\l&{l_2} + m{m_2} + n{n_2} = 0\\ &\Rightarrow \frac{l}{{{m_1}{n_2} - {m_2}{n_1}}} = \frac{m}{{{n_1}{l_2} - {n_2}{l_1}}} = \frac{n}{{{l_1}{m_2} - {l_2}{m_1}}}\\ &\Rightarrow \frac{{{l^2}}}{{{{\left( {{m_1}{n_2} - {m_2}{n_1}} \right)}^2}}} = \frac{{{m^2}}}{{{{\left( {{n_1}{l_2} - {n_2}{l_1}} \right)}^2}}} = \frac{{{n^2}}}{{{{\left( {{l_1}{m_2} - {l_2}{m_1}} \right)}^2}}}\\ &\Rightarrow \frac{{{l^2} + {m^2} + {n^2}}}{{{{\left( {{m_1}{n_2} - {m_2}{n_1}} \right)}^2} + {{\left( {{n_1}{l_2} - {n_2}{l_1}} \right)}^2} + {{\left( {{l_1}{m_2} - {l_2}{m_1}} \right)}^2}}}\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 4 \right)\end{align}\]

Since, \(l,m,n\) are direction cosines of the line.

Hence,

\[{l^2} + {m^2} + {n^2} = 1\;\;\;\;\;\;\;\;\;\; \ldots \left( 5 \right)\]

As we know that,

\[\left( {{l_1}^2 + {m_1}^2 + {n_1}^2} \right)\left( {{l_2}^2 + {m_2}^2 + {n_2}^2} \right) - \left( {{l_1}{l_2} + {m_1}{m_2} + {n_1}{n_2}} \right) = {\left( {{m_1}{n_2} - {m_2}{n_1}} \right)^2} + {\left( {{n_1}{l_2} - {n_2}{l_1}} \right)^2} + {\left( {{l_1}{m_2} - {l_2}{m_1}} \right)^2}\]

Putting the values from (\(1\)), (\(2\)) and (\(3\)), we get

\[\begin{align} &\Rightarrow 1.1 - 0 = {\left( {{m_1}{n_2} - {m_2}{n_1}} \right)^2} + {\left( {{n_1}{l_2} - {n_2}{l_1}} \right)^2} + {\left( {{l_1}{m_2} - {l_2}{m_1}} \right)^2}\\ &\Rightarrow {\left( {{m_1}{n_2} - {m_2}{n_1}} \right)^2} + {\left( {{n_1}{l_2} - {n_2}{l_1}} \right)^2} + {\left( {{l_1}{m_2} - {l_2}{m_1}} \right)^2} = 1\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 6 \right)\end{align}\]

Putting the values from equation (\(5\)) and (\(6\)) in equation (\(4\)), we get

\[\frac{{{l^2} + {m^2} + {n^2}}}{{{{\left( {{m_1}{n_2} - {m_2}{n_1}} \right)}^2} + {{\left( {{n_1}{l_2} - {n_2}{l_1}} \right)}^2} + {{\left( {{l_1}{m_2} - {l_2}{m_1}} \right)}^2}}} = 1\]

Hence,

\[(\frac{{{l^2}}}{{{{\left( {{m_1}{n_2} - {m_2}{n_1}} \right)}^2}}} = \frac{{{m^2}}}{{{{\left( {{n_1}{l_2} - {n_2}{l_1}} \right)}^2}}} = \frac{{{n^2}}}{{{{\left( {{l_1}{m_2} - {l_2}{m_1}} \right)}^2}}} = 1\]

Therefore,

\[\begin{align}l& = {m_1}{n_2} - {m_2}{n_1}\\m &= {n_1}{l_2} - {n_2}{l_1}\\n &= {l_1}{m_2} - {l_2}{m_1}\end{align}\]

Hence, the direction cosines of the required line are \({m_1}{n_2} - {m_2}{n_1},\;\;{n_1}{l_2} - {n_2}{l_1},\;\;{l_1}{m_2} - {l_2}{m_1}\) proved.

Chapter 11 Ex.11.ME Question 3

Find the angle between the lines whose direction ratios are \(a,b,c\) and \(b - c,c - a,a - b\).

Solution

The angle \({\rm{\theta }}\) between the lines with direction cosines \(a,b,c\) and \(\left( {b - c} \right),\left( {c - a} \right),\left( {a - b} \right)\) is given by,

\[\begin{align}{\rm{cos\theta }} &= \left| {\frac{{a\left( {b - c} \right) + b\left( {c - a} \right) + c\left( {a - b} \right)}}{{\sqrt {{a^2} + {b^2} + {c^2}} .\sqrt {{{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2} + {{\left( {a - b} \right)}^2}} }}} \right|\\{\rm{\theta }} &= {\cos ^{ - 1}}\left| {\frac{{ab - ac + bc - ab + ac - bc}}{{\sqrt {{a^2} + {b^2} + {c^2}} .\sqrt {{{\left( {b - c} \right)}^2} + {{\left( {c - a} \right)}^2} + {{\left( {a - b} \right)}^2}} }}} \right|\\ &= {\cos ^{ - 1}}0\\ &= 90^\circ\end{align}\]

Thus, the required angle is \(90^\circ \)

Chapter 11 Ex.11.ME Question 4

Find the equation of a line parallel to \(x-\)axis and passing through the origin.

Solution

The line parallel to\(x-\)axis and passing through the origin is \(x-\)axis itself.

Let A be a point on \(x-\)axis.

Therefore, the coordinates of A are given by \(\left( {a,0,0} \right)\), where \(a \in R\)

Hence, the direction ratios of OA are \(a,0,0\)

The equation of OA is given by,

\[\begin{align} &\Rightarrow \frac{{x - a}}{0} = \frac{{y - 0}}{0} = \frac{{z - 0}}{0}\\ &\Rightarrow \frac{x}{1} = \frac{y}{0} = \frac{z}{0} = a\end{align}\]

Hence, the equation of line parallel to x-axis and passing origin is \(\frac{x}{1} = \frac{y}{0} = \frac{z}{0}\)

Chapter 11 Ex.11.ME Question 5

If the coordinates of the points \(A, B, C, D\) be \(\left( {1,2,3} \right),\;\left( {4,5,7} \right),\;\left( { - 4,3, - 6} \right)\) and \(\left( {2,9,2} \right)\) respectively, then find the angle between the lines \(AB\) and \(CD\).

Solution

The coordinates of \(A, B, C\) and \(D\) are \(\left( {1,2,3} \right),\;\left( {4,5,7} \right),\;\left( { - 4,3, - 6} \right)\) and \(\left( {2,9,2} \right)\) respectively.

Hence,

\[\begin{align}{a_1} = \left( {4 - 1} \right) = 3\\{b_1} = \left( {5 - 2} \right) = 3\\{c_1} = \left( {7 - 3} \right) = 4\end{align}\]

\[\begin{align}{a_2} &= \left[ {2 - \left( { - 4} \right)} \right] = 6\\{b_2} &= \left( {9 - 3} \right) = 6\\{c_2} &= \left[ {2 - \left( { - 6} \right)} \right] = 8\end{align}\]

Therefore,

\[ \Rightarrow \frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}} = \frac{1}{2}\]

Hence, \(AB\parallel CD\)

Thus, the angle between \(AB\) and \(CD\) is either \(0^\circ \) or \(180^\circ \).

Chapter 11 Ex.11.ME Question 6

If the line \(\frac{{x - 1}}{{ - 3}} = \frac{{y - 2}}{{2k}} = \frac{{z - 3}}{2}\) and \(\frac{{x - 1}}{{3k}} = \frac{{y - 1}}{1} = \frac{{z - 6}}{{ - 5}}\) are perpendicular, find the value of k.

Solution

Here,

\[\begin{align}{a_1} &= - 3\\{b_1} &= 2k\\{c_1} &= 2\end{align}\]

\[\begin{align}{a_2} &= 3k\\{b_2} &= 1\\{c_2} &= - 5\end{align}\]

Two lines with direction ratios, \({a_1},{b_1},{c_1}\) and \({a_2},{b_2},{c_2}\) are perpendicular, if

\({a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0\)

Therefore,

\[\begin{align} &\Rightarrow - 3\left( {3k} \right) + 2k \times 1 + 2\left( { - 5} \right) = 0\\ &\Rightarrow - 9k + 2k - 10 = 0\\ &\Rightarrow 7k = - 10\\ &\Rightarrow k = \frac{{ - 10}}{7}\end{align}\]

Hence, for \(k = - \frac{{10}}{7}\), the given lines are perpendicular to each other.

Chapter 11 Ex.11.ME Question 7

Find the vector equation of the plane passing through \(\left( {1,2,3} \right)\) and perpendicular to the plane \(\vec r.\left( {\hat i + 2\hat j - 5\hat k} \right) + 9 = 0\).

Solution

Here,

\(\vec r = \left( {\hat i + 2\hat j + 3\hat k} \right)\) and \(\vec N = \left( {\hat i + 2\hat j - 5\hat k} \right)\)

The equation of a line passing through a point and perpendicular to the given plane is given by

\(\vec l = \vec r + \lambda \vec N;\;\;\lambda \in R\)

Hence,

\( \Rightarrow \overrightarrow l = \left( {\hat i + 2\hat j + 3\hat k} \right) + \lambda \left( {\hat i + 2\hat j - 5\hat k} \right)\)

Chapter 11 Ex.11.ME Question 8

Find the equation of the plane passing through \(\left( {a,b,c} \right)\) and parallel to the plane \(\vec r.\left( {\hat i + \hat j + \hat k} \right) = 2\)

Solution

Any plane parallel to the plane, \(\vec r.\left( {\hat i + \hat j + \hat k} \right) = 2\) , is of the form

\[\vec r.\left( {\hat i + \hat j + \hat k} \right) = \lambda \;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\]

Since, the plane passes through the point \(\left( {a,b,c} \right)\).

Therefore, the position vector \(\vec r\) of this point is \(\vec r = a\hat i + b\hat j + c\hat k\)

Hence, equation (\(1\)) becomes

\[\begin{align} &\Rightarrow \left( {a\hat i + b\hat j + c\hat k} \right).\left( {\hat i + \hat j + \hat k} \right) = \lambda \\ &\Rightarrow a + b + c = \lambda\end{align}\]

Putting \(\lambda = a + b + c\) in equation (\(1\)), we get

\[\vec r.\left( {\hat i + \hat j + \hat k} \right) = a + b + c\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\]

This is vector equation of the required plane.

Putting \(\vec r = x\hat i + y\hat j + z\hat k\) in equation (2), we get

\[\begin{align} &\Rightarrow \left( {x\hat i + y\hat j + z\hat k} \right).\left( {\hat i + \hat j + \hat k} \right) = a + b + c\\ &\Rightarrow x + y + z = a + b + c\end{align}\]

Chapter 11 Ex.11.ME Question 9

Find the shortest distance between lines \(\vec r = 6\hat i + 2\hat j + 2\hat k + \lambda \left( {\hat i - 2\hat j + 2\hat k} \right)\) and \(\vec r = - 4\hat i - \hat k + \mu \left( {3\hat i - 2\hat j - 2\hat k} \right)\).

Solution

The given lines are

\[\begin{align}\vec r &= 6\hat i + 2\hat j + 2\hat k + \lambda \left( {\hat i - 2\hat j + 2\hat k} \right) \qquad \ldots \left( 1 \right)\\\vec r &= - 4\hat i - \hat k + \mu \left( {3\hat i - 2\hat j - 2\hat k} \right)\qquad \ldots \left( 2 \right)\end{align}\]

The shortest distance between two lines \(\vec r = {\vec a_1} + \lambda {\vec b_1}\) and \(\vec r = {\vec a_2} + \lambda {\vec b_2}\) is given by

\(d = \left| {\frac{{\left( {{{\vec b}_1} \times {{\vec b}_2}} \right).\left( {{{\vec a}_2} - {{\vec a}_1}} \right)}}{{\left| {{{\vec b}_1} \times {{\vec b}_2}} \right|}}} \right|\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\)

Comparing, \(\vec r = {\vec a_1} + \lambda {\vec b_1}\) and \(\vec r = {\vec a_2} + \lambda {\vec b_2}\) to equation (\(1\)) and (\(2\)), we get

\({\vec a_1} = 6\hat i + 2\hat j + 2\hat k\) and \({\vec a_2} = - 4\hat i - \hat k\)

\({\vec b_1} = \hat i - 2\hat j + 2\hat k\) and \({\vec b_2} = 3\hat i - 2\hat j - 2\hat k\)

Therefore,

\[\begin{align}{{\vec a}_2} - {{\vec a}_1} &= \left( { - 4\hat i - \hat k} \right) - \left( {6\hat i + 2\hat j + 2\hat k} \right)\\ &= - 10\hat i - 2\hat j - 3\hat k\\{{\vec b}_1} \times {{\vec b}_2} &= \left( {\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\1&{ - 2}&2\\3&{ - 2}&{ - 2}\end{array}} \right)\\ &= \left( {4 + 4} \right)\hat i - \left( { - 2 - 6} \right)\hat j + \left( { - 2 + 6} \right)\hat k\\ &= 8\hat i + 8\hat j + 4\hat k\end{align}\]

Putting all these values in equation (\(1\)), we get

\[\begin{align}d &= \left| {\frac{{\left( {{\rm{8}}\hat i + 8\hat j + 4\hat k} \right).\left( { - 10\hat i - 2\hat j - 3\hat k} \right)}}{{\left| {\left( {{\rm{8}}\hat i + 8\hat j + 4\hat k} \right)} \right|}}} \right|\\ &= \left| {\frac{{ - 80 - 16 - 12}}{{\sqrt {{{\left( 8 \right)}^2} + {{\left( 8 \right)}^2} + {{\left( 4 \right)}^2}} }}} \right|\\ &= \left| {\frac{{ - 108}}{{\sqrt {144} }}} \right|\\ &= \frac{{108}}{{12}}\\ &= 9\end{align}\]

Hence, the shortest distance between the two given lines is 9 units.

Chapter 11 Ex.11.ME Question 10

Find the coordinates of the point where the line through \(\left( {5,1,6} \right)\) and \(\left( {3,4,1} \right)\) crosses the YZplane.

Solution

The equation of the line passing through the points, \(\left( {{x_1},{y_1},{z_1}} \right)\) and \(\left( {{x_2},{y_2},{z_2}} \right)\) is

\[\frac{{x - {x_1}}}{{{x_2} - {x_1}}} = \frac{{y - {y_1}}}{{{y_2} - {y_1}}} = \frac{{z - {z_1}}}{{{z_2} - {z_1}}}\]

The line passing through the points \(\left( {5,1,6} \right)\) and \(\left( {3,4,1} \right)\) is given by,

\[\frac{{x - 5}}{{3 - 5}} = \frac{{y - 1}}{{4 - 1}} = \frac{{z - 6}}{{1 - 6}} \Rightarrow \frac{{x - 5}}{{ - 2}} = \frac{{y - 1}}{3} = \frac{{z - 6}}{{ - 5}} = k(say)\]

\( \Rightarrow x = 5 - 2k,y = 3k + 1,z = 6k - 5\)

Any point on the line is of the form \(\left( {5 - 2k,3k + 1,6 - 5k} \right)\)

Any point on the line passes through YZ-plane

\[\begin{align} &\Rightarrow 5 - 2k = 0\\ &\Rightarrow k = \frac{5}{2}\\ &\Rightarrow 3k + 1 = 3 \times \left( {\frac{5}{2}} \right) + 1 = \frac{{17}}{2}\\ &\Rightarrow 6 - 5k = 6 - 5 \times \left( { - \frac{5}{2}} \right) = - \frac{{13}}{2}\end{align}\]

Hence, the required point is \(\left( {0,\frac{{17}}{2},\frac{{ - 13}}{2}} \right)\)

Chapter 11 Ex.11.ME Question 11

Find the coordinates of the point where the line through \(\left( {5,1,6} \right)\) and \(\left( {3,4,1} \right)\) crosses the \(ZX-\)plane.

Solution

The equation of the line passing through the points \(\left( {{x_1},{y_1},{z_1}} \right)\) and \(\left( {{x_2},{y_2},{z_2}} \right)\) is

\(\frac{{x - {x_1}}}{{{x_2} - {x_1}}} = \frac{{y - {y_1}}}{{{y_2} - {y_1}}} = \frac{{z - {z_1}}}{{{z_2} - {z_1}}}\)

The line passing through the points \(\left( {5,1,6} \right)\) and \(\left( {3,4,1} \right)\) is given by,

\(\frac{{x - 5}}{{3 - 5}} = \frac{{y - 1}}{{4 - 1}} = \frac{{z - 6}}{{1 - 6}} \Rightarrow \frac{{x - 5}}{{ - 2}} = \frac{{y - 1}}{3} = \frac{{z - 6}}{{ - 5}} = k(say)\)

\( \Rightarrow x = 5 - 2k,y = 3k + 1,z = 6k - 5\)

Any point on the line is of the form \(\left( {5 - 2k,3k + 1,6 - 5k} \right)\)

Any point on the line passes through \(ZX-\)plane

\[\begin{align} &\Rightarrow 3k + 1 = 0\\ &\Rightarrow k = - \frac{1}{3}\\ &\Rightarrow 5 - 2k = 5 - 2\left( { - \frac{1}{3}} \right) = \frac{{17}}{3}\\ &\Rightarrow 6 - 5k = 6 - 5\left( { - \frac{1}{3}} \right) = \frac{{23}}{2}\end{align}\]

Hence, the required point is \(\left( {\frac{{17}}{3},0,\frac{{23}}{2}} \right)\)

Chapter 11 Ex.11.ME Question 12

Find the coordinates of the point where the line through \(\left( {3, - 4, - 5} \right)\) and \(\left( {2, - 3,1} \right)\) crosses the plane \(2x + y + z = 7\).

Solution

The equation of the line through the point \(\left( {{x_1},{y_1},{z_1}} \right)\) and \(\left( {{x_2},{y_2},{z_2}} \right)\) is

\[\frac{{x - {x_1}}}{{{x_2} - {x_1}}} = \frac{{y - {y_1}}}{{{y_2} - {y_1}}} = \frac{{z - {z_1}}}{{{z_2} - {z_1}}}\]

Since the line passes through the points \(\left( {3, - 4, - 5} \right)\) and \(\left( {2, - 3,1} \right)\) its equation is given by,

\[\begin{align} &\Rightarrow \frac{{x - 3}}{{2 - 3}} = \frac{{y + 4}}{{ - 3 + 4}} = \frac{{z + 5}}{{1 + 5}}\\ &\Rightarrow \frac{{x - 3}}{{ - 1}} = \frac{{y + 4}}{1} = \frac{{z + 5}}{6} = k(say)\end{align}\]

\( \Rightarrow x = 3 - k,y = k - 4,z = 6k - 5\)

Thus, any point on the line is of the form \(\left( {3 - k,k - 4,6k - 5} \right)\)

This point lies on the plane, \(2x + y + z = 7\)

\[\begin{align} &\Rightarrow 2\left( {3 - k} \right) + \left( {k - 4} \right) + \left( {6k - 5} \right) = 7\\& \Rightarrow 5k - 3 = 7\\ &\Rightarrow k = 2\end{align}\]

Hence, the coordinates of the required point are

\[\begin{align} &\Rightarrow \left( {3 - 2,2 - 4,6 \times 2 - 5} \right)\\ &\Rightarrow \left( {1, - 2,7} \right)\end{align}\]

Chapter 11 Ex.11.ME Question 13

Find the equation of the plane passing through the points \(\left( { - 1,3,2} \right)\) and perpendicular to each of the planes \(x + 2y + 3z = 5\) and \(3x + 3y + z = 0\).

Solution

The equation of the plane passing through the point \(\left( { - 1,3,2} \right)\) is

\[a\left( {x + 1} \right) + b\left( {y - 3} \right) + c\left( {z - 2} \right) = 0\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\]

where \(a,b,c\) are direction ratios of normal to the plane.

We know that two planes,

\({a_1}x + {b_1}y + {c_1}z + {d_1} = 0\) and \({a_2}x + {b_2}y + {c_2}z + {d_2} = 0\) are perpendicular, if \({a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0\)

Since, plane (\(1\)) is perpendicular to the plane, \(x + 2y + 3z = 5\)

Therefore,

\[\begin{align} &\Rightarrow a.1 + b.2 + c.3 = 0\\ &\Rightarrow a + 2b + 3c = 0\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}\]

Also, plane (\(1\)) is perpendicular to the plane, \(3x + 3y + z = 0\)

\[\begin{align}& \Rightarrow a.3 + b.3 + c.1 = 0\\ &\Rightarrow 3a + 3b + c = 0\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}\]

From equation (\(2\)) and (\(3\)), we get

\[\begin{align} &\Rightarrow \frac{a}{{2 \times 1 - 3 \times 3}} = \frac{b}{{3 \times 3 - 1 \times 1}} = \frac{c}{{1 \times 3 - 2 \times 3}}\\ &\Rightarrow \frac{a}{{ - 7}} = \frac{b}{8} = \frac{c}{3} = k{\rm{ }}(say)\\& \Rightarrow a = - 7k,b = 8k,c = - 3k\end{align}\]

Putting the values of \(a,b\) and \(c\) in equation (1), we get

\[\begin{align} &\Rightarrow - 7k\left( {x + 1} \right) + 8k\left( {y - 3} \right) - 3k\left( {z - 2} \right) = 0\\& \Rightarrow \left( { - 7x - 7} \right) + \left( {8y - 24} \right) - 3z + 6 = 0\\ &\Rightarrow - 7x + 8y - 3z - 25 = 0\\& \Rightarrow 7x - 8y + 3z + 25 = 0\end{align}\]

Chapter 11 Ex.11.ME Question 14

If the points \(\left( {1,1,p} \right)\) and \(\left( { - 3,0,1} \right)\) be equidistant from the plane \(\vec r.\left( {3\hat i + 4\hat j - 12\hat k} \right) + 13 = 0\) then find the value of \(p\).

Solution

Here,

\[\begin{align}{{\vec a}_1} &= \hat i + \hat j + p\hat k\\{{\vec a}_2} &= - 3\hat i + \hat k\end{align}\]

The equation of the given plane is \(\vec r.\left( {3\hat i + 4\hat j - 12\hat k} \right) + 13 = 0\)

The perpendicular distance between a point whose vector is \(\vec a\) and the plane \(\vec r.\vec N = d\) is given by

\(D = \frac{{\left| {\vec a.\vec N - d} \right|}}{{\left| {\vec N} \right|}}\)

Here,

\(\vec N = 3\hat i + 4\hat j - 12\hat k\) and \(d = - 13\)

Hence, the distance between the point \(\left( {1,1,p} \right)\) and the given plane is

\[\begin{align} &\Rightarrow {D_1} = \frac{{\left| {\left( {\hat i + \hat j + p\hat k} \right).\left( {3\hat i + 4\hat j - 12\hat k} \right) - \left( { - 13} \right)} \right|}}{{\left| {3\hat i + 4\hat j - 12\hat k} \right|}}\\ &\Rightarrow {D_1} = \frac{{\left| {3 + 4 - 12p + 13} \right|}}{{\sqrt {{3^2} + {4^2} + {{\left( { - 12} \right)}^2}} }}\\ &\Rightarrow {D_1} = \frac{{\left| {20 - 12p} \right|}}{{13}}\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}\]

Similarly, the distance between the point \(\left( { - 3,0,1} \right)\) and the given plane is

\[\begin{align}& \Rightarrow {D_2} = \frac{{\left| {\left( { - 3\hat i + \hat k} \right).\left( {3\hat i + 4\hat j - 12\hat k} \right) - \left( { - 13} \right)} \right|}}{{\left| {3\hat i + 4\hat j - 12\hat k} \right|}}\\ &\Rightarrow {D_2} = \frac{{\left| { - 9 - 12 + 13} \right|}}{{\sqrt {{3^2} + {4^2} + {{\left( { - 12} \right)}^2}} }}\\ &\Rightarrow {D_2} = \frac{8}{{13}}\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}\]

From the given condition, \({D_1} = {D_2}\)

\[\begin{align} &\Rightarrow \frac{{\left| {20 - 12p} \right|}}{{13}} = \frac{8}{{13}}\\ &\Rightarrow \left| {20 - 12p} \right| = 8\\& \Rightarrow 20 - 12p = 8{\rm{ }}or{\rm{ }} - \left( {20 - 12p} \right) = 8\\ &\Rightarrow 12p = 12\,{\rm{ or}}\, 12p = 28\\ &\Rightarrow p = 1\,{\rm{ or}}\,p = \frac{7}{3}\end{align}\]

Thus, the value of \(p = 1\) or \(p = \frac{7}{3}\).

Chapter 11 Ex.11.ME Question 15

Find the equation of the plane passing through the line of intersection of the planes\(\vec r.\left( {\hat i + \hat j + \hat k} \right) = 1\) and \(\vec r.\left( {2\hat i + 3\hat j - \hat k} \right) + 4 = 0\) and parallel to x-axis.

Solution

The given planes are \(\vec r.\left( {\hat i + \hat j + \hat k} \right) = 1\) and \(\vec r.\left( {2\hat i + 3\hat j - \hat k} \right) + 4 = 0\)

The equation of any plane passing through the line of intersection of these planes is given by

\[\begin{align}&\left[ {\vec r.\left( {\hat i + \hat j + \hat k} \right) - 1} \right] + \lambda \left[ {\vec r.\left( {2\hat i + 3\hat j - \hat k} \right) + 4} \right] = 0\\&\vec r.\left[ {\left( {2\lambda + 1} \right)\hat i + \left( {3\lambda + 1} \right)\hat j + \left( {1 - \lambda } \right)\hat k} \right] + \left( {4\lambda - 1} \right) = 0 \qquad \ldots \left( 1 \right)\end{align}\]

Here,

and \({c_1} = \left( {1 - \lambda } \right)\)

Since, the required plane is parallel to x-axis.

Therefore, its normal is perpendicular to x-axis.

The direction ratios of x-axis are 1, 0 and 0.

\({a_1} = \left( {2\lambda + 1} \right),{b_1} = \left( {3\lambda + 1} \right)\)

Therefore,

\({a_2} = 1,{b_2} = 0\) and \({c_2} = 0\)

Hence,

\[\begin{align} &\Rightarrow 1.\left( {2\lambda + 1} \right) + 0\left( {3\lambda + 1} \right) + 0\left( {1 - \lambda } \right) = 0\\& \Rightarrow 2\lambda + 1 = 0\\ &\Rightarrow \lambda = - \frac{1}{2}\end{align}\]

Putting, \(\lambda = - \frac{1}{2}\) in equation (1), we get

\[\begin{align}& \Rightarrow \vec r\left[ { - \frac{1}{2}\hat j + \frac{3}{2}\hat k} \right] + \left( { - 3} \right) = 0\\& \Rightarrow \vec r\left( {\hat j - 3\hat k} \right) + 6 = 0\end{align}\]

Thus, its Catersian equation is \(y - 3z + 6 = 0\)

Chapter 11 Ex.11.ME Question 16

If \(O\) be the origin and the coordinates of P be \(\left( {1,2, - 3} \right)\), then find the equation of the plane passing through \(P\) and perpendicular to \(OP\).

Solution

The given points are \({\rm{O}}\left( {0,0,0} \right)\) and \(P\left( {1,2, - 3} \right)\)

The direction ratios of \(OP\) are

\[\begin{align}a &= \left( {1 - 0} \right) = 1\\b &= \left( {2 - 0} \right) = 2\\c &= \left( { - 3 - 0} \right) = - 3\end{align}\]

The equation of the plane passing through the point\(\left( {{x_1},{y_1},{z_1}} \right)\) is

\[a\left( {x - {x_1}} \right) + b\left( {y - {y_1}} \right) + c\left( {z - {z_1}} \right) = 0\]

where, \(a,b\) and \(c\) are the direction ratios of normal.

Here, the direction ratios of normal are \(1,2\) and \( - 3\) and the point P is \(\left( {1,2, - 3} \right)\).

Hence, the equation of the required plane is

\[\begin{align}& \Rightarrow 1\left( {x - 1} \right) + 2\left( {y - 2} \right) - 3\left( {z + 3} \right) = 0\\& \Rightarrow x + 2y - 3z - 14 = 0\end{align}\]

Chapter 11 Ex.11.ME Question 17

Find the equation of the plane which contains the line of intersection of the planes \(\vec r.\left( {\hat i + 2\hat j + 3\hat k} \right) - 4 = 0,\;\vec r.\left( {2\hat i + \hat j - \hat k} \right) + 5 = 0\) and which is perpendicular to the plane\(\vec r.\left( {5\hat i + 3\hat j - 6\hat k} \right) + 8 = 0\).

Solution

The equations of the given planes are

\[\begin{align}&\vec r.\left( {\hat i + 2\hat j + 3\hat k} \right) - 4 = 0\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\&\vec r.\left( {2\hat i + \hat j - \hat k} \right) + 5 = 0\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}\]

The equation of the required plane is,

\[\begin{align}&\left[ {\vec r.\left( {\hat i + 2\hat j + 3\hat k} \right) - 4} \right] + \lambda \left[ {\vec r.\left( {2\hat i + \hat j - \hat k} \right) + 5} \right] = 0\\&\vec r\left[ {\left( {2\lambda + 1} \right)\hat i + \left( {\lambda + 2} \right)\hat j + \left( {3 - \lambda } \right)\hat k} \right] + \left( {5\lambda - 4} \right) = 0\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}\]

The plane in equation (\(3\)) is perpendicular to the plane, \(\vec r.\left( {5\hat i + 3\hat j - 6\hat k} \right) + 8 = 0\)

Therefore,

\[\begin{align} &\Rightarrow 5\left( {2\lambda + 1} \right) + 3\left( {\lambda + 2} \right) - 6\left( {3 - \lambda } \right) = 0\\ &\Rightarrow 19\lambda - 7 = 0\\ &\Rightarrow \lambda = \frac{7}{{19}}\end{align}\]

Putting \(\lambda = \frac{7}{{19}}\) in equation (\(3\)), we get

\[\begin{align} &\Rightarrow \vec r.\left[ {\frac{{33}}{{19}}\hat i + \frac{{45}}{{19}}\hat j + \frac{{50}}{{19}}\hat k} \right]\frac{{ - 41}}{{19}} = 0\\ &\Rightarrow \vec r.\left( {33\hat i + 45\hat j + 50\hat k} \right) - 41 = 0\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 4 \right)\end{align}\]

The Cartesian equation of this plane is given by

\[\begin{align} &\Rightarrow \left( {x\hat i + y\hat j + z\hat k} \right).\left( {33\hat i + 45\hat j + 50\hat k} \right) - 41 = 0\\ &\Rightarrow 33x + 45y + 50z - 41 = 0\end{align}\]

Chapter 11 Ex.11.ME Question 18

Find the distance of the point \(\left( { - 1, - 5, - 10} \right)\) from the point of intersection of the line

\(\vec r = 2\hat i - \hat j + 2\hat k + \lambda \left( {3\hat i + 4\hat j + 2\hat k} \right)\) and the plane \(\vec r.\left( {\hat i - \hat j + \hat k} \right) = 5\).

Solution

The equation of the given line is

\[\vec r = 2\hat i - \hat j + 2\hat k + \lambda \left( {3\hat i + 4\hat j + 2\hat k} \right)\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\]

The equation of the given plane is

\[\vec r.\left( {\hat i - \hat j + \hat k} \right) = 5\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\]

Putting the value of r from equation (1) in equation (2), we get

\[\begin{align}&\Rightarrow \left[ {2\hat i - \hat j + 2\hat k + \lambda \left( {3\hat i + 4\hat j + 2\hat k} \right)} \right].\left( {\hat i - \hat j + \hat k} \right) = 5\\ &\Rightarrow \left[ {\left( {3\lambda + 2} \right)\hat i + \left( {4\lambda - 1} \right)\hat j + \left( {2\lambda + 2} \right)\hat k} \right].\left( {\hat i - \hat j + \hat k} \right) = 5\\ &\Rightarrow \left( {3\lambda + 2} \right) - \left( {4\lambda - 1} \right) + \left( {2\lambda + 2} \right) = 5\\ &\Rightarrow \lambda = 0\end{align}\]

Putting this value in equation (1), we get the equation of the line as \(\vec r = 2\hat i - \hat j + 2\hat k\)

This means that the position vector of the point of intersection of the line and plane is

\(\vec r = 2\hat i - \hat j + 2\hat k\)

This shows that the point of intersection of the given line and plane is given by the coordinates \(\left( {2, - 1,2} \right)\) and .

The required distance between the points\(\left( { - 1, - 5, - 10} \right)\)\(\left( {2, - 1,2} \right)\) and \(\left( { - 1, - 5, - 10} \right)\) is

\[\begin{align}d &= \sqrt {{{\left( { - 1 - 2} \right)}^2} + {{\left( { - 5 + 1} \right)}^2} + {{\left( { - 10 - 2} \right)}^2}} \\ &= \sqrt {9 + 16 + 144} \\ &= \sqrt {169} \\ &= 13\end{align}\]

Chapter 11 Ex.11.ME Question 19

Find the vector equation of the line passing through \(\left( {1,2,3} \right)\) and parallel to the planes \(\vec r.\left( {\hat i - \hat j + 2\hat k} \right) = 5\) and \(\vec r.\left( {3\hat i + \hat j + \hat k} \right) = 6\)

Solution

Let the required line be parallel to vector \(\vec b\) given by,

\[\vec b = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\]

The postion vector of the point \(\left( {1,2,3} \right)\) is

\(\vec a = \hat i + 2\hat j + 3\hat k\)

The equation of line passing through \(\left( {1,2,3} \right)\) and parallel to \(\vec b\) is given by,

\[\begin{align} &\Rightarrow \vec r = \vec a + \lambda \vec b\\ &\Rightarrow \overrightarrow r .\left( {\hat i - \hat j + 2\hat k} \right) + \lambda \left( {{b_1}\hat i + {b_2}\hat j + {b_3}\hat k} \right)\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}\]

The equations of the given planes are

\[\begin{align}\vec r.\left( {\hat i - \hat j + 2\hat k} \right) = 5\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\\\vec r.\left( {3\hat i + \hat j + \hat k} \right) = 6\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}\]

The line in equation (1) and plane in equation (2) are parallel.

Therefore, the normal to the plane of equation (2) and the given line are perpendicular.

\[\begin{align}&\Rightarrow \left( {\hat i - \hat j + 2\hat k} \right).\lambda \left( {{b_1}\hat i + {b_2}\hat j + {b_3}\hat k} \right) = 0\\ &\Rightarrow \lambda \left( {{b_1} - {b_2} + 2{b_3}} \right) = 0\\ &\Rightarrow {b_1} - {b_2} + 2{b_3} = 0\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 4 \right)\end{align}\]

Similarly,

\[\begin{align}&\Rightarrow \left( {3\hat i + \hat j + \hat k} \right).\lambda \left( {{b_1}\hat i + {b_2}\hat j + {b_3}\hat k} \right) = 0\\ &\Rightarrow \lambda \left( {3{b_1}\hat i + {b_2} + {b_3}} \right) = 0\\ &\Rightarrow 3{b_1} + {b_2} + {b_3} = 0\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 5 \right)\end{align}\]

From equation (4) and (5), we obtain

\[\begin{align} &\Rightarrow \frac{{{b_1}}}{{\left( { - 1} \right) \times 1 - 1 \times 2}} = \frac{{{b_2}}}{{2 \times 3 - 1 \times 1}} = \frac{{{b_3}}}{{1 \times 1 - 3\left( { - 1} \right)}}\\ &\Rightarrow \frac{{{b_1}}}{{ - 3}} = \frac{{{b_2}}}{5} = \frac{{{b_3}}}{4}\end{align}\]

Thus,

The direction ratios of \(\vec b\) are \( - 3,5\) and \(4\).

Hence,

\[\begin{align}\overrightarrow b &= {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\\ &= - 3\hat i + 5\hat j + 4\hat k\end{align}\]

Putting, the value of \(\vec b\) in eqation (\(1\)), we get

\[\vec r = \left( {\hat i + 2\hat j + 3\hat k} \right) + \lambda \left( { - 3\hat i + 5\hat j + 4\hat k} \right)\]

Chapter 11 Ex.11.ME Question 20

Find the vector equation of the line passing through the point \(\left( {1,2, - 4} \right)\) and perpendicular to the two lines: \(\frac{{x - 8}}{3} = \frac{{y + 19}}{{ - 16}} = \frac{{z - 10}}{7}\) and \(\frac{{x - 15}}{3} = \frac{{y - 29}}{8} = \frac{{z - 5}}{{ - 5}}\).

Solution

Here,

\[\begin{align}\vec b &= {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\\\vec a &= \hat i + 2\hat j - 4\hat k\end{align}\]

The equation of the line passing through \(\left( {1,2, - 4} \right)\) and parallel to vector \(\vec b\) is given by

\[\begin{align} &\Rightarrow \vec r = \vec a + \lambda \vec b\\& \Rightarrow \overrightarrow r = \left( {\hat i + 2\hat j - 4\hat k} \right) + \lambda \left( {{b_1}\hat i + {b_2}\hat j + {b_3}\hat k} \right)\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\end{align}\]

The equations of the lines are

\[\begin{align}&\frac{{x - 8}}{3} = \frac{{y + 19}}{{ - 16}} = \frac{{z - 10}}{7}\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\\&\frac{{x - 15}}{3} = \frac{{y - 29}}{8} = \frac{{z - 5}}{{ - 5}}\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 3 \right)\end{align}\]

Since, lines of the equations (\(1\)) and (\(2\)) are perpendicular to each other

\[ \Rightarrow 3{b_1} - 16{b_2} + 7{b_3} = 0\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 4 \right)\]

Also,

Lines (\(1\)) and (\(3\)) are perpendicular to each other

\[ \Rightarrow 3{b_1} + 8{b_2} - 5{b_3} = 0\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 5 \right)\]

From equations (\(4\)) and (\(5\)), we obtain

\[\begin{align} &\Rightarrow \frac{{{b_1}}}{{\left( { - 16} \right) \times \left( { - 5} \right) - 8 \times 7}} = \frac{{{b_2}}}{{7 \times 3 - 3 \times \left( { - 5} \right)}} = \frac{{{b_3}}}{{3 \times 8 - 3 \times \left( { - 16} \right)}}\\ &\Rightarrow \frac{{{b_1}}}{{24}} = \frac{{{b_2}}}{{36}} = \frac{{{b_3}}}{{72}} \\&\Rightarrow \frac{{{b_1}}}{2} = \frac{{{b_2}}}{3} = \frac{{{b_3}}}{6}\end{align}\]

Hence,

The direction ratios of \(\vec b\) are \(2,3\) and \(6\)

Therefore,

\(\vec b = 2\hat i + 3\hat j + 6\hat k\)

Putting \(\vec b = 2\hat i + 3\hat j + 6\hat k\) in equation (\(1\)), we get

\[ \Rightarrow \vec r = \left( {\hat i + 2\hat j - 4\hat k} \right) + \lambda \left( {2\hat i + 3\hat j + 6\hat k} \right)\]

Chapter 11 Ex.11.ME Question 21

Prove that if a plane has the intercepts \(a,b,c\) and is at a distance of \(p\) units from the origin, then \(\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = \frac{1}{{{p^2}}}\).

Solution

The equation of the plane having intercepts \(a,b,c\) with \(x,y,z\) axes respectively is given by,

\[\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\]

The distance \(p\) of the plane from the origin is given by,

\[\begin{align}p &= \left| {\frac{{\frac{0}{a} + \frac{0}{b} + \frac{0}{c} - 1}}{{\sqrt {\left( {\frac{1}{{{a^2}}}} \right) + \left( {\frac{1}{{{b^2}}}} \right) + \left( {\frac{1}{{{c^2}}}} \right)} }}} \right|\\ &= \frac{1}{{\sqrt {\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}} }}\\{p^2} &= \frac{1}{{\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}}}\\\frac{1}{{{p^2}}} &= \frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}}\end{align}\]

Hence, \(\frac{1}{{{a^2}}} + \frac{1}{{{b^2}}} + \frac{1}{{{c^2}}} = \frac{1}{{{p^2}}}\) proved.

Chapter 11 Ex.11.ME Question 22

Distance between the two planes: \(2x + 3y + 4z = 4\) and \(4x + 6y + 8z = 12\) is

(A) \(2\) units

(B) \(4\) units

(C) \(8\) units

(D) \(\frac{2}{{\sqrt {29} }}\) units

Solution

The equations of the planes are

\[ \Rightarrow 2x + 3y + 4z = 4\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\]

\[\begin{align} &\Rightarrow 4x + 6y + 8z = 12\\ &\Rightarrow 2x + 3y + 4z = 4\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}\]

Since given planes are parallel, and we know that the distance between two parallel planes \(ax + by + cz = {d_1}\) and \(ax + by + cz = {d_2}\) is given by,

\[\begin{align}D &= \left| {\frac{{{d_2} - {d_1}}}{{\sqrt {{a^2} + {b^2} + {c^2}} }}} \right|\\ &= \left| {\frac{{6 - 4}}{{\sqrt {{{\left( 2 \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 4 \right)}^2}} }}} \right|\\ &= \frac{2}{{\sqrt {29} }}\end{align}\]

Hence, the distance between the given plane is \(\frac{2}{{\sqrt {29} }}\) units.

Therefore, the correct answer is \(D\).

Chapter 11 Ex.11.ME Question 23

The planes: \(2x - y + 4z = 5\) and \(5x - 2.5y + 10z = 6\) are

(A) Perpendicular

(B) Parallel

(C) intersect \(y-\)axis

(D) passes through \(\left( {0,0,\frac{5}{4}} \right)\)

Solution

The equations of the planes are

\[\begin{align}2x - y + 4z = 5\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; \ldots \left( 1 \right)\\5x - 2.5y + 10z = 6\;\;\;\;\;\;\;\;\;\; \ldots \left( 2 \right)\end{align}\]

Here,

\[\begin{align}\frac{{{a_1}}}{{{a_2}}} = \frac{2}{5}\\\frac{{{b_1}}}{{{b_2}}} = \frac{{ - 1}}{{ - 2.5}} = \frac{2}{5}\\\frac{{{c_1}}}{{{c_2}}} = \frac{4}{{10}} = \frac{2}{5}\end{align}\]

Therefore,

\[ \Rightarrow \frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\]

Hence, the given planes are parallel.

Therefore, the correct answer is \(B.\)

  
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