# NCERT Solutions For Class 12 Maths Chapter 11 Exercise 11.2

Go back to  'Three Dimensional Geometry'

## Chapter 11 Ex.11.2 Question 1

Show that the three lines with direction cosines $$\frac{{12}}{{13}},\frac{{ - 3}}{{13}},\frac{{ - 4}}{{13}};$$$$\frac{4}{{13}},\frac{{12}}{{13}},\frac{3}{{13}};$$$$\frac{3}{{13}},\frac{{ - 4}}{{13}},\frac{{12}}{{13}}$$ are mutually perpendicular.

### Solution

Two lines with direction cosines $${l_1},{m_1},{n_1}$$ and $${l_2},{m_2},{n_2}$$ are perpendicular to each other, if $${l_1}{l_2} + {m_1}{m_2} + {n_1}{n_2} = 0$$

For the lines with direction cosines, $$\frac{{12}}{{13}},\frac{{ - 3}}{{13}},\frac{{ - 4}}{{13}}$$ and $$\frac{4}{{13}},\frac{{12}}{{13}},\frac{3}{{13}}$$, we get

\begin{align}{l_1}{l_2} + {m_1}{m_2} + {n_1}{n_2}& = \frac{{12}}{{13}} \times \frac{4}{{13}} + \left( {\frac{{ - 3}}{{13}}} \right) \times \frac{{12}}{{13}} + \left( {\frac{{ - 4}}{{13}}} \right) \times \frac{3}{{13}}\\& = \frac{{48}}{{169}} - \frac{{36}}{{169}} - \frac{{12}}{{169}}\\&=0\end{align}

Hence, the lines are perpendicular.

For the lines with direction cosines, $$\frac{4}{{13}},\frac{{12}}{{13}},\frac{3}{{13}}$$ and $$\frac{3}{{13}},\frac{{ - 4}}{{13}},\frac{{12}}{{13}}$$, we get

\begin{align}{l_1}{l_2} + {m_1}{m_2} + {n_1}{n_2} &= \frac{4}{{13}} \times \frac{3}{{13}} + \frac{{12}}{{13}} \times \left( {\frac{{ - 4}}{{13}}} \right) + \frac{3}{{13}} \times \frac{{12}}{{13}}\\& = \frac{{12}}{{169}} - \frac{{48}}{{169}} + \frac{{36}}{{169}}\\&=0\end{align}

Hence, the lines are perpendicular.

For the lines with direction cosines, $$\frac{3}{{13}},\frac{{ - 4}}{{13}},\frac{{12}}{{13}}$$ and $$\frac{{12}}{{13}},\frac{{ - 3}}{{13}},\frac{{ - 4}}{{13}}$$, we get

\begin{align}{l_1}{l_2} + {m_1}{m_2} + {n_1}{n_2} &= \left( {\frac{3}{{13}}} \right) \times \left( {\frac{{12}}{{13}}} \right) + \left( {\frac{{ - 4}}{{13}}} \right) \times \left( {\frac{{ - 3}}{{13}}} \right) + \left( {\frac{{12}}{{13}}} \right) \times \left( {\frac{{ - 4}}{{13}}} \right)\\&= \frac{{36}}{{169}} + \frac{{12}}{{169}} - \frac{{48}}{{169}}\\=0\end{align}

Hence, the lines are perpendicular.

So, the all three lines are mutually perpendicular.

## Chapter 11 Ex.11.2 Question 2

Show that the line through the points $$\left( {{\rm{1}}, - {\rm{1}},{\rm{2}}} \right){\rm{,}}\left( {{\rm{3}},{\rm{4}}, - {\rm{2}}} \right)$$ is perpendicular to the line through the points $$\left( {0,{\rm{3}},{\rm{2}}} \right)$$ and $$\left( {{\rm{3}},{\rm{5}},{\rm{6}}} \right)$$.

### Solution

Let AB be the line joining the points $$\left( {{\rm{1}}, - {\rm{1}},{\rm{2}}} \right)$$ and $$\left( {{\rm{3}},{\rm{4}}, - {\rm{2}}} \right)$$; and CD be the line through the points $$\left( {0,{\rm{3}},{\rm{2}}} \right)$$ and $$\left( {{\rm{3}},{\rm{5}},{\rm{6}}} \right)$$

Hence,

\begin{align}{a_1} &= \left( {3 - 1} \right) = 2\\{b_1} &= \left[ {4 - \left( { - 1} \right)} \right] = 5\\{c_1} &= \left( { - 2 - 2} \right) =- 4\\\\{a_2} &= \left( {3 - 0} \right) = 3\\{b_2} &= \left( {5 - 3} \right) = 2\\{c_2}& = \left( {6 - 2} \right) = 4\end{align}

If, $$AB \bot CD ; \Rightarrow {a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0$$

Here,

\begin{align}{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} &= 2 \times 3 + 5 \times 2 + \left( { - 4} \right) \times 4\\&= 6 + 10 - 16\\&= 0\end{align}

Hence, AB and CD are perpendicular to each other.

## Chapter 11 Ex.11.2 Question 3

Show that the line through the points $$\left( {{\rm{4}},{\rm{7}},{\rm{8}}} \right),\left( {{\rm{2}},{\rm{3}},{\rm{4}}} \right)$$ is parallel to the line through the points $$\left( { - 1, - 2,1} \right),\left( {1,2,5} \right)$$.

### Solution

Let AB be the line through the points $$\left( {{\rm{4}},{\rm{7}},{\rm{8}}} \right)$$ and $$\left( {{\rm{2}},{\rm{3}},{\rm{4}}} \right)$$; CD be the line through the points $$\left( { - 1, - 2,1} \right)$$ and $$\left( {1,2,5} \right)$$.

Hence,

\begin{align}{a_1} &= \left( {2 - 4} \right) =- 2\\{b_1} &= \left( {3 - 7} \right) =- 4\\{c_1} &= \left( {4 - 8} \right) =- 4\\\\{a_2} &= \left[ {1 - \left( { - 1} \right)} \right] = 2\\{b_2} &= \left[ {2 - \left( { - 2} \right)} \right] = 4\\{c_2}& = \left( {5 - 1} \right) = 4\end{align}

If, $$AB \bot CD$$; $$\frac{{{a_1}}}{{{a_2}}} = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}} = 0$$

Here,

\begin{align}\frac{{{a_1}}}{{{a_2}}} &= \frac{{ - 2}}{2} =- 1\\\frac{{{b_1}}}{{{b_2}}}& = \frac{{ - 4}}{4} =- 1\\\frac{{{c_1}}}{{{c_2}}} &= \frac{{ - 4}}{4} =- 1\\\Rightarrow \; \frac{{{a_1}}}{{{a_2}}}& = \frac{{{b_1}}}{{{b_2}}} = \frac{{{c_1}}}{{{c_2}}}\end{align}

Hence, AB is parallel to CD.

## Chapter 11 Ex.11.2 Question 4

Find the equation of the line which passes through point $$\left( {1,2,3} \right)$$ and is parallel to the vector $$3\hat i + 2\hat j - 2\hat k$$.

### Solution

It is given that the line passes through the point $$A\left( {1,2,3} \right)$$.

Therefore, the position vector through $$A\left( {1,2,3} \right)$$ is

\begin{align}\vec a &= \hat i + 2\hat j + 3\hat k\\\vec b &= 3\hat i + 2\hat j - 2\hat k\end{align}

So, line passes through point $$A\left( {1,2,3} \right)$$ and parallel to $$\overrightarrow b$$ is given by $$\vec r = \vec a + \lambda \vec b$$, where l is a real number.

Hence,

$$\vec r = \hat i + 2\hat j + 3\hat k + \lambda \left( {3\hat i + 2\hat j - 2\hat k} \right)$$

This is the required equation of the line.

## Chapter 11 Ex.11.2 Question 5

Find the equation of the line in vector and in Cartesian form that passes through the point with positive vector $$2\hat i - \hat j + 4\hat k$$ and is in the direction $$\hat i + 2\hat j - \hat k$$

### Solution

It is given that

\begin{align}\vec a &= 2\hat i - \hat j + 4\hat k\\\vec b &= \hat i + 2\hat j - \hat k\end{align}

Since, the vector equation of the line is given by $$\vec r = \vec a + \lambda \vec b$$, where $$\lambda$$ is some real number.

Hence,

$$\vec r = 2\hat i - \hat j + 4\hat k + \lambda \left( {\hat i + 2\hat j - \hat k} \right)$$

Since, $$\vec r$$ is the position vector of any point $$\left( {x,y,z} \right)$$ on the line

Therefore,

\begin{align}x\hat i - y\hat j + z\hat k &= 2\hat i - \hat j + 4\hat k + \lambda \left( {\hat i + 2\hat j - \hat k} \right)\\&= \left( {2 + \lambda } \right)\hat i + \left( { - 1 + 2\lambda } \right)\hat j + \left( {4 - \lambda } \right)\hat k\end{align}

Eliminating l , we get the Cartesian form equation as



Thus, the equation of the line in vector form is $$\vec r = 2\hat i - \hat j + 4\hat k + \lambda \left( {\hat i + 2\hat j - \hat k} \right)$$ and

cartesian form is $$\frac{{x - 2}}{1} = \frac{{y + 1}}{2} = \frac{{z - 4}}{{ - 1}}$$

## Chapter 11 Ex.11.2 Question 6

Find the Cartesian equation of the line which passes through the point $$\left( { - {\rm{2}},{\rm{4}}, - {\rm{5}}} \right)$$ and parallel to the line given by $$\frac{{x + 3}}{3} = \frac{{y - 4}}{5} = \frac{{z + 8}}{6}$$.

### Solution

It is given that the required line passes through the point $$\left( { - {\rm{2}},{\rm{4}}, - {\rm{5}}} \right)$$ and is parallel to

$$\frac{{x + 3}}{3} = \frac{{y - 4}}{5} = \frac{{z + 8}}{6}$$

Therefore, its direction ratios are $$3k,5k$$ and $$6k$$, where $$k \ne 0$$

It is known that the equation of the line through the point $$\left( {{x_1},{y_1},{z_1}} \right)$$ and with direction ratios $$a,b,c$$  is given by $$\frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c}$$

Hence, the equation of the required line is

\begin{align}&\Rightarrow \; \frac{{x + 2}}{{3k}} = \frac{{y - 4}}{{5k}} = \frac{{z + 5}}{{6k}}\\&\Rightarrow \;\frac{{x + 2}}{3} = \frac{{y - 4}}{5} = \frac{{z + 5}}{6} = k\end{align}

Thus, the cartesian equation of the line is $$\frac{{x + 2}}{3} = \frac{{y - 4}}{5} = \frac{{z + 5}}{6}$$.

## Chapter 11 Ex.11.2 Question 7

The Cartesian equation of a line is $$\frac{{x - 5}}{3} = \frac{{y + 4}}{7} = \frac{{z - 6}}{2}$$. Write its vector form.

### Solution

It is given that the Cartesian equation of the line is $$\frac{{x - 5}}{3} = \frac{{y + 4}}{7} = \frac{{z - 6}}{2}$$

Hence,

The given line passes through the point $$\left( {{\rm{5}}, - {\rm{4}},{\rm{6}}} \right)$$

Therefoe,

The position vector of the point is $$\vec a = 5\hat i - 4\hat j + 6\hat k$$

Also, the direction ratios of the given line are $${\rm{3}},{\rm{7}}$$ and $${\rm{2}}$$

This means that the line is in the direction of the vector, $$\vec b = 3\hat i + 7\hat j + 2\hat k$$

As we known that the line through positive vector $$\vec a$$ and in the direction of the vector $$\vec b$$ is given by the equation, $$\vec r = \vec a + \lambda \vec b;\;\;\lambda\in R$$

Hence,

$$\Rightarrow \vec r = \left( {5\hat i - 4\hat j + 6\hat k} \right) + \lambda \left( {3\hat i + 7\hat j + 2\hat k} \right)$$

This is the required equation of the given line in vector form.

## Chapter 11 Ex.11.2 Question 8

Find the vector and the Cartesian equation of the lines that passes through the origin and $$\left( {5, - 2,3} \right)$$.

### Solution

The required line passes through the origin.

Therefore, its position vector is $$\vec a = 0\;\;\;\;\;\;\;\;\;\; \ldots \left( {\rm{1}} \right)$$

The direction ratios of the line passing through origin and $$\left( {5, - 2,3} \right)$$ are

\begin{align}\left( {{\rm{5}} - 0} \right) &= {\rm{5}}\\\left( { - {\rm{2}} - 0} \right)& =- {\rm{2}}\\\left( {{\rm{3}}-0} \right) &= {\rm{3 }}\end{align}

Hence, the line is parallel to the vector given by the equation, $$\vec b = 5\hat i - 2\hat j + 3\hat k{\rm{ }}$$

The equation of the line in vector form through a point with position vector $$\vec a$$ and parallel to $$\vec b$$ is,

\begin{align}&\Rightarrow \;\vec r = \vec a + \lambda \vec b;\;\;\lambda\in R\\&\Rightarrow \; \vec r = 0 + \lambda \left( {5\hat i - 2\hat j + 3\hat k} \right)\\&\Rightarrow \;\vec r = \lambda \left( {5\hat i - 2\hat j + 3\hat k} \right)\end{align}

The equation of the line through the point $$\left( {{x_1},{y_1},{z_1}} \right)$$, and direction ratios $$a,b,c$$ is given by, $$\frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c}$$

Hence, the equation of the required line in the Cartesian form is

\begin{align}&\Rightarrow \;\frac{{x - 0}}{5} = \frac{{y - 0}}{{ - 2}} = \frac{{z - 0}}{3}\\>&\Rightarrow\; \frac{x}{5} = \frac{y}{2} = \frac{z}{3}\end{align}

## Chapter 11 Ex.11.2 Question 9

Find the vector and the cartesian equations of the line that passes through the points $$\left( {{\rm{3}}, - {\rm{2}}, - {\rm{5}}} \right),\left( {{\rm{3}}, - {\rm{2}},{\rm{6}}} \right)$$.

### Solution

Let the line passing through the points, $$P\left( {{\rm{3}}, - {\rm{2}}, - {\rm{5}}} \right)$$ and $$Q\left( {{\rm{3}}, - {\rm{2}},{\rm{6}}} \right)$$ be PQ. Since PQ passes through $$P\left( {{\rm{3}}, - {\rm{2}}, - {\rm{5}}} \right)$$, its position vector is given by

$$\vec a = 3\hat i - 2\hat j - 5\hat k$$

The direction ratios of PQ are given by

\begin{align}\left( {3 - 3} \right) &= 0\\\left( { - 2 + 2} \right) &= 0\\\left( {6 + 5} \right)& = 11\end{align}

The equation of the vector in the direction of PQ is

\begin{align}\vec b &= 0.\hat i - 0.\hat j + 11\hat k{\rm{ }}\\&= 11\hat k{\rm{ }}\end{align}

The equation of PQ in vector form is given by,

\begin{align}\vec r &= \vec a + \lambda b,\;\;\lambda\in R\\&= \left( {3\hat i - 2\hat j + 5\hat k} \right) + 11\lambda \hat k\end{align}

The equation of PQ in Cartesian form is

\begin{align}&\Rightarrow\; \frac{{x - {x_1}}}{a} = \frac{{y - {y_1}}}{b} = \frac{{z - {z_1}}}{c}\\&\Rightarrow\; \frac{{x - 3}}{5} = \frac{{y + 2}}{2} = \frac{{z + 5}}{3}\end{align}

## Chapter 11 Ex.11.2 Question 10

Find the angle between the following pairs of lines:

(i) $$\vec r = 2\hat i - 5\hat j + \hat k + \lambda \left( {3\hat i + 2\hat j + 6\hat k} \right)$$ and $$\vec r = 7\hat i - 6\hat k + \mu \left( {\hat i + 2\hat j + 2\hat k} \right)$$

(ii) $$\vec r = 3\hat i + \hat j - 2\hat k + \lambda \left( {\hat i - \hat j - 2\hat k} \right)$$ and $$\vec r = 2\hat i - \hat j - 56\hat k + \mu \left( {3\hat i - 5\hat j - 4\hat k} \right)$$

### Solution

Let q be the angle between the given lines.

Then the angle between the given pairs of lines is given by

$$\cos {\rm{\theta }} = \left| {\frac{{{{\vec b}_1}.{{\vec b}_2}}}{{\left| {{{\vec b}_1}} \right|.\left| {{{\vec b}_2}} \right|}}} \right|$$

(i) The given lines are parallel to the vectors, $${\vec b_1} = 3\hat i + 2\hat j + 6\hat k$$ and $${\vec b_2} = \hat i + 2\hat j + 2\hat k$$, respectively.

Therefore,

\begin{align}\left| {{{\vec b}_1}} \right| &= \sqrt {{3^2} + {2^2} + {6^2}}= \sqrt {49}= 7\\\left| {{{\vec b}_2}} \right| &= \sqrt {{1^2} + {2^2} + {2^2}}= \sqrt 9= 3\\{{\vec b}_1}.{{\vec b}_2} &= \left( {3i + 2j + 6k} \right).\left( {i + 2j + 2k} \right)\\&= 3 \times 1 + 2 \times 2 + 6 \times 2\\&= 3 + 4 + 12\\&= 19\end{align}

Hence,

\begin{align}\cos {\rm{\theta }} &= \left| {\frac{{{{\vec b}_1}.{{\vec b}_2}}}{{\left| {{{\vec b}_1}} \right|.\left| {{{\vec b}_2}} \right|}}} \right|\\\cos {\rm{\theta }} &= \left| {\frac{{19}}{{7 \times 3}}} \right| = \frac{{19}}{{21}}\\{\rm{\theta }}& = {\cos ^{ - 1}}\left( {\frac{{19}}{{21}}} \right)\end{align}

(ii) The given lines are parallel to the vectors, $${\vec b_1} = \hat i - \hat j - 2\hat k$$ and $${\vec b_2} = 3\hat i - 5\hat j - 4\hat k$$, respectively.

Therefore,

\begin{align}\left| {{{\vec b}_1}} \right|& = \sqrt {{{\left( 1 \right)}^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 2} \right)}^2}}= \sqrt 6 \\\left| {{{\vec b}_2}} \right| &= \sqrt {{{\left( 3 \right)}^2} + {{\left( { - 5} \right)}^2} + {{\left( { - 4} \right)}^2}}= \sqrt {50}= 5\sqrt 2 \\{{\vec b}_1}.{{\vec b}_2} &= \left( {\hat i - \hat j - 2\hat k} \right).\left( {3\hat i - 5\hat j - 4\hat k} \right)\\&= 1 \times 3 - 1 \times \left( { - 5} \right) - 2 \times \left( { - 4} \right)\\&= 3 + 5 + 8\\&= 16\end{align}

Hence,

\begin{align}\cos {\rm{\theta }} &= \left| {\frac{{{{\vec b}_1}.{{\vec b}_2}}}{{\left| {{{\vec b}_1}} \right|.\left| {{{\vec b}_2}} \right|}}} \right|\\\cos {\rm{\theta }} &= \left| {\frac{{16}}{{\left( {\sqrt 6 } \right).\left( {5\sqrt 2 } \right)}}} \right| = \left| {\frac{{16}}{{\sqrt 2 .\sqrt 3 .5\sqrt 2 }}} \right| = \left| {\frac{{16}}{{10\sqrt 3 }}} \right|\\\cos {\rm{\theta }} &= \frac{8}{{5\sqrt 3 }}\\{\rm{\theta }} &= {\cos ^{ - 1}}\left( {\frac{8}{{5\sqrt 3 }}} \right)\end{align}

## Chapter 11 Ex.11.2 Question 11

Find the angle between the following pair of lines:

(i) $$\frac{{x - 2}}{2} = \frac{{y - 1}}{5} = \frac{{z + 3}}{{ - 3}}$$ and $$\frac{{x + 2}}{{ - 1}} = \frac{{y - 4}}{8} = \frac{{z - 5}}{4}$$

(ii) $$\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$$ and $$\frac{{x - 5}}{4} = \frac{{y - 2}}{1} = \frac{{z - 3}}{8}$$

### Solution

(i) Let $${\vec b_1}$$ and $${\vec b_2}$$ be the vectors parallel to the line pair of lines $$\frac{{x - 2}}{2} = \frac{{y - 1}}{5} = \frac{{z + 3}}{{ - 3}}$$ and $$\frac{{x + 2}}{{ - 1}} = \frac{{y - 4}}{8} = \frac{{z - 5}}{4}$$ respectively.

Hence, $${\vec b_1} = 2\hat i + 5\hat j - 3\hat k$$ and $${\vec b_2} =- i + 8\hat j + 4\hat k$$

Therefore,

\begin{align}\left| {{{\vec b}_1}} \right| &= \sqrt {{{\left( 2 \right)}^2} + {{\left( 5 \right)}^2} + {{\left( { - 3} \right)}^2}}= \sqrt {38} \\\left| {{{\vec b}_2}} \right| &= \sqrt {{{\left( { - 1} \right)}^2} + {{\left( 8 \right)}^2} + {{\left( 4 \right)}^2}}= \sqrt {81}= 9\\{{\vec b}_1}.{{\vec b}_2}& = \left( {2\hat i + 5\hat j - 3\hat k} \right).\left( { - \hat i + 8\hat j + 4\hat k} \right)\\&= 2 \times \left( { - 1} \right) + 5 \times 8 + \left( { - 3} \right) \times 4\\&=- 2 + 40 - 12\\&= 26\end{align}

The angle q between the given pair of lines is given by the relation,

\begin{align}\cos {\rm{\theta }} &= \left| {\frac{{{{\vec b}_1}.{{\vec b}_2}}}{{\left| {{{\vec b}_1}} \right|.\left| {{{\vec b}_2}} \right|}}} \right|\\\cos {\rm{\theta }} &= \left| {\frac{{26}}{{\sqrt {38}\times 9}}} \right| = \frac{{26}}{{9\sqrt {38} }}\\{\rm{\theta }} &= {\cos ^{ - 1}}\left( {\frac{{26}}{{9\sqrt {38} }}} \right)\end{align}

(ii) Let $${\vec b_1}$$ and $${\vec b_2}$$ be the vectors parallel to the given pair of lines $$\frac{x}{2} = \frac{y}{2} = \frac{z}{1}$$ and $$\frac{{x - 5}}{4} = \frac{{y - 2}}{1} = \frac{{z - 3}}{8}$$, respectively.

Hence, $${\vec b_1} = 2\hat i + 2\hat j + \hat k$$ and $${\vec b_2} = 4i + \hat j + 8\hat k$$

Therefore,

\begin{align}\left| {{{\vec b}_1}} \right| &= \sqrt {{{(2)}^2} + {{(2)}^2} + {{(1)}^2}}= \sqrt 9= 3\\\left| {{{\vec b}_2}} \right| &= \sqrt {{{(4)}^2} + {{(1)}^2} + {{(8)}^2}}= \sqrt {81}= 9\\{{\vec b}_1}.{{\vec b}_2} &= \left( {2\hat i + 2\hat j + \hat k} \right).\left( {4\hat i + \hat j + 8\hat k} \right)\\ &= 2 \times 4 + 2 \times 1 + 1 \times 8\\ &= 8 + 2 + 8\\ &= 18\end{align}

If q is the angle between the pair of lines, then $$\cos {\rm{\theta }} = \left| {\frac{{{{\vec b}_1}.{{\vec b}_2}}}{{\left| {{{\vec b}_1}} \right|.\left| {{{\vec b}_2}} \right|}}} \right|$$

\begin{align}\cos {\rm{\theta }} &= \left| {\frac{{{{\vec b}_1}.{{\vec b}_2}}}{{\left| {{{\vec b}_1}} \right|.\left| {{{\vec b}_2}} \right|}}} \right|\\\cos {\rm{\theta }} &= \left| {\frac{{18}}{{3 \times 9}}} \right| = \frac{2}{3}\\{\rm{\theta }} &= {\cos ^{ - 1}}\left( {\frac{2}{3}} \right)\end{align}

## Chapter 11 Ex.11.2 Question 12

Find the values of p so the line $$\frac{{1 - x}}{3} = \frac{{7y - 14}}{{2p}} = \frac{{z - 3}}{2}$$ and $$\frac{{7 - 7x}}{{3p}} = \frac{{y - 5}}{1} = \frac{{6 - z}}{5}$$ are at right angles.

### Solution

The given equations can be written in the standard form as $$\frac{{1 - x}}{3} = \frac{{7y - 14}}{{2p}} = \frac{{z - 3}}{2}$$ and $$\frac{{7 - 7x}}{{3p}} = \frac{{y - 5}}{1} = \frac{{6 - z}}{5}$$

The direction ratios of the lines are given by

$${a_1} =- 3,\;{b_1} = \frac{{2p}}{7}$$ and $${c_1} = 2$$

$${a_2} = \frac{{ - 3p}}{7},\;{b_2} = 1$$ and $${c_2} =- 5$$

Since, both the lines are perpendicular to each other,

Therefore,

\begin{align}{a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}& = 0\\\left( { - 3} \right) \times \left( {\frac{{ - 3p}}{7}} \right) + \left( {\frac{{2p}}{7}} \right) \times 1 + 2 \times \left( { - 5} \right) &= 0\\\frac{{9p}}{7} + \frac{{2p}}{7} - 10& = 0\\\frac{{11}}{7}p &= 10\\11p& = 10 \times 7\\p& = \frac{{70}}{{11}}\end{align}

Hence the value of $$p = \frac{{70}}{{11}}$$

## Chapter 11 Ex.11.2 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

The equations of the given lines are $$\frac{{x - 5}}{7} = \frac{{y + 2}}{{ - 5}} = \frac{z}{1}$$ and $$\frac{x}{1} = \frac{y}{2} = \frac{z}{3}$$

Here,

$${a_1} = 7,\;{b_1} =- 5$$ and $${c_1} = 1$$

$${a_2} = 1,\;{b_2} = 2$$ and $${c_2} = 3$$

Two lines with direction ratios, $${a_1},{b_1},{c_1}$$ and $${a_2},{b_2},{c_2}$$ are perpendicular to each other, if $${a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2} = 0$$

Since,

\begin{align}7 \times 1 + \left( { - 5} \right) \times 2 + 1 \times 3 &= 7 - 10 + 3\\&= 0\end{align}

Hence, the given lines are perpendicular to each other.

## Chapter 11 Ex.11.2 Question 14

Find the shortest distance between the lines

$$\vec r = \left( {\hat i + 2\hat j + \hat k} \right) + \lambda \left( {\hat i - \hat j + \hat k} \right)$$ and $$\vec r = 2\hat i - \hat j - \hat k + \mu \left( {2\hat i + \hat j + 2\hat k} \right)$$

### Solution

Given lines are $$\vec r = \left( {\hat i + 2\hat j + \hat k} \right) + \lambda \left( {\hat i - \hat j + \hat k} \right)$$ and $$\vec r = 2\hat i - \hat j - \hat k + \mu \left( {2\hat i + \hat j + 2\hat k} \right)$$

Hence,

$${\vec a_1} = \left( {\hat i + 2\hat j + \hat k} \right)$$ and $${\vec b_1} = \left( {\hat i - \hat j + \hat k} \right)$$

and $${\vec b_2} = \left( {2\hat i + \hat j + 2\hat k} \right)$$

Shortest distance between the lines $$\vec r = {\vec a_1} + \lambda {\vec b_1}$$ and $$\vec r = {\vec a_2} + \mu {\vec b_2}$$ is given by,

$$d = \left| {\frac{{\left( {{{\vec b}_1} \times {{\vec b}_2}} \right).\left( {{{\vec a}_2} - {{\vec a}_1}} \right)}}{{\left| {{{\vec b}_1} \times {{\vec b}_2}} \right|}}} \right|\;\;\;\;\;\;\;\;\;\;\; \ldots \left( {\rm{1}} \right)$$

Here,

\begin{align}{{\vec a}_2} - {{\vec a}_1} &= \left( {2\hat i - \hat j - \hat k} \right) - \left( {\hat i + 2\hat j + \hat k} \right) = \hat i - 3\hat j - 2\hat k\\{{\vec b}_1} \times {{\vec b}_2} &= \left| {\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\1&{ - 1}&1\\2&1&2\end{array}} \right|\\&= \left( { - 2 - 1} \right)\hat i - \left( {2 - 2} \right)\hat j + \left( {1 + 2} \right)\hat k\\&=- 3\hat i + 3\hat k\\\left| {{{\vec b}_1} \times {{\vec b}_2}} \right| &= \sqrt {{{\left( { - 3} \right)}^2} + {{\left( 3 \right)}^2}} \\&= \sqrt {9 + 9} \\&= \sqrt {18} \\&= 3\sqrt 2\end{align}

Putting all the values in equation (1), we get

\begin{align}d &= \left| {\frac{{\left( { - 3\hat i + 3\hat k} \right).\left( {\hat i - 3\hat j - 2\hat k} \right)}}{{3\sqrt 2 }}} \right|\\&= \left| {\frac{{ - 3.1 + 3\left( { - 2} \right)}}{{3\sqrt 2 }}} \right|\\&= \left| {\frac{{ - 9}}{{3\sqrt 2 }}} \right|\\&= \frac{3}{{\sqrt 2 }}\\&= \frac{3}{{\sqrt 2 }} \times \frac{{\sqrt 2 }}{{\sqrt 2 }}\\&= \frac{{3\sqrt 2 }}{2}\end{align}

Hence, the shortest distance between the two lines is $$\frac{{3\sqrt 2 }}{2}$$ units.

## Chapter 11 Ex.11.2 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

The given lines are $$\frac{{x + 1}}{7} = \frac{{y + 1}}{{ - 6}} = \frac{{z + 1}}{1}$$ and $$\frac{{x - 3}}{1} = \frac{{y - 5}}{{ - 2}} = \frac{{z - 7}}{1}$$

The shortest distance between the two lines,

$$\frac{{x - {x_1}}}{{{a_1}}} = \frac{{y - {y_1}}}{{{b_1}}} = \frac{{z - {z_1}}}{{{c_1}}}$$ and $$\frac{{x - {x_2}}}{{{a_2}}} = \frac{{y - {y_2}}}{{{b_2}}} = \frac{{z - {z_2}}}{{{c_2}}}$$ is given by,

$$d = \frac{{\left| {\begin{array}{*{20}{c}}{{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\end{array}} \right|}}{{\sqrt {{{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}^2} + {{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}^2} + {{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2}} }}\;\;\;\;\;\;\;\;\;\;\; \ldots \left( {\rm{1}} \right)$$

Here,

$${x_1} =- 1,\;{y_1} =- 1,\;{z_1} =- 1$$ and $${x_2} = 3,\;{y_2} = 5,\;{z_2} = 7$$

$${a_1} = 7,\;{b_1} =- 6,\;{c_1} = 1$$ and $${a_2} = 1,\;{b_2} =- 2,\;{c_2} = 1$$

Hence,

\begin{align}\left| {\begin{array}{*{20}{c}}{{x_2} - {x_1}}&{{y_2} - {y_1}}&{{z_2} - {z_1}}\\{{a_1}}&{{b_1}}&{{c_1}}\\{{a_2}}&{{b_2}}&{{c_2}}\end{array}} \right| &= \left| {\begin{array}{*{20}{c}}4&6&8\\7&{ - 6}&1\\1&{ - 2}&1\end{array}} \right|\\&= 4\left( { - 6 + 2} \right) - 6\left( {1 + 7} \right) + 8\left( { - 14 + 6} \right)\\&=- 16 - 36 - 64\\&=- 116\end{align}

Also,

\begin{align}\sqrt {{{\left( {{b_1}{c_2} - {b_2}{c_1}} \right)}^2} + {{\left( {{c_1}{a_2} - {c_2}{a_1}} \right)}^2} + {{\left( {{a_1}{b_2} - {a_2}{b_1}} \right)}^2}}&= \sqrt {{{\left( { - 6 + 2} \right)}^2} + {{\left( {1 + 7} \right)}^2} + {{\left( { - 14 + 6} \right)}^2}} \\&= \sqrt {16 + 36 + 64} \\&= \sqrt {116}\end{align}

Putting all the values in equation (1), we get

\begin{align}d &= \frac{{ - 116}}{{\sqrt {116} }}\\&=- \sqrt {116} \\&=- 2\sqrt {29} \\\left| d \right| &= 2\sqrt {29}\end{align}

Therefore, the distance between the given lines is $$2\sqrt {29}$$ units.

## Chapter 11 Ex.11.2 Question 16

Find the shortest distance between the lines whose vector equations are

$$\vec r = \left( {\hat i + 2\hat j + 3\hat k} \right) + \lambda \left( {\hat i - 3\hat j + 2\hat k} \right)$$ and $$\vec r = 4\hat i + 5\hat j + 6\hat k + \mu \left( {2\hat i + 3\hat j + \hat k} \right)$$.

### Solution

The given lines are $$\vec r = \left( {\hat i + 2\hat j + 3\hat k} \right) + \lambda \left( {\hat i - 3\hat j + 2\hat k} \right)$$ and $$\vec r = 4\hat i + 5\hat j + 6\hat k + \mu \left( {2\hat i + 3\hat j + \hat k} \right)$$

Hence,

$${\vec a_1} = \left( {\hat i + 2\hat j + 3\hat k} \right)$$ and $${\vec b_1} = \left( {\hat i - 3\hat j + 2\hat k} \right)$$

$${\vec a_2} = \left( {4\hat i + 5\hat j + 6\hat k} \right)$$ and $${\vec b_2} = \left( {2\hat i + 3\hat j + \hat k} \right)$$

Shortest distance between the lines $$\vec r = {\vec a_1} + \lambda {\vec b_1}$$ and $$\vec r = {\vec a_2} + \mu {\vec b_2}$$ is given by,

$$d = \left| {\frac{{\left( {{{\vec b}_1} \times {{\vec b}_2}} \right).\left( {{{\vec a}_2} - {{\vec a}_1}} \right)}}{{\left| {{{\vec b}_1} \times {{\vec b}_2}} \right|}}} \right|\;\;\;\;\;\;\;\;\;\;\; \ldots \left( {\rm{1}} \right)$$

Here,

\begin{align}{{\vec a}_2} - {{\vec a}_1} &= \left( {4\hat i + 5\hat j + 6\hat k} \right) - \left( {\hat i + 2\hat j + 3\hat k} \right) = 3\hat i + 3\hat j + 3\hat k\\{{\vec b}_1} \times {{\vec b}_2}& = \left| {\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\1&{ - 3}&2\\2&3&1\end{array}} \right|\\&= \left( { - 3 - 6} \right)\hat i - \left( {1 - 4} \right)\hat j + \left( {3 + 6} \right)\hat k\\&=- 9\hat i + 3\hat j + 9\hat k\\\left| {{{\vec b}_1} \times {{\vec b}_2}} \right| &= \sqrt {{{\left( { - 9} \right)}^2} + {{\left( 3 \right)}^2} + {{\left( 9 \right)}^2}} \\&= \sqrt {81 + 9 + 81} \\&= \sqrt {171} \\&= 3\sqrt {19}\end{align}

Putting all the values in equation (1), we get

\begin{align}d &= \left| {\frac{{\left( { - 9\hat i + 3\hat j + 9\hat k} \right).\left( {3\hat i + 3\hat j + 3\hat k} \right)}}{{3\sqrt {19} }}} \right|\\&= \left| {\frac{{ - 9 \times 3 + 3 \times 3 + 9 \times 3}}{{3\sqrt {19} }}} \right|\\&= \left| {\frac{{ - 27 + 9 + 27}}{{3\sqrt {19} }}} \right|\\&= \left| {\frac{9}{{3\sqrt {19} }}} \right|\\&= \frac{3}{{\sqrt {19} }}\end{align}

Hence, the shortest distance between the two lines is $$\frac{3}{{\sqrt {19} }}$$ units.

## Chapter 11 Ex.11.2 Question 17

Find the shortest distance between the lines whose vector equations are

$$\vec r = \left( {1 - t} \right)\hat i + \left( {t - 2} \right)\hat j + \left( {3 - 2t} \right)\hat k$$ and $$\vec r = \left( {s + 1} \right)\hat i + \left( {2s - 1} \right)\hat j - \left( {2s + 1} \right)\hat k$$.

### Solution

The given lines are $$\vec r = \left( {1 - t} \right)\hat i + \left( {t - 2} \right)\hat j + \left( {3 - 2t} \right)\hat k$$ and $$\vec r = \left( {s + 1} \right)\hat i + \left( {2s - 1} \right)\hat j - \left( {2s + 1} \right)\hat k$$

i.e., $$\vec r = \left( {\hat i - 2\hat j + 3\hat k} \right) + t\left( { - \hat i + \hat j - 2\hat k} \right)$$ and $$\vec r = \left( {\hat i - \hat j + \hat k} \right) + s\left( {\hat i + 2\hat j - 2\hat k} \right)$$

Hence,

$${\vec a_1} = \left( {\hat i - 2\hat j + 3\hat k} \right)$$ and $${\vec b_1} = \left( { - \hat i + \hat j - 2\hat k} \right)$$

$${\vec a_2} = \left( {\hat i - \hat j - \hat k} \right)$$ and $${\vec b_2} = \left( {\hat i + 2\hat j - 2\hat k} \right)$$

Shortest distance between the lines $$\vec r = {\vec a_1} + \lambda {\vec b_1}$$ and $$\vec r = {\vec a_2} + \mu {\vec b_2}$$ is given by,

$$d = \left| {\frac{{\left( {{{\vec b}_1} \times {{\vec b}_2}} \right).\left( {{{\vec a}_2} - {{\vec a}_1}} \right)}}{{\left| {{{\vec b}_1} \times {{\vec b}_2}} \right|}}} \right|\;\;\;\;\;\;\;\;\;\;\; \ldots \left( {\rm{1}} \right)$$

Here,

\begin{align}{{\vec a}_2} - {{\vec a}_1} &= \left( {\hat i - \hat j - \hat k} \right) - \left( {\hat i - 2\hat j + 3\hat k} \right) = \hat j - 4\hat k\\{{\vec b}_1} \times {{\vec b}_2}& = \left| {\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\{ - 1}&1&{ - 2}\\1&2&{ - 2}\end{array}} \right|\\&= \left( { - 2 + 4} \right)\hat i - \left( {2 + 2} \right)\hat j + \left( { - 2 - 1} \right)\hat k\\&= 2\hat i - 4\hat j - 3\hat k\\\left| {{{\vec b}_1} \times {{\vec b}_2}} \right|& = \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( { - 3} \right)}^2}} \\&= \sqrt {4 + 16 + 9} \\&= \sqrt {29}\end{align}

Putting all the values in equation (1), we get

\begin{align}d &= \left| {\frac{{\left( {2\hat i - 4\hat j - 3\hat k} \right).\left( {\hat j - 4\hat k} \right)}}{{\sqrt {29} }}} \right|\\&= \left| {\frac{{ - 4 \times 1 - 3 \times \left( { - 4} \right)}}{{\sqrt {29} }}} \right|\\&= \left| {\frac{{ - 4 + 12}}{{\sqrt {29} }}} \right|\\&= \frac{8}{{\sqrt {29} }}\end{align}

Hence, the shortest distance between the lines is $$\frac{8}{{\sqrt {29} }}$$units.

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