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

Last updated date: 13th Jul 2024
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## NCERT Solutions for Class 12 Maths Chapter 11 Miscellaneous Exercise - Free PDF Download

Class 12 Maths NCERT Solutions for Chapter 11 Three Dimensional Geometry includes solutions to all Miscellaneous Exercise problems. Three Dimensional Geometry Class 12 NCERT Solutions Miscellaneous Exercises are based on the ideas presented in Maths Chapter 11. This activity is crucial for both the CBSE Board examinations and competitive tests. To perform well on the board exam, download the latest CBSE Class 12 Maths Syllabus in PDF format, which are solved by experts to help you understand easily.

Competitive Exams after 12th Science

## Access NCERT Class 12 Maths Chapter 11 Three Dimensional Geometry

### 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.

## Class 12 Maths Chapter 11: Exercises Breakdown

 Exercise Number of Questions Exercise 11.1 5 Exercise 11.2 15

## CBSE Class 12 Maths Chapter 11 Other Study Materials

 S.No Important Links for Chapter 11  Three Dimensional Geometry 1 2 4 Class 12 Three Dimensional Geometry NCERT Exemplar Solution 6 Class 12 Three Dimensional Geometry RS Aggarwal Solutions

## Chapter-Specific NCERT Solutions for Class 12 Maths

Given below are the chapter-wise NCERT Solutions for Class 12 Maths. Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

## FAQs on NCERT Solutions for Class 12 Maths Miscellaneous Exercise Chapter 11- Three Dimensional Geometry

1. Can you provide some examples of problems covered in the miscellaneous exercise?

The miscellaneous exercise covers a wide range of topics in 3D geometry, including:

• Direction cosines and ratios of lines

• Finding the equation of a line (Cartesian and vector forms)

• Identifying co-planar and skew lines

• Calculating the angle between two lines

• Finding the shortest distance between two lines (parallel and skew)

• Equation of a plane (normal form, intercept form, etc.)

• Determining the angle between two planes

• Finding the distance between a point and a plane

2. Where can I find additional resources for practising miscellaneous exercise problems?

• The NCERT textbook itself might provide solutions to some problems within the miscellaneous exercise section.

• Vedantu offers comprehensive solutions and explanations for these problems. You can find them through a web search using terms like "NCERT Solutions Class 12 Maths Chapter 11 Miscellaneous Exercise 3D Geometry."

3. Three Dimensional Geometry sample questions?

1. Find the direction cosines of the line joining the points A(2, 1, 3) and B(4, -1, 2).

2. Find the equation of the plane passing through the points P(1, 2, 3), Q(2, 1, 4), and R(3, 0, 5). (This question might involve using vectors to represent direction)

3. Determine if the lines with direction ratios 3, 2, -1 and 1, 2, 2 are coplanar or skew.

4. Find the shortest distance between the line x = 1 + 2t, y = 3 - t, z = 4t and the plane x + y + z = 6. (This question might involve using distance formula)