Chapter 11: Three dimensional Geometry

Ans: We are given the equations of lines as 

$\dfrac{x-5}{7}=\dfrac{y+2}{-5}=\dfrac{z}{1}$ and$\dfrac{x}{1}=\dfrac{y}{2}=\dfrac{z}{3}$

The direction ratios of the given lines are 

$7,-5,1$ and $1,2,3$

Two lines will be perpendicular to each other if 

${{a}_{1}}{{a}_{2}}+{{b}_{1}}{{b}_{2}}+{{c}_{1}}{{c}_{2}}=0$

$ 7\times 1+\left( -5 \right)\times 2+1\times 3 \\ $

$ =7-10+3 \\ $

$ =0 \\ $

Therefore, the given lines are perpendicular to each other.

Ans: The equations of given lines are

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

It is known that the shortest distance between the lines $\vec{r}={{\vec{a}}_{1}}+\lambda {{\vec{b}}_{1}}$ and $\vec{r}={{\vec{a}}_{1}}+\mu {{\vec{b}}_{2}}$ is given by

\[d=\left| \dfrac{\left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right)\cdot \left( {{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}} \right)}{\left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|} \right|\]…(1)

On comparing the equations, we obtain,

$ {{{\vec{a}}}_{1}}=\left( \hat{i}+2\hat{j}+\hat{k} \right) \\ $

$ {{{\vec{b}}}_{1}}=\left( \hat{i}-\hat{j}+\hat{k} \right) \\ $

$ {{{\vec{a}}}_{2}}=\left( 2\hat{i}-\hat{j}-\hat{k} \right) \\ $

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

$ {{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}}=\left( 2\hat{i}-\hat{j}-\hat{k} \right)-\left( \hat{i}+2\hat{j}+\hat{k} \right) \\ $

$ \left( \hat{i}-3\hat{j}-2\hat{k} \right) \\ $

$\begin{align} & {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}}  \\ 1 & -1 & 1  \\ 2 & 1 & 2  \\ \end{matrix} \right| \\ & =\left( -2-1 \right)\hat{i}-\left( 2-2 \right)\hat{j}+\left( 1+2 \right)\hat{k} \\ & =-3\hat{i}+3\hat{k} \\ \end{align}$

$ \Rightarrow \left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|=\sqrt{{{\left( -3 \right)}^{2}}+{{3}^{2}}} \\ $

$ =\sqrt{9+9}=\sqrt{18} \\ $

$ =3\sqrt{2} \\ $

Substituting all values obtained in equation (1),

$ d=\left| \dfrac{\left( -3\hat{i}+3\hat{k} \right)\cdot \left( \hat{i}-3\hat{j}-2\hat{k} \right)}{3\sqrt{2}} \right| \\$ 

$ =\left| \dfrac{\left( -3 \right)1+3\left( -2 \right)}{3\sqrt{2}} \right| \\ $

$ =\left| \dfrac{-9}{3\sqrt{2}} \right| \\ $

$ =\dfrac{3}{\sqrt{2}} \\ $

Therefore, the shortest distance between the lines given is $\dfrac{3}{\sqrt{2}}$ units.

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

It is known that the shortest distance between the lines

$\dfrac{x-{{x}_{1}}}{{{a}_{1}}}=\dfrac{y-{{y}_{1}}}{{{b}_{1}}}=\dfrac{z-{{z}_{1}}}{{{c}_{1}}}$ and $\dfrac{x-{{x}_{2}}}{{{a}_{2}}}=\dfrac{y-{{y}_{2}}}{{{b}_{2}}}=\dfrac{z-{{z}_{2}}}{{{c}_{2}}}$ is given by

\[d=\left| \dfrac{\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}}  \\ {{a}_{1}} & {{b}_{1}} & {{c}_{1}}  \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}}  \\ \end{matrix} \right|}{\sqrt{{{\left( {{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}} \right)}^{2}}+{{\left( {{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}} \right)}^{2}}+{{\left( {{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}} \right)}^{2}}}} \right|\].…(1)

Comparing the equations, we obtain,

${{x}_{1}}=-1,{{y}_{1}}=-1,{{z}_{1}}=-1 \\ $

$ {{a}_{1}}=7,{{b}_{1}}=-6,{{c}_{1}}=1 \\ $

$ {{x}_{2}}=3,{{y}_{2}}=5,{{z}_{2}}=7 \\ $

$ {{a}_{2}}=1,{{b}_{2}}=-2,{{c}_{2}}=1 \\ $

Then, \[\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}}  \\ {{a}_{1}} & {{b}_{1}} & {{c}_{1}}  \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}}  \\ \end{matrix} \right|=\left| \begin{matrix} 4 & 6 & 8  \\ 7 & -6 & 1  \\ 1 & -2 & 1  \\ \end{matrix} \right|\]

$=4\left( -16+2 \right)-6\left( 7-1 \right)+8\left( -14+6 \right) \\ $

$ =-16-36-64 \\ $

$ =-116 \\ $

Also,

$ \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{116} \\ $

$ =2\sqrt{29} \\ $

Substituting all the values in equation (1),

$ d=\left| \dfrac{-116}{2\sqrt{29}} \right| \\ $

$ =\left| \dfrac{-58}{\sqrt{29}} \right| \\ $

$ =\left| -2\sqrt{29} \right| \\ $

$ =2\sqrt{29} \\ $

Since the distance is always non-negative, the distance between the given lines is $-2\sqrt{29}$ units.

Ans: The equations of 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 $\vec{r}=\left( 4\hat{i}+5\hat{j}+6\hat{k} \right)+\mu \left( 2\hat{i}+3\hat{j}+\hat{k} \right)$.

It is known that the shortest distance between the lines $\vec{r}={{\vec{a}}_{1}}+\lambda {{\vec{b}}_{1}}$ and $\vec{r}={{\vec{a}}_{1}}+\mu {{\vec{b}}_{2}}$ is given by

\[d=\left| \dfrac{\left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right)\cdot \left( {{{\vec{a}}}_{2}}-{{{\vec{a}}}_{1}} \right)}{\left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|} \right|\]…(1)

On comparing the equations, we obtain,

$ {{{\vec{a}}}_{1}}=\left( \hat{i}+2\hat{j}+3\hat{k} \right) \\ $

$ {{{\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) \\ $

$ {{{\vec{b}}}_{2}}=\left( 2\hat{i}+3\hat{j}+\hat{k} \right) \\ $

$ {{{\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) \\ $

$ =\left( 3\hat{i}+3\hat{j}+3\hat{k} \right) \\ $

$\begin{align} & {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}}=\left| \begin{matrix} {\hat{i}} & {\hat{j}} & {\hat{k}}  \\ 1 & -3 & 2  \\ 2 & 3 & 1  \\ \end{matrix} \right| \\  & =\left( -3-6 \right)\hat{i}-\left( 1-4 \right)\hat{j}+\left( 3+6 \right)\hat{k} \\ & =-9\hat{i}+3\hat{j}+9\hat{k} \\ \end{align}$

$ \Rightarrow \left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|=\sqrt{{{\left( -9 \right)}^{2}}+{{3}^{2}}+{{9}^{2}}} \\ $

$ =\sqrt{81+9+81}=\sqrt{171} \\ $

$ =3\sqrt{19} \\ $

Substituting all values obtained in equation (1),

$ d=\left| \dfrac{\left( -9\hat{i}+3\hat{j}+9\hat{k} \right)\cdot \left( 3\hat{i}+3\hat{j}+3\hat{k} \right)}{3\sqrt{19}} \right| \\ $

$ =\left| \dfrac{\left( -9 \right)3+3\times 3+9\times 3}{3\sqrt{19}} \right| \\ $

$ =\left| \dfrac{9}{3\sqrt{19}} \right| \\ $

$ =\dfrac{3}{\sqrt{19}} \\ $

Therefore, the shortest distance between the lines given is \[\dfrac{3}{\sqrt{19}}\] units.

Ans: Firstly, let’s consider the given equations

$\Rightarrow   \vec{r}=(1-t)\hat{i} +(t-2)\hat{j}+(3-2t)\hat{k}$

$\vec{r}=\hat{i}-t\hat{i}+t\hat{j}-2\hat{j}+3\hat{k}-2t\hat{k}$

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

$\Rightarrow \vec{r}=(s+1)\hat{i}+(2s-1)\hat{j}-(2s+1)\hat{k}$

$\vec{r}=(s+1)\hat{i} +(2s-1)\hat{j}-(2s+1)\hat{k}$

$\vec{r}=s\hat{i}+\hat{i}+2s\hat{j}-\hat{j}-2s\hat{k}-\hat{i}$

$\vec{r}=\hat{i}-\hat{j}-\hat{k}+s(\hat{i}+2\hat{j}-2\hat{k})$

So, now we need to find the shortest distance between

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

We know that shortest distance between two lines $\vec{r}=\vec{a_{1}}+\lambda \vec{b_{1}}$ and $\vec{r}=\vec{a_{2}}+ \mu  \vec{b_{2}} $

$d=\left | \dfrac{(\vec{b_1}\times \vec{b_2})\cdot (\vec{a_2}\times \vec{a_1}) }{\left | \vec{b_1} \times \vec{b_2} \right |} \right |......(1)$

By comparing the equations we get,

$\vec{a_{1}}=\hat{i}-2\hat{j}+3\hat{k}$ ,  $\vec{b_{1}} =-\hat{i}+\hat{j}-2\hat{k}$

$\vec{a_{2}} =\hat{i}-\hat{j}-\hat{k}$,  $\vec{b_{2}} =\hat{i}+2\hat{j}-2\hat{k}$

Since,

$(X_{1}\hat{i}+Y_{1}\hat{j}+Z_{1}\hat{k})-(X_{2}\hat{i}+Y_{2}\hat{j}+Z_{2}\hat{k})$

$\Rightarrow (X_{1}-X_{2})\hat{i}+(Y_{1}-Y_{2})\hat{j}+(Z_{1}-Z_{2})\hat{k}$

So, $\vec{a_{2}}-\vec{a_{1}}=(\hat{i}-\hat{j}-\hat{k})-(\hat{i}-2\hat{j}+3\hat{k})=\hat{j}-4\hat{k}$ …..(2)

And,

$\vec{b_{1}}\times\vec{b_{2}}=(-\hat{i}+\hat{j}-2\hat{k})\times(\hat{i}+2\hat{j}-2\hat{k})$

$\begin{vmatrix} \hat{i} &  \hat{j}& \hat{k} \\ -1 &  1&  -2\\ 1&  2&  -2\\ \end{vmatrix}$

$\vec{b_{1}}\times\vec{b_{2}}=2\hat{i}-4\hat{j}-3\hat{k}$ …..(3)

$\Rightarrow \left | \vec{b_{1}}\times\vec{b_{2}} \right |=\sqrt{2^{2}+(-4)^{2}+(-3)^{2}}=\sqrt{4+16+9}=29$

Now multiplying (2) and (3) we get,

$\left ( a_{1}\hat{i} + b_{1}\hat{j} + c_{1}\hat{k} \right )\cdot \left ( a_{2}\hat{i} + b_{2}\hat{j} + c_{2}\hat{k} \right ) = a_{1}a_{2}+b_{1}b_{2}+c_{1}c_{2}$

$(\vec{b_{1}}\times\vec{b_{2}})\cdot (\vec{a_{1}}\times\vec{a_{2}})=(2\hat{i}-4\hat{j}-3\hat{k})\cdot (\hat{j}-4\hat{k})$

=-4+18=8 ….(5)

By substituting all the values in (1), we obtain

The shortest distance between the two lines,

$d=\left | \dfrac{8}{\sqrt{29}} \right |=\dfrac{8}{\sqrt{29}}$

$\therefore$ The shortest distance is $\dfrac{8}{\sqrt{29}}$