# Miscellaneous Exercise Three Dimensional Geometry Solution - NCERT Class 12

Go back toÂ  'Three Dimensional Geometry'

## 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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