Ex 11.2 Class 12 Maths Ncert Solutions

We know that
If \( l_{1}, \mathrm{~m}_{1}, \mathrm{n}_{1} \) and \( \mathrm{l}_{2}, \mathrm{~m}_{2}, \mathrm{n}_{2} \) are the direction cosines of two lines; and \( \theta \) is the acute angle between the two lines; then \( \cos \theta=\left|l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}\right| \)
If two lines are perpendicular, then the angle between the two is \( \theta=90^{\circ} \)
\( \Rightarrow \text {For perpendicular lines,}\left|l_{1} l_{2}+m_{1} m_{2}+n_{1} n_{2}\right|=\cos 90^{\circ}=0 \), i.e.
\( \left|l_{1} l_{2}+\mathrm{m}_{1} \mathrm{~m}_{2}+\mathrm{n}_{1} \mathrm{n}_{2}\right|=0 \)
So, in order to check if the three lines are mutually perpendicular, we compute \( \left|l_{1} l_{2}+m_{1} m_{2}+\mathrm{n}_{1} \mathrm{n}_{2}\right| \) for all the pairs of the three lines.
Now let the direction cosines of \( L_{1}, L_{2} \) and \( L_{3} \) be \( l_{1}, m_{1}, n_{1} ; l_{2}, m_{2}, n_{2} \) and \( l_{3}, \mathrm{~m}_{3}, \mathrm{n}_{3} \).
First, consider
\(\Rightarrow\left|l_{1} l_{2}+\mathrm{m}_{1} \mathrm{~m}_{2}+\mathrm{n}_{1} \mathrm{n}_{2}\right|=\left|\left(\frac{12}{13} \times \frac{4}{13}\right)+\left(\frac{-3}{13} \times \frac{12}{13}\right)+\left(\frac{-4}{13} \times \frac{3}{13}\right)\right|\)
\(=\frac{48}{13}+\left(\frac{-36}{13}\right)+\left(\frac{-12}{13}\right)\)
\(\Rightarrow=\frac{48+(-48)}{13}=0\)
\(\Rightarrow \mathrm{L}_{1} \perp \mathrm{L}_{2}\ldots \) (i)
Next, consider
\(\Rightarrow\)
\(\left|l_{2} l_{3}+\mathrm{m}_{2} \mathrm{~m}_{3}+\mathrm{n}_{2} \mathrm{n}_{3}\right|=\left|\left(\frac{4}{13} \times \frac{3}{13}\right)+\left(\frac{12}{13} \times \frac{-4}{13}\right)+\left(\frac{3}{13} \times \frac{12}{13}\right)\right|\)
\(=\frac{12}{13}+\left(\frac{-48}{13}\right)+\frac{36}{13}\)
\(\Rightarrow=\frac{(-48)+48}{13}=0\)
\(\Rightarrow \mathrm{L}_{2} \perp \mathrm{L}_{3}\ldots \text { (ii) }\)
Now, consider
\(\Rightarrow\left|l_{3} l_{1}+\mathrm{m}_{3} \mathrm{~m}_{1}+\mathrm{n}_{3} \mathrm{n}_{1}\right|=\left|\left(\frac{3}{13} \times \frac{12}{13}\right)+\left(\frac{-4}{13} \times \frac{-3}{13}\right)+\left(\frac{12}{13} \times \frac{-4}{13}\right)\right|\)
\(=\frac{36}{13}+\frac{12}{13}+\left(\frac{-48}{13}\right)\)
\(\Rightarrow=\frac{48+(-48)}{13}=0\)
\(\Rightarrow L_{1} \perp L_{3} \ldots \text { (iii) }\)
\( \therefore \) By (i), (ii) and (iii), we have
\( \mathrm{L}_{1}, \mathrm{~L}_{2} \) and \( \mathrm{L}_{3} \) are mutually perpendicular.

We know that
Two lines with direction ratios \( a_{1}, b_{1}, c_{1} \) and \( \mathrm{a}_{2}, \mathrm{~b}_{2}, \mathrm{c}_{2} \) are perpendicular if the angle between them is \( \theta=90^{\circ} \), i.e. \( a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0 \)
Also, we know that the direction ratios of the line segment joining \( \left(\mathrm{x}_{1},\mathrm{y}_{1},\right.\left.\mathrm{z}_{1}\right) \) and \( \left(\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right) \) is taken as \( \mathrm{x}_{2}-\mathrm{x}_{1}, \mathrm{y}_{2}-\mathrm{y}_{1}, \mathrm{z}_{2}-\mathrm{z}_{1}\left(\right. \) or \( \mathrm{x}_{1}-\mathrm{x}_{2}, \mathrm{y}_{1}-\mathrm{y}_{2} \), \( \left.\mathrm{Z}_{1}-\mathrm{Z}_{2}\right) \).
\( \Rightarrow \) The direction ratios of the line through the points \( (1,-1,2) \) and \( (3,4,- \) 2 ) is:
\(a_{1}=3-1=2, b_{1}=4-(-1)=4+1=5, c_{1}=-2-2=-4\)
and the direction ratios of the line through the points \( (0,3,2) \) and \( (3,5,6) \) is:
\(a_{2}=3-0=3, b_{2}=5-3=2, c_{2}=6-2=4\)
Now, consider
\(a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=2 \times 3+5 \times 2+(-4) \times 4=6+10+(-16)=16+(-16)\)
\(=0\)
\( \Rightarrow \)The line through the points \( (1,-1,2),(3,4,-2) \) is perpendicular to the line through the points \( (0,3,2) \) and \( (3,5,6) \).

We know that
Two lines with direction ratios \( a_{1}, b_{1}, c_{1} \) and \( a_{2}, b_{2}, c_{2} \) are parallel if the angle between them is \( \theta=0^{\circ} \), i. e.
\(\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}\)
Also, we know that the direction ratios of the line segment joining ( \( \mathrm{x}_{1}, \mathrm{y}_{1} \),\( \mathrm{z}_{1} \) ) and ( \( \left.\mathrm{x}_{2}, \mathrm{y}_{2}, \mathrm{z}_{2}\right) \) is taken as \( \mathrm{x}_{2}-\mathrm{x}_{1}, \mathrm{y}_{2}-\mathrm{y}_{1}, \mathrm{z}_{2}-\mathrm{z}_{1}\left(\right. \) or \( \mathrm{x}_{1}-\mathrm{x}_{2}, \mathrm{y}_{1}-\mathrm{y}_{2} \), \( \mathrm{z}_{1}-\mathrm{z}_{2} \) ).
\( \Rightarrow \) The direction ratios of the line through the points \( (4,7,8) \) and \( (2,3,4) \) is:
\(a_{1}=2-4=-2, b_{1}=3-7=-4, c_{1}=4-8=-4\)
And the direction ratios of the line through the points \( (-1,-2,1) \) and (1, 2,5 ) is:
\(a_{2}=1-(-1)=1+1=2, b_{2}=2-(-2)=2+2=4, c_{2}=5-1=4\)
Consider \( \frac{a_{1}}{a_{2}}=\frac{-2}{2}=-1, \frac{b_{1}}{b_{2}}=\frac{-4}{4}=-1 \frac{c_{1}}{c_{2}}=\frac{-4}{4}=-1 \)
\(\Rightarrow \frac{a_{1}}{a_{2}}=\frac{b_{1}}{b_{2}}=\frac{c_{1}}{c_{2}}=-1\)
\( \therefore \) The line through the points \( (4,7,8),(2,3,4) \) is parallel to the line through the points \( (-1,-2,1),(1,2,5) \).

We know that
Vector equation of a line that passes through a given point whose position vector is \( \vec{a} \) and parallel to a given vector \( \vec{b} \) is \( \vec{r}=\vec{a}+\lambda \vec{b} \).
So, here the position vector of the point \( (1,2,3) \) is given by \( \vec{a}=\hat{i}+2 \hat{j}- \) \( 3 \hat{k} \) and the parallel vector is \( 3 \hat{i}+2 \hat{j}-2 \hat{k} \).
\( \therefore \) The vector equation of the required line is:
\( \vec{r}=\hat{i}+2 \hat{j}-3 \hat{k}+\lambda(3 \hat{i}+2 \hat{j}-2 \hat{k}) \), where \( \lambda \) is a real number.

We know that
Vector equation of a line that passes through a given point whose position vector is \( \vec{a} \) and parallel to a given vector \( \vec{b} \) is \( \vec{r}=\vec{a}+\lambda \vec{b} \).
Here, \( \vec{a}=2 \hat{i}-\hat{j}+4 \hat{k} \) and \( \vec{b}=\hat{i}+2 \hat{j}-\hat{k} \)
\( \Rightarrow \) The vector equation of the required line is:
\(\Rightarrow \overrightarrow{\mathrm{r}}=2 \hat{i}-\hat{j}+4 \hat{\mathrm{k}}+\lambda(\hat{i}+2 \hat{j}-\hat{\mathrm{k}})\)
Also, we know that
The Cartesian equation of a line through a point \( \left(x_{1}, y_{1}, z_{1}\right) \) and having direction cosines \( l, \mathrm{~m}, \mathrm{n} \) is \( \frac{x-x_{1}}{\mathrm{l}}=\frac{y-y_{1}}{\mathrm{~m}}=\frac{z-z_{1}}{\mathrm{n}} \)
Also, we know that if the direction ratios of the line are \( a, b \), \( c \), then
\(\Rightarrow l=\frac{a}{\sqrt{a^{2}+b^{2}=c^{2}}}, \mathrm{~m}=\frac{b}{\sqrt{a^{2}+b^{2}=c^{2}}}, \mathrm{n}=\frac{c}{\sqrt{a^{2}+b^{2}=c^{2}}}\)
\( \Rightarrow \) The Cartesian equation of a line through a point \( \left(x_{1}, y_{1}, z_{1}\right) \) and having direction ratios \( \mathrm{a}, \mathrm{b}, \mathrm{c} \) is \( \frac{x-x_{1}}{\mathrm{a}}=\frac{y-y_{1}}{\mathrm{~b}}=\frac{z-z_{1}}{\mathrm{c}} \)
Here, \( x_{1}=2, y_{1}=-1, z_{1}=4 \) and \( \mathrm{a}=1, \mathrm{~b}=2, \mathrm{c}=-1 \)
\( \Rightarrow \) The Cartesian equation of the required line is:
\(\frac{x-2}{1}=\frac{y-(-1)}{2}=\frac{z-4}{-1} \Rightarrow \frac{x-2}{1}=\frac{y+1}{2}=\frac{z-4}{-1}\)

We know that
The Cartesian equation of a line through a point \( \left(x_{1}, y_{1}, z_{1}\right) \) and having direction ratios a, b, c is \( \frac{x-x_{1}}{\mathrm{a}}=\frac{y-y_{1}}{\mathrm{~b}}=\frac{z-z_{1}}{\mathrm{c}} \)
Here, the point \( \left(x_{1}, y_{1}, z_{1}\right) \) is \( (-2,4,-5) \) and the direction ratios are: \( \mathrm{a}=3, \mathrm{~b}=5, \mathrm{c}=6 \)
\( \Rightarrow \) The Cartesian equation of the required line is:
\(\Rightarrow \frac{x-(-2)}{3}=\frac{y-4}{5}=\frac{z-(-5)}{6} \Rightarrow \frac{x+2}{3}=\frac{y-4}{5}=\frac{z+5}{6}\)

We know that
The Cartesian equation of a line through a point \( \left(x_{1}, y_{1}, z_{1}\right) \) and having direction cosines \( l, \mathrm{~m}, \mathrm{n} \) is \( \frac{x-x_{1}}{\mathrm{l}}=\frac{y-y_{1}}{\mathrm{~m}}=\frac{z-z_{1}}{\mathrm{n}} \)
Comparing this standard form with the given equation, we get \( x_{1}=5, y_{1}=-4, z_{1}=6 \) and \( l=3, \mathrm{~m}=7, \mathrm{n}=2 \)
\( \Rightarrow \) The point through which the line passes has the position vector \( \vec{a}= \) \( 5 \hat{i}-4 \hat{j}+6 \hat{k} \) and the vector parallel to the line is given by \( \vec{b}=3 \hat{i}-7 \hat{j}+ \) \( 2 \hat{\mathrm{k}} \)
Now, \( \because \) Vector equation of a line that passes through a given point whose position vector is \( \vec{a} \) and parallel to a given vector \( \vec{b} \) is \( \vec{r}=\vec{a}+\lambda \vec{b}\).
\( \therefore \) The vector equation of the required line is:
\(\vec{r}=(5 \hat{i}-4 \hat{j}+6 \hat{k})+\lambda(3 \hat{i}+7 \hat{j}+2 \hat{k})\)

We know that
The vector equation of as line which passes through two points whose position vectors are \( \vec{a} \) and \( \vec{b} \) is \( \vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a}) \).
Here, the position vectors of the two points \( (0,0,0) \) and \( (5,-2,3) \) are \( \vec{a}= \) \( 0 \hat{i}-0 \hat{j}+0 \hat{k} \) and \( \vec{b}=5 \hat{i}-2 \hat{j}+3 \hat{k} \) respectively.
So, the vector equation of the required line is:
\(\overrightarrow{\mathrm{r}}=0 \hat{i}-0 \hat{j}+0 \hat{\mathrm{k}}+\lambda[(5 \hat{i}-2 \hat{j}+3 \hat{\mathrm{k}})(0 \hat{i}-0 \hat{j}+0 \hat{\mathrm{k}})]\)
\(\Rightarrow \overrightarrow{\mathrm{r}}=\lambda(5 \hat{i}-2 \hat{j}+3 \hat{\mathrm{k}})\)
Now, we know that
Cartesian equation of a line that passes through two points \( \left(x_{1}, y_{1}, z_{1}\right) \) and \( \left(x_{2}, y_{2}, z_{2}\right) \) is
\(\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}\)
So, the Cartesian equation of the line that passes through the origin \( (0,0 \), \( 0) \) and \( (5,-2,3) \) is
\(\frac{x-0}{5-0}=\frac{y-0}{-2-0}=\frac{z-0}{3-0} \Rightarrow \frac{x}{5}=\frac{y}{-2}=\frac{z}{3}\)

We know that
The vector equation of as line which passes through two points whose position vectors are \( \vec{a} \) and \( \vec{b} \) is \( \vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a}) \).
Here, the position vectors of the two points \( (3,-2,-5) \) and \( (3,-2,6) \) are \( \vec{a}=3 \hat{i}-2 \hat{j}+5 \hat{k} \) and \( \vec{b}=3 \hat{i}-2 \hat{j}+6 \hat{k} \), respectively.
So, the vector equation of the required line is:
\(\Rightarrow \overrightarrow{\mathrm{r}}=3 \hat{i}-2 \hat{j}+5 \hat{\mathrm{k}}+\lambda[(3 \hat{i}-2 \hat{j}+6 \hat{k})(3 \hat{i}-2 \hat{j}+5 \hat{k})]\)
\(\Rightarrow \overrightarrow{\mathrm{r}}=3 \hat{i}-2 \hat{j}+5 \hat{k}+\lambda(3 \hat{i}-2 \hat{j}+6 \hat{k}-3 \hat{i}-2 \hat{j}+5 \hat{k})\)
\(\Rightarrow \overrightarrow{\mathrm{r}}=3 \hat{i}-2 \hat{j}+5 \hat{\mathrm{k}}+\lambda(11 \hat{\mathrm{k}})\)
Now, we also know that
Cartesian equation of a line that passes through two points \( \left(x_{1}, y_{1}, z_{1}\right) \) and \( \left(x_{2}, y_{2}, z_{2}\right) \) is
\(\frac{x-x_{1}}{x_{2}-x_{1}}=\frac{y-y_{1}}{y_{2}-y_{1}}=\frac{z-z_{1}}{z_{2}-z_{1}}\)
So, the Cartesian equation of the line that passes through the origin ( \( 3,- \) \( 2,-5) \) and \( (3,-2,6) \) is
\(\frac{x-3}{3-3}=\frac{y-(-2)}{(-2)-(-2)}=\frac{z-(-5)}{6-(-5)}\)
\(\Rightarrow \frac{x-3}{0}=\frac{y+2}{0}=\frac{z+5}{0}\)

We know that
If \( \theta \) is the acute angle between \( \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}_{1}}+\lambda \overrightarrow{\mathrm{b}_{1}} \) and \( \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}_{2}}+\mu \overrightarrow{\mathrm{b}_{2}} \) then
\( \cos \theta=\left|\frac{\overrightarrow{b_{1}}, \overrightarrow{b_{2}}}{ \mid\overrightarrow{b_{1}} \mid, \mid \overrightarrow{b_{2}\mid}}\right| \ldots \) (i)
\( \overrightarrow{\mathrm{r}}=2 \hat{i}-5 \hat{j}+\hat{\mathrm{k}}+\lambda(3 \hat{i}+2 \hat{j}+6 \hat{\mathrm{k}}) \ \text{and}\) \( \overrightarrow{\mathrm{r}}=7 \hat{i}-6 \hat{j}+\mu(\hat{i}+2 \hat{j}+2 \hat{\mathrm{k}}) \)
Here \( \overrightarrow{b_{1}}=3 \hat{i}+2 \hat{j}+6 \hat{k} \) and \( \overrightarrow{b_{2}}=\hat{i}+2 \hat{j}+2 \hat{k} \)
So, from (i), we have
\(\cos \theta=\left|\frac{(3 \hat{i}+2 \hat{j}+6 \widehat{k}),(\hat{i}+2 \hat{j}+2 \widehat{k})}{|3 \hat{i}+2 \hat{j}+6 \widehat{k}|,|\hat{i}+2 \hat{j}+2 \widehat{k}|}\right| \ldots(ii)\)
\(\Rightarrow \because|a \hat{i}+b \hat{j}+c \hat{k}|=\sqrt{a^{2}+b^{2}=c^{2}}\)
\(\Rightarrow \therefore|3 \hat{i}+2 \hat{j}+6 \hat{k}|=\sqrt{3^{2}+2^{2}+6^{2}}=\sqrt{9+4+36}=\sqrt{49}=7\)
And \( |\hat{i}+2 \hat{j}+2 \hat{k}|=\sqrt{1^{2}+2^{2}+2^{2}}=\sqrt{1+4+4}=\sqrt{9}=3 \)
Now\( \because\left(a_{1} \hat{i}+b_{1} \hat{j}+c_{1} \hat{k}\right) \cdot\left(a_{2} \hat{i}+b_{2} \hat{j}+c_{2} \hat{k}\right)=a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2} \)
\(\Rightarrow \therefore(3 \hat{i}+2 \hat{j}+6 \hat{k}) \cdot(\hat{i}+2 \hat{j}+2 \hat{k}) \)
\(= 3 \times 1+2 \times 2+6 \times 2\)
\( =3+4+12=19\)
\( \Rightarrow \) By (ii), we have
\( \Rightarrow \cos \theta=\frac{19}{7 \times 3}=\frac{19}{21} \)
\( \Rightarrow \theta=\cos ^{-1}\left(\frac{19}{21}\right) \)
(ii)\( \vec{r}=3 \hat{i}+\hat{j}-2 \hat{k}+\lambda(\hat{i}-\hat{j}-2 \hat{k}) \ \text{and}\) \( \vec{r}=2 \hat{i}-\hat{j}-56 \hat{k}+\mu(3 \hat{i}-5 \hat{j}-4 \hat{k}) \)
Here \( \overrightarrow{b_{1}}=\hat{i}-\hat{j}-2 \hat{k} \) and \( \overrightarrow{b_{2}}=3 \hat{i}-5 \hat{j}-4 \hat{k} \)
So, from (i), we have
\(\cos \theta=\left|\frac{(\hat{i}-\hat{j}-2 \widehat{\mathrm{k}}),(3 \hat{i}-5 \hat{j}-4 \widehat{\mathrm{k}})}{|\hat{i}-\hat{j}-2 \widehat{\mathrm{k}}|,|3 \hat{i}-5 \hat{j}-4 \widehat{\mathrm{k}}|}\right| \ldots(iii)\)
\(\Rightarrow \because|\mathrm{a} \hat{i}+\mathrm{b} \hat{j}+\mathrm{c}\hat{k}|=\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}=\mathrm{c}^{2}}\)
\(\Rightarrow \therefore|\hat{i}-\hat{j}-2 \hat{\mathrm{k}}|=\sqrt{1^{2}+(-1)^{2}+2^{2}}=\sqrt{1+1+4}=\sqrt{6}=\sqrt{3} \times \sqrt{2}\)
And \( |3 \hat{i}-5 \hat{j}-4 \hat{k}|=\sqrt{3^{2}+(-5)^{2}+(-4)^{2}}=\sqrt{9+25+16} \)
\(=\sqrt{50}=\sqrt{3} \times \sqrt{2}\)
\(\text { Now } \because\left(a_{i} \hat{i}+b_{1} \hat{j}+c_{1} \hat{k}\right) \cdot\left(a_{2} \hat{i}+b_{2} \hat{j}+c_{2} \hat{k}\right)=a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}\)
\( \Rightarrow \therefore(\hat{i}-\hat{j}-2 \hat{k}) \cdot(3 \hat{i}-5 \hat{j}-4 \hat{k})=1 \times 3+(-1) \times(-5)+(-2) \times(-4)\)
\(=3+5+8=16\)
\(\Rightarrow \cos \theta=\frac{16}{\sqrt{3} \times \sqrt{2} \times 5 \sqrt{2}}=\frac{16}{5 \times 2 \sqrt{3}}=\frac{8}{5 \sqrt{3}}\)
\(\Rightarrow \theta=\cos ^{-1}\left(\frac{8}{5 \sqrt{3}}\right)\)

We know that
If \( \frac{x-x_{1}}{l_{1}}=\frac{y-y_{1}}{\mathrm{~m}_{1}}=\frac{z-z_{1}}{\mathrm{n}_{1}} \) and \( \frac{x-x_{1}}{l_{2}}=\frac{y-y_{1}}{\mathrm{~m}_{2}}=\frac{z-z_{1}}{\mathrm{n}_{2}} \) are the equations of two lines, then the acute angle between the two lines is given by
\(\cos \theta=\left|l_{1} l_{2}+\mathrm{m}_{1} \mathrm{~m}_{2}+\mathrm{n}_{1} \mathrm{n}_{2}\right|\ldots\) (i)
(i) \( \frac{x-2}{2}=\frac{y-1}{5}-\frac{z+3}{-3} \) and \( \frac{x+2}{-1}=\frac{y-4}{8}-\frac{z-5}{4} \)
Here, \( a_{1}=2, b_{1}=5, c_{1}=-3 \) and \( a_{2}=-1, b_{2}=8, c_{2}=4 \)
Now, \( \because l=\frac{\mathrm{a}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}=\mathrm{c}^{2}}}, \mathrm{~m}=\frac{\mathrm{b}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}=\mathrm{c}^{2}}}, \mathrm{n}=\frac{\mathrm{c}}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}=\mathrm{c}^{2}}} \ldots \) (ii)
Here, \( \sqrt{a_{1}{ }^{2}+b_{1}{ }^{2}=c_{1}{ }^{2}}=\sqrt{2^{2}+5^{2}=(-3)^{2}}=\sqrt{4+25+9}=\sqrt{38} \)
And \( \sqrt{\mathrm{a}_{2}{ }^{2}+\mathrm{b}_{2}{ }^{2}=\mathrm{c}_{2}{ }^{2}}=\sqrt{(-1)^{2}+8^{2}=4^{2}}=\sqrt{1+64+16}=\sqrt{81}=9 \)
So, from (ii), we have
\( \Rightarrow \)
\( l_{1}=\frac{a_{1}}{\sqrt{a_{1}{ }^{2}+b_{1}{ }^{2}=c_{1}{ }^{2}}}=\frac{2}{3} m_{1}=\frac{b_{1}}{\sqrt{a_{1}{ }^{2}+b_{1}{ }^{2}=c_{1}{ }^{2}}}=\frac{2}{3}, n_{1}=\frac{c_{1}}{\sqrt{a_{1}{ }^{2}+b_{1}{ }^{2}=c_{1}{ }^{2}}}=\frac{1}{3} \)
and \( l_{2}=\frac{\mathrm{a}_{2}}{\sqrt{\mathrm{a}_{2}{ }^{2}+\mathrm{b}_{2}{ }^{2} \mathrm{c}_{2}{ }^{2}}}=\frac{4}{9}, \mathrm{~m}_{2}=\frac{\mathrm{b}_{2}}{\sqrt{\mathrm{a}_{2}{ }^{2}+\mathrm{b}_{2}{ }^{2}=\mathrm{c}_{2}{ }^{2}}}=\frac{1}{9}, \mathrm{n}_{2}=\frac{\mathrm{c}_{2}}{\sqrt{\mathrm{a}_{2}{ }^{2}+\mathrm{b}_{2}{ }^{2}=\mathrm{c}_{2}{ }^{2}}}=\frac{8}{9} \)
\( \therefore \) From (i), we have
\( \Rightarrow \cos \theta=\left|\left(\frac{2}{3} \times \frac{4}{9}\right)+\left(\frac{2}{3} \times \frac{1}{9}\right)+\left(\frac{1}{3} \times \frac{8}{9}\right)\right|=\left|\frac{8+2+8}{27}\right|=\frac{18}{27}=\frac{2}{3} \)
\( \Rightarrow \theta=\cos ^{-1}\left(\frac{2}{3}\right) \)

For any two lines to be at right angles, the angle between them should be \( \theta=90^{\circ} \).
\( \Rightarrow a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0 \), where \( a_{1}, b_{1}, c_{1} \) and \( a_{2}, b_{2}, c_{2} \) are the direction ratios of two lines.
The standard form of a pair of Cartesian lines is:
\( \frac{x-x_{1}}{\mathrm{a}_{1}}=\frac{y-y_{1}}{\mathrm{~b}_{1}}=\frac{z-z_{1}}{\mathrm{c}_{1}} \) and \( \frac{x-x_{1}}{\mathrm{a}_{2}}=\frac{y-y_{1}}{\mathrm{~b}_{2}}=\frac{z-z_{1}}{\mathrm{c}_{2}} \ldots\) (i)
Now, first we rewrite the given equations according to the standard form,
i.e. \( \frac{(-x-1)}{3}=\frac{7(y-2)}{2 \mathrm{p}}=\frac{z-3}{2} \) and \( \frac{-7(x-1)}{3 \mathrm{p}}=\frac{y-5}{1}=\frac{-(z-6)}{5} \) i.e.
\( \frac{x-1}{-3}=\frac{y-2}{2 \mathrm{p} / 7}=\frac{z-3}{2} \) and \( \frac{x-1}{-\frac{3 \mathrm{p}}{7} }=\frac{y-5}{1}=\frac{z-6}{-5} \ldots\) (ii)
Now, comparing (i) and (ii), we get
\( a_{1}=-3, b_{1}=\frac{2 p}{7}, c_{1}=2 \) and \( a_{2}=\frac{-3 p}{7}, b_{2}=1, c_{2}=-5 \)
Now, as both the lines are at right angles,
\(\text { so } a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0\)
\(\Rightarrow(-3) \times\left(\frac{-3 p}{7}\right)+\left(\frac{2 p}{7}\right) \times 1+2 \times(-5)=0\)
\(\Rightarrow \frac{9 p}{7}+\frac{2 p}{7}-10=0\)
\(\Rightarrow \frac{9 p+2 p}{7}=10\)
\(\Rightarrow \frac{11 p}{7}=10\)
\(\Rightarrow 11 p=70\)
\(\Rightarrow p=\frac{70}{11}\)

We know that
Two lines with direction ratios \( a_{1}, b_{1}, c_{1} \) and \( a_{2}, b_{2}, c_{2} \) are perpendicular if the angle between them is \( \theta=90^{\circ} \), i.e. \( a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=0 \)
Also, here the direction ratios are:
\(\mathrm{a}_{1}=7, \mathrm{~b}_{1}=-5, \mathrm{c}_{1}=1 \text { and } \mathrm{a}_{2}=1, \mathrm{~b}_{2}=2, \mathrm{c}_{2}=3\)
Now, Consider
\(a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}=7 \times 1+(-5) \times 2+1 \times 3=7-10+3=-3+3=0\)
\( \therefore \) The two lines are perpendicular to each other.

We know that
\(\text{Shortest distance between two lines} \overrightarrow{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}} \) \( \text{and} \ \overrightarrow{r}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}} \) is
\(\mathrm{d}=\left|\frac{\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) \cdot\left(\overrightarrow{a_{1}}-\overrightarrow{a_{2}}\right)}{\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|}\right|\ldots\) (i)
Here, \( \overrightarrow{a_{1}}=\hat{i}+2 \hat{j}-\hat{k}, \overrightarrow{b_{1}}=\hat{i}-\hat{j}+\hat{k} \) and
\(\overrightarrow{a_{2}}=2 \hat{i}-\hat{j}-\hat{k}, \overrightarrow{b_{2}}=2 \hat{i}+\hat{j}+2 \hat{k}\)
Now, \( \because\left(x_{1} \hat{i}+y_{1} \hat{j}+z_{1} \hat{\mathrm{k}}\right)-\left(x_{2} \hat{i}+y_{2} \hat{j}+z_{2} \hat{\mathrm{k}}\right)=\left(x_{1}-x_{2}\right) \hat{i}\)
\(+\left(y_{1}-y_{2}\right) \hat{j}+\left(z_{1}-z_{2}\right) \hat{\mathrm{k}}\)
\(\therefore \overrightarrow{\mathrm{a}_{2}}-\overrightarrow{\mathrm{a}_{1}}=(2 \hat{i}-\hat{j}-\hat{\mathrm{k}})-(\hat{i}+2 \hat{j}-\hat{\mathrm{k}})=\hat{i}-3 \hat{j}-2 \hat{\mathrm{k}} \ldots \text { (ii) }\)
Now, \( \overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}=(\hat{i}-\hat{j}+\hat{\mathrm{k}}) \times(2 \hat{i}+\hat{j}+2 \hat{\mathrm{k}}) \)
\(\begin{array}{l}
\Rightarrow=\left|\begin{array}{ccc}
\hat{i} & \hat{j} & \hat{\mathrm{k}} \\
1 & -1 & 1 \\
2 & 1 & 2
\end{array}\right|=-3 \hat{i}+3 \hat{\mathrm{k}} \end{array}\)
\(\Rightarrow \overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}=-3 \hat{i}+3 \hat{\mathrm{k}} \ldots\) (iii)
\(\Rightarrow\left|\overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}\right|=\sqrt{(-3)^{2}+3^{2}}=\sqrt{9+9}=\sqrt{18}=3 \sqrt{2}\ldots\) (iv)
Now, \( \because\left(a_{1} \hat{i}+b_{1} \hat{j}+c_{1} \hat{\mathrm{k}}\right)-\left(a_{2} \hat{i}+b_{2} \hat{j}+c_{2} \hat{\mathrm{k}}\right)=\mathrm{a}_{1} \mathrm{a}_{2}+\mathrm{b}_{1} \mathrm{~b}_{2}+\mathrm{c}_{1} \mathrm{c}_{2} \)
\(\therefore\left(\overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}\right) \cdot\left(\overrightarrow{\mathrm{a}_{2}}-\overrightarrow{\mathrm{a}_{1}}\right)=(-3 \hat{i}+3 \hat{\mathrm{k}}) \cdot(\hat{i}-3 \hat{j}-2 \hat{\mathrm{k}})\)
\(\left(\overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}\right) \cdot\left(\overrightarrow{\mathrm{a}_{2}}-\overrightarrow{\mathrm{a}_{1}}\right)=-3-6=-9\ldots\) (v)
Now, using (i), we have
The shortest distance between the two lines, \( d=\left|\frac{-9}{3 \sqrt{2}}\right|=\frac{-9}{3 \sqrt{2}} \)
[From (iv) and (v)]
\(\Rightarrow=\frac{3}{\sqrt{2}}\)
Rationalizing the fraction by multiplying the numerator and denominator by \( \sqrt{2} \),
\(\Rightarrow \mathrm{d}=\frac{3}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}=\frac{3 \sqrt{2}}{2}\)

We know that
Shortest distance between the lines:
\( \frac{x-x_{1}}{\mathrm{a}_{1}}=\frac{y-y_{1}}{\mathrm{~b}_{1}}=\frac{z-z_{1}}{\mathrm{c}_{1}} \) and \( \frac{x-x_{2}}{\mathrm{a}_{2}}=\frac{y-y_{2}}{\mathrm{~b}_{2}}=\frac{z-z_{2}}{\mathrm{c}_{2}} \) is
\( \mathrm{d}=\frac{\left|\begin{array}{ccc}x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\ a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2}\end{array}\right|}{\sqrt{\left(b_{1} c_{2}-b_{2} c_{1}\right)^{2}+\left(c_{1} a_{2}-c_{2} a_{1}\right)^{2}+\left(a_{1} b_{2}-a_{2} b_{1}\right)^{2}}} \ldots\) (i)
The standard form of a pair of Cartesian lines is:
\( \frac{x-x_{1}}{\mathrm{a}_{1}}=\frac{y-y_{1}}{\mathrm{~b}_{1}}=\frac{z-z_{1}}{\mathrm{c}_{1}} \) and \( \frac{x-x_{2}}{\mathrm{a}_{2}}=\frac{y-y_{2}}{\mathrm{~b}_{2}}=\frac{z-z_{2}}{\mathrm{c}_{2}} \)
And the given equations are: \( \frac{x+1}{7}=\frac{y+1}{-6}=\frac{z+1}{1} \) and \( \frac{x-3}{1}=\frac{y-5}{-2}=\frac{z-7}{1} \)
Comparing the given equations with the standard form, we get
\(x_{1}=-1, y_{1}=-1, z_{1}=-1 ; x_{2}=3, y_{2}=5, z_{2}=7\)
\(a_{1}=7, b_{1}=-6, c_{1}=1 ; a_{2}=1, b_{2}=-2, c_{2}=1\)
Now, consider
\(\begin{array}{l}
\Rightarrow\left|\begin{array}{ccc}
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
\mathrm{a}_{1} & \mathrm{~b}_{1} & \mathrm{c}_{1} \\
\mathrm{a}_{2} & \mathrm{~b}_{2} & \mathrm{c}_{2}
\end{array}\right| \\
\Rightarrow=\left|\begin{array}{ccc}
3-(-1) & 5-(-1) & 7-(-1) \\
7 & -6 & 1 \\
1 & -2 & 1
\end{array}\right|=\left|\begin{array}{ccc}
3+1 & 5+1 & 7+1 \\
7 & -6 & 1 \\
1 & -2 & 1
\end{array}\right| \\
\Rightarrow=\left|\begin{array}{ccc}
4 & 6 & 8 \\
7 & -6 & 1 \\
1 & -2 & 1
\end{array}\right| \\
\Rightarrow=4(-6+2)-6(7-1)+8(-14+6) \\
\Rightarrow=4(-4)-6 \times 6+8(-8) \\
\Rightarrow=-16-36-64=-116
\end{array}\)
Next, consider
\(\Rightarrow \sqrt{\left(\mathrm{b}_{1} \mathrm{c}_{2}-\mathrm{b}_{2} \mathrm{c}_{1}\right)^{2}+\left(\mathrm{c}_{1} \mathrm{a}_{2}-\mathrm{c}_{2} \mathrm{a}_{1}\right)^{2}+\left(\mathrm{a}_{1} \mathrm{~b}_{2}-\mathrm{a}_{2} \mathrm{~b}_{1}\right)^{2}}\)
\(\Rightarrow=\sqrt{((-6 \times 1)-(-2 \times 1))^{2}+((1 \times 1)-(1 \times 7))^{2}+((7 \times-2)-(1 \times-6))^{2}}\)
\(\Rightarrow=\sqrt{(-6+2)^{2}+(1-7)^{2}+(-14+6)^{2}}=\sqrt{(-4)^{2}+(-6)^{2}+(-8)^{2}}\)
\(\Rightarrow=\sqrt{16+36+64}=\sqrt{116}\)
\( \Rightarrow \) From (i), we have
\(\Rightarrow \mathrm{d}=\left|\frac{-116}{\sqrt{116}}\right|=\frac{116}{\sqrt{116}}=\frac{-116}{\sqrt{116}}=2 \sqrt{29}\)

We know that
\( \text{Shortest distance between two lines}\overrightarrow{r}=\overrightarrow{a_{1}}+\lambda \overrightarrow{b_{1}} \) \( \text{and }\overrightarrow{r}=\overrightarrow{a_{2}}+\mu \overrightarrow{b_{2}} \) is
\(\mathrm{d}=\left|\frac{\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) \cdot\left(\overrightarrow{a_{1}}-\overrightarrow{a_{2}}\right)}{\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|}\right|\ldots\) (i)
Here, \( \overrightarrow{a_{1}}=\hat{i}+2 \hat{j}+3 \hat{k}, \overrightarrow{b_{1}}=\hat{i}-3 \hat{j}+2 \hat{k} \) and
\(\overrightarrow{\mathrm{a}_{2}}=4 \hat{i}+5 \hat{j}+6 \hat{\mathrm{k}}, \overrightarrow{\mathrm{b}_{2}}=2 \hat{i}+3 \hat{j}+\hat{k}\)
Now, \( \because\left(x_{1} \hat{i}+y_{1} \hat{j}+z_{1} \hat{\mathrm{k}}\right)-\left(x_{2} \hat{i}+y_{2} \hat{j}+z_{2} \hat{\mathrm{k}}\right)=\left(x_{1}-x_{2}\right) \hat{i}\)
\(+\left(y_{1}-y_{2}\right) \hat{j}+\left(z_{1}-z_{2}\right) \hat{\mathrm{k}}\)
\(\therefore \overrightarrow{\mathrm{a}_{2}}-\overrightarrow{\mathrm{a}_{1}}=(4 \hat{i}+5 \hat{j}+6 \hat{\mathrm{k}})-(\hat{i}+2 \hat{j}+3 \hat{\mathrm{k}})=3 \hat{i}+3 \hat{j}+3 \hat{\mathrm{k}}\ldots\) (ii)
Now, \( \overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}=(\hat{i}-3 \hat{j}+2 \hat{\mathrm{k}}) \times(2 \hat{i}+3 \hat{j}+\hat{\mathrm{k}}) \)
\( \Rightarrow=\left|\begin{array}{ccc}\hat{i} & \hat{j} & \hat{k} \\ 1 & -3 & 2 \\ 2 & 3 & 1\end{array}\right|=-9 \hat{i}+3 \hat{j}+9 \hat{k} \)
\(\Rightarrow \overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}=-9 \hat{i}+3 \hat{j}+9 \hat{\mathrm{k}}\ldots\) (iii)
\(\Rightarrow\left|\overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}\right|=\sqrt{(-9)^{2}+3^{2}+9^{2}}=\sqrt{81+9+81}\)\(=\sqrt{171}=3 \sqrt{19}\ldots\) (iv)
Now, \( \because\left(a_{1} \hat{i}+b_{1} \hat{j}+c_{1} \hat{k}\right)-\left(a_{2} \hat{i}+b_{2} \hat{j}+c_{2} \hat{k}\right)=a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2} \)
\(\therefore\left(\overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}\right) \cdot\left(\overrightarrow{\mathrm{a}_{2}}-\overrightarrow{\mathrm{a}_{1}}\right)=(-9 \hat{i}+3 \hat{j}+9 \hat{\mathrm{k}}) \cdot(3 \hat{i}+3 \hat{j}+3 \hat{\mathrm{k}})\)
\(=-27+9+27=9\ldots\) (v)
Now, using (i), we have
\( \text{The shortest distance between the two lines,}\)
\( \mathrm{d}=\left|\frac{9}{3 \sqrt{19}}\right|=\frac{9}{3 \sqrt{19}}=\frac{3}{\sqrt{19}} \)

Firstly, consider
\(=\overrightarrow{\mathrm{r}}=(1+\mathrm{t}) \hat{i}+(\mathrm{t}-2) \hat{j}+(3-2 \mathrm{t}) \hat{\mathrm{k}}\)
\(\Rightarrow \overrightarrow{\mathrm{r}}=\hat{i}-\mathrm{t} \hat{i}+\mathrm{t} \hat{j}+2 \hat{j}+3 \hat{\mathrm{k}}-2 \mathrm{t} \hat{k}\)
\(\Rightarrow \overrightarrow{\mathrm{r}}=\hat{i}-2 \hat{j}+3 \hat{\mathrm{k}}+\mathrm{t}(-\hat{i}+\hat{j}-2 \hat{k})\)
\(=\overrightarrow{\mathrm{r}}=(\mathrm{s}+1) \hat{i}+(2 s-1) \hat{j}+(2 s-1) \hat{k}\)
\(\Rightarrow \overrightarrow{\mathrm{r}}=\mathrm{s} \hat{i}+\hat{i}+2 \mathrm{~s} \hat{j}-\hat{j}-2 \mathrm{~s} \hat{\mathrm{k}}-\hat{\mathrm{k}}\)
\(\Rightarrow \overrightarrow{\mathrm{r}}=\hat{i}-\hat{j}-\hat{\mathrm{k}}+\mathrm{s}(\hat{i}+2 \hat{j}-2 \hat{k})\)
So, we need to find the shortest distance between \( \overrightarrow{\mathrm{r}}=\hat{i}-2 \hat{j}+3 \hat{k}+\mathrm{t}(-\hat{i}+\hat{j}-2 \hat{\mathrm{k}}) \) \(\text{and} \overrightarrow{\mathrm{r}}=\hat{i}-\hat{j}-\hat{\mathrm{k}}+\mathrm{s}(\hat{i}+2 \hat{j}-2 \hat{\mathrm{k}}) \).
Now, we know that
Shortest distance between two lines \( \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}_{1}}+\lambda \overrightarrow{\mathrm{b}_{1}} \) and \( \overrightarrow{\mathrm{r}}=\overrightarrow{\mathrm{a}_{2}}+\mu \overrightarrow{\mathrm{b}_{2}} \) is
\(\left|\frac{\left(\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right) \cdot\left(\overrightarrow{a_{1}}-\overrightarrow{a_{2}}\right)}{\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|}\right|\ldots\) (i)
Here, \( \overrightarrow{a_{1}}=\hat{i}-2 \hat{j}+3 \hat{k}, \overrightarrow{b_{1}}=-\hat{i}+\hat{j}-2 \hat{k} \) and
\(\overrightarrow{a_{2}}=\hat{i}-\hat{j}-\hat{k}, \overrightarrow{b_{2}}=\hat{i}+2 \hat{j}-2 \hat{k}\)
Now, \( \because\left(x_{1} \hat{i}+y_{1} \hat{j}+z_{1} \hat{\mathrm{k}}\right)-\left(x_{2} \hat{i}+y_{2} \hat{j}+z_{2} \hat{\mathrm{k}}\right)=\left(x_{1}-x_{2}\right) \hat{i} \)
\( +\left(y_{1}-y_{2}\right) \hat{j}+\left(z_{1}-z_{2}\right) \hat{k} \)
\(\therefore \overrightarrow{\mathrm{a}_{2}}-\overrightarrow{\mathrm{a}_{1}}=(\hat{i}-\hat{j}-\hat{\mathrm{k}})-(\hat{i}-2 \hat{j}+3 \hat{\mathrm{k}})=\hat{j}-4 \hat{\mathrm{k}}\ldots\) (ii)
Now, \( \overrightarrow{b_{1}} \times \overrightarrow{b_{2}}=(-\hat{i}+\hat{j}-2 \hat{k}) \times(\hat{i}+2 \hat{j}-2 \hat{k}) \)
\(\begin{array}{l}
\Rightarrow\left|\begin{array}{rrr}
\hat{i} & \hat{j} & \hat{\mathrm{k}} \\
-1 & 1 & -2 \\
1 & 2 & -2
\end{array}\right|=2 \hat{i}-4 \hat{j}-3 \hat{\mathrm{k}} \\
\Rightarrow \overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}=2 \hat{i}-4 \hat{j}-3 \hat{\mathrm{k}\ldots (\text{iii})}\end{array}\)
\(\Rightarrow\left|\overrightarrow{b_{1}} \times \overrightarrow{b_{2}}\right|=\sqrt{2^{2}+(-4)^{2}+(-3)^{2}}=\sqrt{4+16+9}=\sqrt{29} \ldots\) (iv)
Now, \( \because\left(a_{1} \hat{i}+b_{1} \hat{j}+c_{1} \hat{k}\right)-\left(a_{2} \hat{i}+b_{2} \hat{j}+c_{2} \hat{k}\right)=a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2} \)
\(\therefore\left(\overrightarrow{\mathrm{b}_{1}} \times \overrightarrow{\mathrm{b}_{2}}\right) \cdot\left(\overrightarrow{\mathrm{a}_{2}}-\overrightarrow{\mathrm{a}_{1}}\right)=(2 \hat{i}-4 \hat{j}-3 \hat{\mathrm{k}}) \cdot(\hat{j}-4 \hat{\mathrm{k}})=-4+12=8\ldots(v)\)
Now, using (i), we have
The shortest distance between the two lines, \( \mathrm{d}=\left|\frac{8}{\sqrt{29}}\right|=\frac{8}{\sqrt{29}} \)