NCERT Solutions for Class 12 Maths Chapter 11 Three Dimensional Geometry Miscellaneous Exercise

Miscellaneous Exercise

1.  find the angle between the lines whose  direction ratios are $\text{a,b,c}$ and $\text{b-c, c-a, a-b,}$ .

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 ${{90}^{\circ }}$.

2. Find the equation of a line parallel to  x-axis  line passing through the origin.

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 $\frac{\text{x}}{\text{1}}\text{=}\frac{\text{y}}{\text{0}}\text{=}\frac{\text{z}}{\text{0}}$.

3.if the lines $\frac{\text{x-1}}{\text{-3}}\text{=}\frac{\text{y-2}}{\text{2k}}\text{=}\frac{\text{z-3}}{\text{2}}$ and $\frac{\text{x-1}}{\text{3k}}\text{=}\frac{\text{y-1}}{\text{1}}\text{=}\frac{\text{z-6}}{\text{-5}}$ are perpendicular Find the value of k

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 $\text{-}\frac{\text{10}}{\text{7}}$

4. Find the shortest distance between these two lines $\overrightarrow{r}=6\widehat{i}+2\widehat{j}+2\widehat{k}+\lambda \left( \widehat{i}-2\widehat{j}+2\widehat{k} \right)$

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

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.

5. Find the vector equation of the line passing through the points $\left( \text{1,2,-4} \right)$ and perpendicular to the two lines $\frac{\text{x-8}}{\text{3}}\text{=}\frac{\text{y+19}}{\text{-16}}\text{=}\frac{\text{z-10}}{\text{7}}$ and $\frac{\text{x-15}}{\text{3}}\text{=}\frac{\text{y-29}}{\text{8}}\text{=}\frac{\text{z-5}}{\text{-5}}$

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

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

Conclusion

NCERT solutions for class 12 maths Three Dimensional Geometry miscellaneous exercise is crucial for understanding various concepts thoroughly. It covers diverse problems that require the application of multiple formulas and techniques. It's important to focus on understanding the underlying principles behind each question rather than just memorizing solutions. Remember to understand the theory behind each concept, practice regularly, and refer to solved examples to master this exercise effectively.

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