Chapter 11: Three dimensional Geometry

Ans: As we know that, for any angle $\text{ }\!\!\theta\!\!\text{ }$, with direction cosines, $\text{a,b,c}$ and $\text{b-c, c-a, a-b}$ can be found by,

$\text{cos }\!\!\theta\!\!\text{ =}\left| \frac{\text{a}\left( \text{b-c} \right)\text{+b}\left( \text{b-c} \right)\text{+c}\left( \text{c-a} \right)}{\sqrt{{{\text{a}}^{\text{2}}}\text{+}{{\text{b}}^{\text{2}}}\text{+}{{\text{c}}^{\text{2}}}\text{+}\sqrt{{{\left( \text{b-c} \right)}^{\text{2}}}\text{+}{{\left( \text{c-a} \right)}^{\text{2}}}\text{+}{{\left( \text{a-b} \right)}^{\text{2}}}}}} \right|$

Solving this we get, $\text{cos }\!\!\theta\!\!\text{ =0}$

$\text{ }\!\!\theta\!\!\text{ =co}{{\text{s}}^{\text{-1}}}\text{0}$

$\Rightarrow \text{ }\!\!\theta\!\!\text{ }={{90}^{\circ }}$

Therefore, the angle between the two lines will be ${{90}^{\circ }}$.

Ans: As it is given that the line is passing through the origin and is also parallel to x-axis is x-axis,

Now,

Let us consider a point on x-axis be $\text{A}$ 

So, the coordinates of $A$ will be $\left( \text{a,0,0} \right)$

Now, the direction ratios of $\text{OA}$ will be,

$\Rightarrow \left( \text{a-0} \right)\text{=a,0,0}$

The equation of $\text{OA}$$\Rightarrow \frac{\text{x-0}}{\text{a}}\text{=}\frac{\text{y-0}}{\text{0}}\text{=}\frac{\text{z-0}}{\text{0}}\Rightarrow \frac{\text{x}}{\text{1}}\text{=}\frac{\text{y}}{\text{0}}\text{=}\frac{\text{z}}{\text{0}}\text{=a}$

Therefore, the equation of the line passing through origin and parallel to x-axis is $\frac{\text{x}}{\text{1}}\text{=}\frac{\text{y}}{\text{0}}\text{=}\frac{\text{z}}{\text{0}}$.

Ans: From the given equation we can say that ${{\text{a}}_{\text{1}}}\text{=-3,}{{\text{b}}_{\text{1}}}\text{=2k,}{{\text{c}}_{\text{1}}}\text{=2}$and ${{\text{a}}_{\text{2}}}\text{=3k,}{{\text{b}}_{\text{2}}}\text{=1,}{{\text{c}}_{\text{2}}}\text{=-5}$.

We know that the two lines are perpendicular, if ${{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}\text{+}{{\text{b}}_{\text{1}}}{{\text{b}}_{\text{2}}}\text{+}{{\text{c}}_{\text{1}}}{{\text{c}}_{\text{2}}}\text{=0}$

$\text{-3}\left( \text{3k} \right)\text{+2k }\!\!\times\!\!\text{ 1+2}\left( \text{-5} \right)\text{=0}$

$\Rightarrow \text{-9k+2k-10=0}$

$\Rightarrow \text{7k=-10}$

$\Rightarrow \text{k=}\frac{\text{-10}}{\text{7}}$

Therefore, the value of $k$is $\text{-}\frac{\text{10}}{\text{7}}$

Ans: According to the question, we need to find the distance between the lines,

$\overrightarrow{r}=6\widehat{i}+2\widehat{j}+2\widehat{k}+\lambda \left( \widehat{i}-2\widehat{j}+2\widehat{k} \right)$

\[\overrightarrow{r}=-4\widehat{i}-\widehat{k}+\mu \left( 3\widehat{i}-2\widehat{j}-2\widehat{k} \right)\]

As we know we can find the shortest distance by,

$d=\left| \frac{\left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right).\left( {{{\vec{a}}}_{1}}-{{{\vec{a}}}_{2}} \right)}{\left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|} \right|$

Now, from the equation of lines we get

\[{{\overrightarrow{\text{a}}}_{1}}\text{=}6\widehat{i}+2\widehat{j}+2\widehat{k}\]

$\overrightarrow{{{\text{b}}_{\text{1}}}}{=\hat{i}-2\hat{j}+2\hat{k}}$

$\overrightarrow{{{\text{a}}_{\text{2}}}}\text{=}-4\widehat{i}-\widehat{k}$

$\overrightarrow{{{\text{b}}_{\text{2}}}}\text{=}3\widehat{i}-2\widehat{j}-2\widehat{k}$

$\Rightarrow \overrightarrow{{{\text{a}}_{\text{2}}}}\text{0}\overrightarrow{{{\text{a}}_{\text{1}}}}\text{=}\left( -4\widehat{i}-\widehat{k} \right)\text{0}\left( 6\widehat{i}+2\widehat{j}+2\widehat{k} \right)\text{=}-10\widehat{i}-2\widehat{j}-3\widehat{k}$

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

$\left( {{{\vec{b}}}_{\text{1}}}\text{ }\times \text{ }{{{\vec{b}}}_{\text{2}}} \right)\text{.}\left( \overrightarrow{{{\text{a}}_{\text{2}}}}\text{0}\overrightarrow{{{\text{a}}_{\text{1}}}} \right)\text{=}\left( 8\widehat{i}+8\widehat{j}+4\widehat{k} \right)\text{.}\left( -10\widehat{i}-2\widehat{j}-3\widehat{k} \right)$

$\text{=-80-16-12}$

$\text{=-108}$

Now, putting these values in $d=\left| \frac{\left( {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right).\left( {{{\vec{a}}}_{1}}-{{{\vec{a}}}_{2}} \right)}{\left| {{{\vec{b}}}_{1}}\times {{{\vec{b}}}_{2}} \right|} \right|$, we get

$\text{d=}\left| \frac{\text{-108}}{\text{12}} \right|\text{=9}$

Therefore, the shortest distance between the above two lines is of $\text{9}$ units.

Ans: According to the question, we get that ${\vec{b}=}{{\text{b}}_{\text{1}}}{\hat{i}+}{{\text{b}}_{\text{2}}}{\hat{j}+}{{\text{b}}_{\text{3}}}{\hat{k}}$ and ${\vec{a}=\hat{i}+2\hat{j}-4\hat{k}}$

We know that the equation of the line passing through point and also parallel to vector, we get

${\vec{r}=\hat{i}+2\hat{j}-4\hat{k}+ }\!\!\lambda\!\!\text{ }\left( {{\text{b}}_{\text{1}}}{\hat{i}+}{{\text{b}}_{\text{2}}}{\hat{j}+}{{\text{b}}_{\text{3}}}{\hat{k}} \right)$ … $\left( \text{1} \right)$

Now, the equation of the two lines will be 

$\frac{\text{x-8}}{\text{3}}\text{=}\frac{\text{y+19}}{\text{-16}}\text{=}\frac{\text{z-10}}{\text{7}}$ … $\left( \text{2} \right)$

$\frac{\text{x-15}}{\text{3}}\text{=}\frac{\text{y-29}}{\text{8}}\text{=}\frac{\text{z-5}}{\text{-5}}$ … $\left( \text{3} \right)$

As we know that line $\left( \text{1} \right)$ and $\left( \text{2} \right)$ are perpendicular to each other, we get

$\text{3}{{\text{b}}_{\text{1}}}\text{-16}{{\text{b}}_{\text{2}}}\text{+7}{{\text{b}}_{\text{3}}}\text{=0}$ … $\left( \text{4} \right)$

Also, we know that the line $\left( \text{1} \right)$ and $\left( \text{3} \right)$ are perpendicular to each other, we get

$\text{3}{{\text{b}}_{\text{1}}}\text{+8}{{\text{b}}_{\text{2}}}\text{-5}{{\text{b}}_{\text{3}}}\text{=0}$ … $\left( \text{5} \right)$

Now, from equation $\left( \text{4} \right)$ and $\left( \text{5} \right)$ we get that

$\frac{{{\text{b}}_{\text{1}}}}{\left( \text{-16} \right)\left( \text{-5} \right)\text{-8}\left( \text{7} \right)}\text{=}\frac{{{\text{b}}_{\text{2}}}}{\text{7}\left( \text{3} \right)\text{-3}\left( \text{-5} \right)}\text{=}\frac{{{\text{b}}_{\text{3}}}}{\text{3}\left( \text{8} \right)\text{-3}\left( \text{-16} \right)}$

$\Rightarrow \frac{{{\text{b}}_{\text{1}}}}{\text{24}}\text{=}\frac{{{\text{b}}_{\text{2}}}}{\text{36}}\text{=}\frac{{{\text{b}}_{\text{3}}}}{\text{72}}\Rightarrow \frac{{{\text{b}}_{\text{1}}}}{\text{2}}\text{=}\frac{{{\text{b}}_{\text{2}}}}{\text{3}}\text{=}\frac{{{\text{b}}_{\text{3}}}}{\text{6}}$

Therefore, direction ratios of ${\vec{b}}$ are $\text{2,3,6}$

Which means ${\vec{b}=2\hat{i}+3\hat{j}+6\hat{k}}$

Putting ${\vec{b}=2\hat{i}+3\hat{j}+6\hat{k}}$ in equation $\left( \text{1} \right)$, we get

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