NCERT Solutions Class 12 Maths Chapter 11 -Three Dimensional Geometry

• Exercise 11.1: This exercise focuses on the basic concepts of direction cosines and direction ratios. It includes problems that require students to find the direction cosines and direction ratios given specific conditions. Important topics include the calculation of direction cosines from given coordinates, understanding the relationship between direction cosines and direction ratios, and applying these concepts in various geometric problems.

• Exercise 11.2:

• Equation of a Line in Space: This part of the exercise involves finding the vector and Cartesian equations of lines in space. This part of the exercise involves problems such as deriving the equation of a line passing through a given point and parallel to a given vector, and finding the equation of a line given two points in space.

• Angle between Two Lines: This section focuses on determining the angle between two lines using their direction ratios or cosines. Problems typically involve calculating the cosine of the angle between two given lines. Using the dot product of direction vectors to find the angle.

• Shortest Distance between Two Lines: Problems in this section deal with finding the shortest distance between two skew lines. Important concepts include using vector algebra to derive the formula for the shortest distance and solving problems using the derived formula.

• Miscellaneous Exercise on Chapter 11: This exercise includes a variety of problems that encompass all the concepts covered in the chapter. Typically, it consists of 5 diverse questions that require comprehensive understanding and application of the chapter's topics. The questions may involve multiple concepts, such as combining direction cosines, equations of lines, angles, and distances in a single problem.

Overall, this chapter is essential for students who want to pursue higher education in Mathematics and Physics. It provides a solid foundation for students to understand the concepts of three-dimensional geometry and their applications in real-world problems.

Access NCERT Solutions for Class 12 Maths  Chapter 11 - Three-Dimensional Geometry

Exercise 11.1

1.if a line makes angles $\text{9}{{\text{0}}^{\text{o}}}\text{,13}{{\text{5}}^{\text{o}}}\text{,4}{{\text{5}}^{\text{o}}}$ with $\text{x,y}$ and $\text{z}$ axes respectively,find its direction cosines.

Ans: Let us consider $\text{l,m}$ and $\text{n}$be the direction cosines of line

Then, 

$\text{l}=\text{cos9}{{\text{0}}^{\text{o}}}=\text{0}$,

$\text{m}=\text{cos13}{{\text{5}}^{\text{o}}}$

$=\text{cos}\left( \text{9}{{\text{0}}^{\text{o}}}\text{+4}{{\text{5}}^{\text{o}}} \right)$

$=\text{sin4}{{\text{5}}^{\text{o}}}$

$=-\dfrac{\text{1}}{\sqrt{\text{2}}}$

And,

$\text{n}=\text{cos4}{{\text{5}}^{\text{o}}}=\dfrac{\text{1}}{\sqrt{\text{2}}}$

Therefore, the direction cosines of the line are $\text{0,-}\dfrac{\text{1}}{\sqrt{\text{2}}}$ and $\dfrac{\text{1}}{\sqrt{\text{2}}}$.

2. Find the direction cosines of the line which makes equal angles with the coordinate axes.

Ans: Let us consider that the line makes an angle $\text{ }\!\!\alpha\!\!\text{ }$ with coordinate axes

Which means $\text{l=cos }\!\!\alpha\!\!\text{ ,m=cos }\!\!\alpha\!\!\text{ ,n=cos }\!\!\alpha\!\!\text{ }$

Now, we know that 

${{\text{l}}^{\text{2}}}\text{+}{{\text{m}}^{\text{2}}}\text{+}{{\text{n}}^{\text{2}}}\Rightarrow \text{co}{{\text{s}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ +co}{{\text{s}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ +co}{{\text{s}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ }$

$\Rightarrow \text{3co}{{\text{s}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ =1}$

$\Rightarrow \text{co}{{\text{s}}^{\text{2}}}\text{ }\!\!\alpha\!\!\text{ =}\dfrac{\text{1}}{\text{3}}\Rightarrow \text{cos }\!\!\alpha\!\!\text{ = }\!\!\pm\!\!\text{ }\dfrac{\text{1}}{\sqrt{\text{3}}}$

Therefore, the direction cosines of the line are $\text{ }\!\!\pm\!\!\text{ }\dfrac{\text{1}}{\sqrt{\text{3}}}\text{, }\!\!\pm\!\!\text{ }\dfrac{\text{1}}{\sqrt{\text{3}}}\text{, }\!\!\pm\!\!\text{ }\dfrac{\text{1}}{\sqrt{\text{3}}}$.

3. If a line has the direction ratios $\text{-18,12,-4}$, then what are its direction cosines?

Ans: We have the direction ratios as $\text{-18,12,-4}$,

Now, the direction cosines will be as

$\text{l=}\dfrac{\text{-18}}{\sqrt{{{\left( \text{-18} \right)}^{\text{2}}}\text{+}{{\left( \text{12} \right)}^{\text{2}}}\text{+}{{\left( \text{-4} \right)}^{\text{2}}}}}\text{,m=}\dfrac{\text{12}}{\sqrt{{{\left( \text{-18} \right)}^{\text{2}}}\text{+}{{\left( \text{12} \right)}^{\text{2}}}\text{+}{{\left( \text{-4} \right)}^{\text{2}}}}}\text{,n=}\dfrac{\text{-4}}{\sqrt{{{\left( \text{-18} \right)}^{\text{2}}}\text{+}{{\left( \text{12} \right)}^{\text{2}}}\text{+}{{\left( \text{-4} \right)}^{\text{2}}}}}$

$\dfrac{\text{-18}}{\text{22}}\text{,}\dfrac{\text{12}}{\text{22}}\text{,}\dfrac{\text{-4}}{\text{22}}\Rightarrow \dfrac{\text{-9}}{\text{11}}\text{,}\dfrac{\text{6}}{\text{11}}\text{,}\dfrac{\text{-2}}{\text{11}}$

Therefore, direction cosines of the line are $\dfrac{\text{-9}}{\text{11}}\text{,}\dfrac{\text{6}}{\text{11}}$ and $\dfrac{\text{-2}}{\text{11}}$.

4. Show that $\left( \text{2,3,4} \right)\text{,}\left( \text{-1,-2,1} \right)\text{,}\left( \text{5,8,7} \right)$ are collinear.

Ans: Let us consider the points be $\text{A}\left( \text{2,3,4} \right)\text{,B}\left( \text{-1,-2,1} \right)$ and $\text{C}\left( \text{5,8,7} \right)$.

Now, as we know that direction cosines can be found by $\left( {{\text{x}}_{\text{2}}}\text{-}{{\text{x}}_{\text{1}}} \right)\text{,}\left( {{\text{y}}_{\text{2}}}\text{-}{{\text{y}}_{\text{1}}} \right)$, and $\left( {{\text{z}}_{\text{2}}}\text{-}{{\text{z}}_{\text{1}}} \right)$

Therefore,

Direction ratios of $\text{AB}$ and $\text{BC}$ be $\text{-3,-5}$,$\text{-3}$ and $\text{6,10}$,$6$ respectively

As we can see that $\text{AB}$ and $\text{BC}$ are proportional, we get that $\text{AB}$ is parallel to $\text{BC}$.

Therefore, the points are collinear.

5. Find the direction cosines of the sides of the triangle whose vertices  are $\left( \text{3,5,-4} \right)\text{,}\left( \text{-1,1,2} \right)\text{,}\left( \text{-5,-5,-2} \right)$.

Ans: Let us consider the points be $\text{A}\left( \text{3,5,-4} \right)\text{,B}\left( \text{-1,1,2} \right)$ and $\text{C}\left( \text{-5,-5,-2} \right)$,

Now, the direction ratios of $\text{AB}$ will be $\text{-4,-4}$ and $\text{6}$,

We get

$\sqrt{{{\left( \text{-4} \right)}^{\text{2}}}\text{+}{{\left( \text{-4} \right)}^{\text{2}}}\text{+}{{\left( \text{6} \right)}^{\text{2}}}}\text{=}\sqrt{\text{68}}\Rightarrow \text{2}\sqrt{\text{17}}$

Now, 

$\text{l=}\dfrac{\text{-4}}{\sqrt{{{\left( \text{-4} \right)}^{\text{2}}}\text{+}{{\left( \text{-4} \right)}^{\text{2}}}\text{+}{{\left( \text{6} \right)}^{\text{2}}}}}\text{,m=}\dfrac{\text{-4}}{\sqrt{{{\left( \text{-4} \right)}^{\text{2}}}\text{+}{{\left( \text{-4} \right)}^{\text{2}}}\text{+}{{\left( \text{6} \right)}^{\text{2}}}}}\text{,n=}\dfrac{\text{6}}{\sqrt{{{\left( \text{-4} \right)}^{\text{2}}}\text{+}{{\left( \text{-4} \right)}^{\text{2}}}\text{+}{{\left( \text{6} \right)}^{\text{2}}}}}$

$\Rightarrow \text{l=}\dfrac{\text{-2}}{\sqrt{\text{17}}}\text{,m=}\dfrac{\text{-2}}{\sqrt{\text{17}}}\text{,n=}\dfrac{\text{3}}{\sqrt{\text{17}}}$

Therefore, the direction cosines of $\text{AB}$ are $\dfrac{\text{-2}}{\sqrt{\text{17}}}\text{,}\dfrac{\text{-2}}{\sqrt{\text{17}}}\text{,}\dfrac{\text{3}}{\sqrt{\text{17}}}$

Similarly, the direction ratios of side $\text{BC}$ will be $\text{-4,-6}$ and $\text{-4}$.

Now,

$\text{l=}\dfrac{\text{-4}}{\sqrt{{{\left( \text{-4} \right)}^{\text{2}}}\text{+}{{\left( \text{-6} \right)}^{\text{2}}}\text{+}{{\left( \text{-4} \right)}^{\text{2}}}}}\text{,m=}\dfrac{\text{-6}}{\sqrt{{{\left( \text{-4} \right)}^{\text{2}}}\text{+}{{\left( \text{-6} \right)}^{\text{2}}}\text{+}{{\left( \text{-4} \right)}^{\text{2}}}}}\text{,n=}\dfrac{\text{-4}}{\sqrt{{{\left( \text{-4} \right)}^{\text{2}}}\text{+}{{\left( \text{-6} \right)}^{\text{2}}}\text{+}{{\left( \text{-4} \right)}^{\text{2}}}}}$

$\text{l=}\dfrac{\text{-4}}{\text{2}\sqrt{\text{17}}}\text{,m=}\dfrac{\text{-6}}{\text{2}\sqrt{\text{17}}}\text{,n=}\dfrac{\text{-4}}{\text{2}\sqrt{\text{17}}}$

Therefore, the direction cosines of $\text{BC}$ is $\dfrac{\text{-2}}{\sqrt{\text{17}}}\text{,}\dfrac{\text{-3}}{\sqrt{\text{17}}}\text{,}\dfrac{\text{-2}}{\sqrt{\text{17}}}$

Similarly, the direction ratios of $\text{CA}$ will be$\text{-8,-10}$ and $\text{2}$.

Now,

$\text{l=}\dfrac{\text{-8}}{\sqrt{{{\left( \text{-8} \right)}^{\text{2}}}\text{+}{{\left( \text{10} \right)}^{\text{2}}}\text{+}{{\left( \text{2} \right)}^{\text{2}}}}}\text{,m=}\dfrac{\text{10}}{\sqrt{{{\left( \text{-8} \right)}^{\text{2}}}\text{+}{{\left( \text{10} \right)}^{\text{2}}}\text{+}{{\left( \text{2} \right)}^{\text{2}}}}}\text{,n=}\dfrac{\text{2}}{\sqrt{{{\left( \text{-8} \right)}^{\text{2}}}\text{+}{{\left( \text{10} \right)}^{\text{2}}}\text{+}{{\left( \text{2} \right)}^{\text{2}}}}}$

$\text{l=}\dfrac{\text{-8}}{\text{2}\sqrt{\text{42}}}\text{,m=}\dfrac{\text{-10}}{\text{2}\sqrt{\text{42}}}\text{,n=}\dfrac{\text{2}}{\text{2}\sqrt{\text{42}}}$.

Therefore, the direction cosines of $\text{CA}$is $\dfrac{\text{-4}}{\sqrt{\text{42}}}\text{,}\dfrac{\text{-5}}{\sqrt{\text{42}}}\text{,}\dfrac{\text{1}}{\sqrt{\text{42}}}$ 

Exercise 11.2 

1. Show that the three lines with direction cosines $\dfrac{\text{12}}{\text{13}}\text{,}\dfrac{\text{-3}}{\text{13}}\text{,}\dfrac{\text{-4}}{\text{13}}\text{;}\dfrac{\text{4}}{\text{13}}\text{,}\dfrac{\text{12}}{\text{13}}\text{,}\dfrac{\text{3}}{\text{13}}\text{;}\dfrac{\text{3}}{\text{13}}\text{,}\dfrac{\text{-4}}{\text{13}}\text{,}\dfrac{\text{12}}{\text{13}}$ are mutually perpendicular,

Ans: As we know, if ${{\text{l}}_{\text{1}}}{{\text{l}}_{\text{2}}}\text{+}{{\text{m}}_{\text{1}}}{{\text{m}}_{\text{2}}}\text{+}{{\text{n}}_{\text{1}}}{{\text{n}}_{\text{2}}}\text{=0}$, the lines are perpendicular

$\left( \text{i} \right)$Now, from direction cosines $\dfrac{\text{12}}{\text{13}}\text{,}\dfrac{\text{-3}}{\text{13}}\text{,}\dfrac{\text{-4}}{\text{13}}$ and $\dfrac{\text{4}}{\text{13}}\text{,}\dfrac{\text{12}}{\text{13}}\text{,}\dfrac{\text{3}}{\text{13}}$, we get

${{\text{l}}_{\text{1}}}{{\text{l}}_{\text{2}}}\text{+}{{\text{m}}_{\text{1}}}{{\text{m}}_{\text{2}}}\text{+}{{\text{n}}_{\text{1}}}{{\text{n}}_{\text{2}}}\text{=}\dfrac{\text{12}}{\text{13}}\text{ }\!\!\times\!\!\text{ }\dfrac{\text{4}}{\text{13}}\text{+}\left( \dfrac{\text{-3}}{\text{13}} \right)\text{ }\!\!\times\!\!\text{ }\dfrac{\text{12}}{\text{13}}\text{+}\left( \dfrac{\text{-4}}{\text{13}} \right)\text{ }\!\!\times\!\!\text{ }\dfrac{\text{3}}{\text{13}}$

$\Rightarrow \dfrac{\text{48}}{\text{169}}\text{-}\dfrac{\text{36}}{\text{169}}\text{-}\dfrac{\text{12}}{\text{169}}$

$\Rightarrow \text{0}$

Therefore, the lines are perpendicular.

$\left( \text{ii} \right)$Similarly, if we take $\frac{\text{4}}{\text{13}}\text{,}\frac{\text{12}}{\text{13}}\text{,}\frac{\text{3}}{\text{13}}$ and $\frac{\text{12}}{\text{13}}\text{,}\frac{\text{-3}}{\text{13}}\text{,}\frac{\text{-4}}{\text{13}}$, we get

${{\text{l}}_{\text{1}}}{{\text{l}}_{\text{2}}}\text{+}{{\text{m}}_{\text{1}}}{{\text{m}}_{\text{2}}}\text{+}{{\text{n}}_{\text{1}}}{{\text{n}}_{\text{2}}}\text{=}\frac{\text{4}}{\text{13}}\text{ }\!\!\times\!\!\text{ }\frac{\text{3}}{\text{13}}\text{+}\left( \frac{\text{12}}{\text{13}} \right)\text{ }\!\!\times\!\!\text{ }\left( \frac{\text{-4}}{\text{13}} \right)\text{+}\frac{\text{3}}{\text{13}}\text{ }\!\!\times\!\!\text{ }\left( \frac{\text{12}}{\text{13}} \right)$

$\Rightarrow \frac{\text{12}}{\text{169}}\text{-}\frac{\text{48}}{\text{169}}\text{-}\frac{\text{36}}{\text{169}}\text{=0}$

Therefore, the lines are perpendicular.

$\left( \text{iii} \right)$Again, if we consider $\frac{\text{-3}}{\text{13}}\text{,}\frac{\text{-4}}{\text{13}}\text{,}\frac{\text{12}}{\text{13}}$ and $\frac{\text{12}}{\text{13}}\text{,}\frac{\text{-3}}{\text{13}}\text{,}\frac{\text{-4}}{\text{13}}$, we get

${{\text{l}}_{\text{1}}}{{\text{l}}_{\text{2}}}\text{+}{{\text{m}}_{\text{1}}}{{\text{m}}_{\text{2}}}\text{+}{{\text{n}}_{\text{1}}}{{\text{n}}_{\text{2}}}\text{=}\frac{\text{3}}{\text{13}}\text{ }\!\!\times\!\!\text{ }\frac{\text{12}}{\text{13}}\text{+}\left( \frac{\text{-4}}{\text{13}} \right)\text{ }\!\!\times\!\!\text{ }\left( \frac{\text{-4}}{\text{13}} \right)\text{+}\frac{\text{12}}{\text{13}}\text{ }\!\!\times\!\!\text{ }\left( \frac{\text{-4}}{\text{13}} \right)$

$\Rightarrow \frac{\text{36}}{\text{169}}\text{-}\frac{\text{12}}{\text{169}}\text{-}\frac{\text{48}}{\text{169}}\text{=0}$

Therefore, the lines are perpendicular.

Therefore, we can say that all the lines are mutually perpendicular.

2.show that the line passing through the points $\left( \text{1,-1,2} \right)\left( \text{3,4,-2} \right)$ is perpendicular to the line through the points $\left( \text{0,3,2} \right)$ and $\left( \text{3,5,6} \right)$?

Ans: Let us consider that $\text{AB}$ and $\text{CD}$ are the lines that pass through the points, $\left( \text{1,1,-2} \right)$, $\left( \text{3,4,-2} \right)$ and $\left( \text{0,3,2} \right)$, $\left( \text{3,5,6} \right)$, respectively.

Now, we have  ${{\text{a}}_{\text{1}}}\text{=}\left( \text{2} \right)\text{,}{{\text{b}}_{\text{1}}}\text{=}\left( \text{5} \right)\text{,}{{\text{c}}_{\text{1}}}\text{=}\left( \text{-4} \right)$ and ${{\text{a}}_{\text{2}}}\text{=}\left( \text{3} \right)\text{,}{{\text{b}}_{\text{2}}}\text{=}\left( \text{2} \right)\text{,}{{\text{c}}_{\text{2}}}\text{=}\left( \text{4} \right)$

As we know that if $\text{AB}\bot \text{CD}$ then ${{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}\text{+}{{\text{b}}_{\text{1}}}{{\text{b}}_{\text{2}}}\text{+}{{\text{c}}_{\text{1}}}{{\text{c}}_{\text{2}}}\text{=0}$

Now, 

${{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}\text{+}{{\text{b}}_{\text{1}}}{{\text{b}}_{\text{2}}}\text{+}{{\text{c}}_{\text{1}}}{{\text{c}}_{\text{2}}}\text{=2 }\!\!\times\!\!\text{ 3+5 }\!\!\times\!\!\text{ 2+}\left( \text{-4} \right)\text{ }\!\!\times\!\!\text{ 4}$

$\Rightarrow \text{2 }\!\!\times\!\!\text{ 3+5 }\!\!\times\!\!\text{ 2-4 }\!\!\times\!\!\text{ 4=6+10-16}$

$\Rightarrow \text{0}$

Therefore, $\text{AB}$ and $\text{CD}$ are perpendicular to each other.

3. Show that the line through the points $\left( \text{4,7,8} \right)\left( \text{2,3,4} \right)$ is parallel to the line through the points $\left( \text{1,-2,1} \right)\left( \text{1,2,5} \right)$.

Ans: Let us consider the lines $\text{AB}$ and $\text{CD}$ that pass through points $\left( \text{4,7,8} \right)$, $\left( \text{2,3,4} \right)$, and$\left( \text{-1,-2,1} \right)$, $\left( \text{1,2,5} \right)$ respectively.

Now, we get

${{\text{a}}_{\text{1}}}\text{=}\left( \text{2-4} \right)\text{,}{{\text{b}}_{\text{1}}}\text{=}\left( \text{3-7} \right)\text{,}{{\text{c}}_{\text{1}}}\text{=}\left( \text{4-8} \right)$ and ${{\text{a}}_{\text{2}}}\text{=}\left( \text{1+1} \right)\text{,}{{\text{b}}_{\text{2}}}\text{=}\left( \text{2+2} \right)\text{,}{{\text{c}}_{\text{2}}}\text{=}\left( \text{5-1} \right)$

${{\text{a}}_{\text{1}}}\text{=}\left( \text{-2} \right)\text{,}{{\text{b}}_{\text{1}}}\text{=}\left( \text{-4} \right)\text{,}{{\text{c}}_{\text{1}}}\text{=}\left( \text{-4} \right)$ and ${{\text{a}}_{\text{2}}}\text{=}\left( \text{2} \right)\text{,}{{\text{b}}_{\text{2}}}\text{=}\left( \text{4} \right)\text{,}{{\text{c}}_{\text{2}}}\text{=}\left( \text{4} \right)$

Now, we know that if $\text{AB}\parallel \text{CD}$ then $\frac{{{\text{a}}_{\text{1}}}}{{{\text{a}}_{\text{2}}}}\text{=}\frac{{{\text{b}}_{\text{1}}}}{{{\text{b}}_{\text{2}}}}\text{=}\frac{{{\text{c}}_{\text{1}}}}{{{\text{c}}_{\text{2}}}}$,

Now,

$\frac{{{\text{a}}_{\text{1}}}}{{{\text{a}}_{\text{2}}}}\text{=}\frac{\text{-2}}{\text{2}}\Rightarrow \text{-1}$,$\frac{{{\text{b}}_{\text{1}}}}{{{\text{b}}_{\text{2}}}}\text{=}\frac{\text{-4}}{\text{4}}\Rightarrow \text{-1}$,$\frac{{{\text{c}}_{\text{1}}}}{{{\text{c}}_{\text{2}}}}\text{=}\frac{\text{-4}}{\text{4}}\text{=-1}$

We got $\frac{{{\text{a}}_{\text{1}}}}{{{\text{a}}_{\text{2}}}}\text{=}\frac{{{\text{b}}_{\text{1}}}}{{{\text{b}}_{\text{2}}}}\text{=}\frac{{{\text{c}}_{\text{1}}}}{{{\text{c}}_{\text{2}}}}$

Therefore, $\text{AB}$ is parallel to $\text{CD}$.

4. Find the equation of the line which passes through point $\left( \text{1,2,3} \right)$ and parallel to the vector $\text{3i+2j-2k}$ .

Ans: Now, let us consider the position vector $\text{A}$ be $\text{a=i+2j+3k}$ and let $\vec{b}$=$3\hat{i}+2\hat{j}-2\hat{k}$

Now, we know that the line passes through $\text{A}$ and is parallel to $\vec{b}$,

As we know $\vec{r}$=$\vec{a}+\lambda \vec{b}$ where $\lambda $ is a constant

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

Therefore, the equation of the line is $\vec{r}$=$\hat{i}+2\hat{j}+3\hat{k}+ \!\!\lambda\!\!\text{ }\left( {3\hat{i}+2\hat{j}-2\hat{k}} \right)$

5. Find the equation of the line in vector and in Cartesian form that passes through the point with positive vector ${2\hat{i}-\hat{j}-4\hat{k}}$ and is in the direction ${\hat{i}+2\hat{j}-\hat{k}}$.

Ans: We know that the line passes through the point with positive vector

Now, let us consider ${\vec{a}=2\hat{i}-\hat{j}+4\hat{k}}$ and ${\vec{b}=\hat{i}+2\hat{j}-\hat{k}}$

Now, line passes through point $\text{A}$ and parallel to ${\vec{b}}$, we get

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

Therefore, the equation of the line in vector form is ${\vec{r}=2\hat{i}-\hat{j}+4\hat{k}+ }\!\!\lambda\!\!\text{ }\left( {\hat{i}+2\hat{j}-\hat{k}} \right)$.

Now, we know

${\vec{r}=x\hat{i}-y\hat{j}+z\hat{k}}\Rightarrow {x\hat{i}-y\hat{j}+z\hat{k}=}\left( \text{ }\!\!\lambda\!\!\text{ +2} \right){\hat{i}+}\left( \text{2 }\!\!\lambda\!\!\text{ -1} \right){\hat{j}+}\left( \text{- }\!\!\lambda\!\!\text{ +4} \right){\hat{k}}$

Therefore, the equation of the line in cartesian form will be $\frac{\text{x-2}}{\text{1}}\text{=}\frac{\text{y+1}}{\text{2}}=\frac{\text{z-4}}{\text{-1}}$.

6. Find the Cartesian equation of the line which passes through the point $\left( \text{-2,4,-5} \right)$ and parallel to the line given by $\frac{\text{x+3}}{\text{3}}\text{=}\frac{\text{y-4}}{\text{5}}\text{=}\frac{\text{z+8}}{\text{6}}.

Ans: We know that the line passes through point $\left( \text{-2,4,-5} \right)$ and also parallel to  $\frac{\text{x+3}}{\text{3}}\text{=}\frac{\text{y-4}}{\text{5}}\text{=}\frac{\text{z+8}}{\text{6}}$

Now, as we can see the direction ratios of the line are $\text{3,5}$ and $\text{6}$.

As we know the required line is parallel to $\frac{\text{x+3}}{\text{3}}\text{=}\frac{\text{y-4}}{\text{5}}\text{=}\frac{\text{z+8}}{\text{6}}$

Therefore, the direction ratios will be $\text{3k,5k}$ and $\text{6k}$

As we know that the equation of the line through the point and with direction ratio is shown in form $\frac{\text{x-}{{\text{x}}_{\text{1}}}}{\text{a}}\text{=}\frac{\text{y-}{{\text{y}}_{\text{1}}}}{\text{b}}\text{=}\frac{\text{z-}{{\text{z}}_{\text{1}}}}{\text{c}}$

Therefore, the equation of the line $\frac{\text{x+2}}{\text{3}}\text{=}\frac{\text{y-4}}{\text{5}}\text{=}\frac{\text{z+5}}{\text{6}}\text{=k}$.

7. The Cartesian equation of a line is $\frac{\text{x-5}}{\text{3}}\text{=}\frac{\text{y+4}}{\text{7}}\text{=}\frac{\text{z-6}}{\text{2}}$ Write its vector form.

Ans: As we can see the cartesian equation of the line, we can tell that the line is passing through $\left( \text{5,4,-6} \right)$, and he direction ratios are $\text{3,7}$ and $\text{2}$.

Now, we got the position vector ${\vec{a}=5\hat{i}-4\hat{j}+6\hat{k}}$

From this we got the direction of the vector be $\vec{b}$=$3\hat{i}+7\hat{j}+2\hat{k}$

Therefore, the vector form of the line will be ${\vec{r}=5\hat{i}-4\hat{j}+6\hat{k}+ }\!\!\lambda\!\!\text{ }\left({3\hat{i}+7\hat{j}+2\hat{k}} \right)$

8. Find the angle between the following pair of lines (i) \[\overrightarrow{r}=2\widehat{i}-5\widehat{j}+\widehat{k}+\lambda \left( 3\widehat{i}+2\widehat{j}+6\widehat{k} \right)\]. and $\overrightarrow{r}=7\hat{i}-6\hat{k}+\mu \text{ }\left( \hat{i}+2\hat{j}+2\hat{k} \right)$ {ii} ${\vec{r}=3\hat{i}+\hat{j}-2\hat{k}+ }\!\!\lambda\!\!\text{ }\left( {\hat{i}-\hat{j}-2\hat{k}} \right)$ and ${\vec{r}=2\hat{i}-\hat{j}-56\hat{k}+ }\!\!\mu\!\!\text{ }\left( {3\hat{i}-5\hat{j}-4\hat{k}} \right)$

(i) Ans:  $\left( \text{i} \right)$ let us consider the angle be $\text{ }\!\!\theta\!\!\text{ }$,

As we know that the angle between the lines can be found by $\text{cos }\!\!\theta\!\!\text{ =}\left| \frac{{{{{\vec{b}}}}_{\text{1}}}{{{{\vec{b}}}}_{\text{2}}}}{\left| {{{{\vec{b}}}}_{\text{1}}} \right|\left| {{{{\vec{b}}}}_{\text{2}}} \right|} \right|$

As the lines are parallel to ${{{\vec{b}}}_{\text{1}}}{=3\hat{i}+2\hat{j}+6\hat{k}}$ and ${{{\vec{b}}}_{\text{2}}}{=\hat{i}+2\hat{j}+2\hat{k}}$, we got

$\left| {{{{\vec{b}}}}_{\text{1}}} \right|\text{=}\sqrt{{{\text{3}}^{\text{2}}}\text{+}{{\text{2}}^{\text{2}}}\text{+}{{\text{6}}^{\text{2}}}}\text{=7}$, $\left| {{{{\vec{b}}}}_{\text{2}}} \right|\text{=}\sqrt{{{\text{1}}^{\text{2}}}\text{+}{{\text{2}}^{\text{2}}}\text{+}{{\text{2}}^{\text{2}}}}\text{=3}$ and ${{{\vec{b}}}_{\text{1}}}{{{\vec{b}}}_{\text{2}}}=\left( {3\hat{i}+2\hat{j}+6\hat{k}} \right)\left( {\hat{i}+2\hat{j}+2\hat{k}} \right)\text{=19}$

Therefore, the angle between the lines will be 

$\text{cos }\!\!\theta\!\!\text{ =}\frac{\text{19}}{\text{7 }\!\!\times\!\!\text{ 3}}$

$\Rightarrow \text{ }\!\!\theta\!\!\text{ =co}{{\text{s}}^{\text{-1}}}\frac{\text{19}}{\text{21}}$

(ii) $\vec{r} = 3\cap{i}+\cap{j}-2\cap{k}+\lambda(\cap{i}-\cap{j}-2\cap{k})$ and $\vec{r}=2\cap{i}-\cap{j}-56\cap{k}+\mu (3\cap{i}-5\cap{j}-4\cap{k})$ 

As the lines are parallel to the vectors ${{{\vec{b}}}_{\text{1}}}{=\hat{i}-\hat{j}-2\hat{k}}$ and ${{{\vec{b}}}_{\text{2}}}{=3\hat{i}-5\hat{j}+-4\hat{k}}$, we get

$\left| {{{{\vec{b}}}}_{\text{1}}} \right|\text{=}\sqrt{{{\text{1}}^{\text{2}}}\text{+}{{\left( \text{-1} \right)}^{\text{2}}}\text{+}{{\left( \text{-2} \right)}^{\text{2}}}}\text{=}\sqrt{\text{6}}$, $\left| {{{{\vec{b}}}}_{\text{2}}} \right|\text{=}\sqrt{{{\text{3}}^{\text{2}}}\text{+}{{\left( \text{-5} \right)}^{\text{2}}}\text{+}{{\left( \text{-2} \right)}^{\text{2}}}}\text{=5}\sqrt{\text{2}}$ and ${{{\vec{b}}}_{\text{1}}}{{{\vec{b}}}_{\text{2}}}\text{=}\left( {\hat{i}-\hat{j}-2\hat{k}} \right)\left( {3\hat{i}-5\hat{j}+-4\hat{k}} \right)\text{=16}$

Therefore, the angle between them will be,

$\text{cos }\!\!\theta\!\!\text{ =}\frac{\text{16}}{\text{10}\sqrt{\text{3}}}$

$\Rightarrow \text{cos }\!\!\theta\!\!\text{ =}\frac{\text{8}}{\text{5}\sqrt{\text{3}}}$

$\Rightarrow \text{ }\!\!\theta\!\!\text{ =co}{{\text{s}}^{\text{-1}}}\frac{\text{8}}{\text{5}\sqrt{\text{3}}}$

9. Find the angle between the following pair of lines $\frac{\text{x-2}}{\text{2}} = \frac{\text{y-1}}{\text{5}}\text{=}\frac{\text{z+3}}{\text{-3}}$ and $\frac{\text{x+2}}{\text{-1}}\text{=}\frac{\text{y-4}}{\text{8}}\text{=}\frac{\text{z-5}}{\text{4}}$ $\frac{\text{x}}{\text{2}}\text{=}\frac{\text{y}}{\text{2}}\text{=}\frac{\text{z}}{\text{1}}$ and $\frac{\text{x-5}}{\text{4}}\text{=}\frac{\text{y-2}}{\text{1}}\text{=}\frac{\text{z-3}}{\text{8}}$

Ans: $\left( \text{i} \right)$ Let us take ${{{\vec{b}}}_{\text{1}}}$ and ${{{\vec{b}}}_{2}}$be the vectors parallel to the lines, we get

${{{\vec{b}}}_{\text{1}}}{=2\hat{i}+5\hat{j}-3\hat{k}}$  and ${{{\vec{b}}}_{2}}{=-\hat{i}+8\hat{j}+4\hat{k}}$

Now $\left| {{{{\vec{b}}}}_{\text{1}}} \right|\text{=}\sqrt{{{\text{2}}^{\text{2}}}\text{+}{{\text{5}}^{\text{2}}}\text{+}{{\left( \text{-3} \right)}^{\text{2}}}}\text{=}\sqrt{\text{38}}$, $\left| {{{{\vec{b}}}}_{\text{2}}} \right|\text{=}\sqrt{{{\left( \text{-1} \right)}^{\text{2}}}\text{+}{{\text{8}}^{\text{2}}}\text{+}{{\text{4}}^{\text{2}}}}\text{=9}$ 

And,

${{{\vec{b}}}_{\text{1}}}{{{\vec{b}}}_{\text{2}}}\text{=}\left( {2\hat{i}+5\hat{j}-3\hat{k}} \right)\left({-\hat{i}+8\hat{j}+4\hat{k}} \right)$

$\text{=2}\left( \text{-1} \right)\text{+5}\left( \text{8} \right)\text{+4}\left( \text{-3} \right)$

$\text{=26}$

We can find the angle by using $\text{cos }\!\!\theta\!\!\text{ =}\left| \frac{{{{{\vec{b}}}}_{\text{1}}}{{{{\vec{b}}}}_{\text{2}}}}{\left| {{{{\vec{b}}}}_{\text{1}}} \right|\left| {{{{\vec{b}}}}_{\text{2}}} \right|} \right|$

Therefore, 

$\text{cos }\!\!\theta\!\!\text{ =}\frac{\text{26}}{\text{9}\sqrt{\text{38}}}$

$\Rightarrow \text{ }\!\!\theta\!\!\text{ =co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{26}}{\text{9}\sqrt{\text{38}}} \right)$

Therefore, the angle will be $\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{26}}{\text{9}\sqrt{\text{38}}} \right)$. 

ii)

Ans:$\left( \text{ii} \right)$ Similarly let us consider ${{{\vec{b}}}_{\text{1}}}$ and ${{{\vec{b}}}_{2}}$be the vectors parallel to lines, we get

${{{\vec{b}}}_{\text{1}}}{=2\hat{i}+2\hat{j}+\hat{k}}$  and ${{{\vec{b}}}_{2}}{=4\hat{i}+\hat{j}+8\hat{k}}$

Now, $\left| {{{{\vec{b}}}}_{\text{1}}} \right|\text{=}\sqrt{{{\text{2}}^{\text{2}}}\text{+}{{\text{2}}^{\text{2}}}\text{+}{{\left( \text{1} \right)}^{\text{2}}}}\text{=3}$, $\left| {{{{\vec{b}}}}_{\text{2}}} \right|\text{=}\sqrt{{{\text{4}}^{\text{2}}}\text{+}{{\text{1}}^{\text{2}}}\text{+}{{\text{8}}^{\text{2}}}}\text{=9}$ and 

${{{\vec{b}}}_{\text{1}}}{{{\vec{b}}}_{\text{2}}}\text{=}\left( {2\hat{i}+2\hat{j}+1\hat{k}} \right)\text{.}\left({4\hat{i}+\hat{j}+8\hat{k}} \right)$

$\text{=2}\left( \text{4} \right)\text{+2}\left( \text{1} \right)\text{+1}\left( \text{8} \right)$

$\text{=18}$

As we know the angle can be found by $\text{cos }\!\!\theta\!\!\text{ =}\left| \frac{{{{{\vec{b}}}}_{\text{1}}}{{{{\vec{b}}}}_{\text{2}}}}{\left| {{{{\vec{b}}}}_{\text{1}}} \right|\left| {{{{\vec{b}}}}_{\text{2}}} \right|} \right|$

Therefore,

$\text{cos }\!\!\theta\!\!\text{ =}\frac{\text{18}}{\text{27}}\text{=}\frac{\text{2}}{\text{3}}$

$\Rightarrow \text{ }\!\!\theta\!\!\text{ =co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{2}}{\text{3}} \right)$

Therefore, the angle is $\text{co}{{\text{s}}^{\text{-1}}}\left( \frac{\text{2}}{\text{3}} \right)$.

10.Find the values of $\text{p}$ so the line $\dfrac{\text{1-x}}{\text{3}}\text{=}\dfrac{\text{7y-14}}{\text{2p}}\text{=}\dfrac{\text{z-3}}{\text{2}}\text{and}\dfrac{\text{7-7x}}{\text{3p}}\text{=}\dfrac{\text{y-5}}{\text{1}}\text{=}\dfrac{\text{6-z}}{\text{2}}$ are at right angles.

Ans: As we know that the correct form of the equation is as follows,

$\dfrac{\text{x-1}}{\text{-3}}\text{=}\dfrac{\text{y-2}}{\dfrac{\text{2p}}{\text{7}}}\text{=}\dfrac{\text{z-3}}{\text{2}}\text{and}\dfrac{\text{x-1}}{\dfrac{\text{-3p}}{\text{7}}}\text{=}\dfrac{\text{y-5}}{\text{1}}\text{=}\dfrac{\text{z-6}}{\text{-5}}$

From this we get the direction ratios as 

${{\text{a}}_{\text{1}}}\text{=-3,}{{\text{b}}_{\text{1}}}\text{=}\dfrac{\text{2p}}{\text{7}}\text{,}{{\text{c}}_{\text{1}}}\text{=2 and }{{\text{a}}_{\text{2}}}\text{=}\dfrac{\text{-3p}}{\text{7}}\text{,}{{\text{b}}_{\text{2}}}\text{=1,}{{\text{c}}_{\text{2}}}\text{=-5}$

As we know the lines are perpendicular, we get

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

$\Rightarrow \dfrac{\text{9p}}{\text{7}}\text{+}\dfrac{\text{2p}}{\text{7}}\text{=10}$

$\Rightarrow \text{11p=70}$

$\Rightarrow \text{p=}\dfrac{\text{70}}{\text{11}}$

Therefore, the value of $\text{p}$ is $\dfrac{\text{70}}{\text{11}}$.

11:show that the lines $\frac{\text{x-5}}{\text{7}}\text{=}\frac{\text{y+2}}{\text{-5}}\text{=}\frac{\text{z}}{\text{1}}\text{and}\frac{\text{x}}{\text{1}}\text{=}\frac{\text{y}}{\text{2}}\text{=}\frac{\text{z}}{\text{3}}$ are perpendicular to each other.

Ans: From the given equation, we get the direction ratios as,

${{\text{a}}_{\text{1}}}\text{=7, }{{\text{b}}_{\text{1}}}\text{=-5, }{{\text{c}}_{\text{1}}}\text{=1}$, ${{\text{a}}_{\text{2}}}\text{=1 ,}{{\text{b}}_{\text{2}}}\text{=2 ,}{{\text{c}}_{\text{2}}}\text{=3}$

As we know, if ${{\text{a}}_{\text{1}}}{{\text{a}}_{\text{2}}}\text{+}{{\text{b}}_{\text{1}}}{{\text{b}}_{\text{2}}}\text{+}{{\text{c}}_{\text{1}}}{{\text{c}}_{\text{2}}}\text{=0}$, the lines are perpendicular to each other

Now,

$\text{7}\left( \text{1} \right)\text{+}\left( \text{-5} \right)\text{2+1}\left( \text{3} \right)\Rightarrow \text{7-10+3=0}$

Therefore, the lines are perpendicular.

12. Find the shortest distance between the lines are $\vec{r}=\hat{i}+2\hat{j}+\hat{k}+\lambda\left( {\hat{i}-\hat{j}+\hat{k}} \right)$ and $\vec{r}=2\hat{i}-\hat{j}-\hat{k}+ \mu\left({2\hat{i}+\hat{j}+2\hat{k}} \right)$.

Ans: We have been given lines, $\vec{r}=\hat{i}+2\hat{j}+\hat{k}+\lambda \left( \hat{i}-\hat{j}-\hat{k} \right)$ and \[\vec{r}=2\hat{i}-\hat{j}-\hat{k}+\mu \text{ }\left( 2\hat{i}+\hat{j}+2\hat{k} \right)\]

As we know that the shortest distance can be found as $\text{d=}\left| \frac{\left( {{{{\vec{b}}}}_{\text{1}}}\text{ }\!\!\times\!\!\text{ }{{{{\vec{b}}}}_{\text{2}}} \right)\left( {{{{\vec{a}}}}_{\text{2}}}\text{-}{{{{\vec{a}}}}_{1}} \right)}{\left| {{{{\vec{b}}}}_{\text{1}}}\text{ }\!\!\times\!\!\text{ }{{{{\vec{b}}}}_{\text{2}}} \right|} \right|$

Now, from the given lines we get that

${{{\vec{a}}}_{\text{1}}}{=\hat{i}+2\hat{j}+\hat{k},}$

${{{\vec{b}}}_{\text{1}}}{=\hat{i}-\hat{j}-\hat{k}}$

${{{\vec{a}}}_{\text{2}}}{=2\hat{i}-\hat{j}-\hat{k},}$

${{{\vec{b}}}_{\text{2}}}{=2\hat{i}+\hat{j}+2\hat{k}}$

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

${=\hat{i}-3\hat{j}-2\hat{k}}$,

${{{\vec{b}}}_{\text{1}}}\text{ }\!\!\times\!\!\text{ }{{{\vec{b}}}_{\text{2}}}\text{=}\left| \begin{matrix} {{\hat{i}}} & {{\hat{j}}} & {{\hat{k}}} \\ \text{1} & \text{-1} & \text{1} \\ \text{2} & \text{-1} & \text{2} \\ \end{matrix} \right|$

$\Rightarrow {{{\vec{b}}}_{\text{1}}}\text{ }\!\!\times\!\!\text{ }{{{\vec{b}}}_{\text{2}}}{=-3\hat{i}+3\hat{k}}$.

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

Now, if we put all the values in theirs places, we get

$\text{d=}\left| \frac{\left( {-3\hat{i}+3\hat{k}} \right)\left( {\hat{i}-3\hat{j}-2\hat{k}} \right)}{\text{3}\sqrt{\text{2}}} \right|\Rightarrow \text{d=}\left| \frac{\text{-3}\left( \text{1} \right)\text{+3}\left( \text{2} \right)}{\text{3}\sqrt{\text{2}}} \right|$

$\text{d=}\left| \frac{\text{-9}}{\text{3}\sqrt{\text{2}}} \right|\Rightarrow \text{d=}\frac{\text{3}\sqrt{\text{2}}}{\text{2}}$

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

13. Find the shortest distance between the lines

$\frac{\text{x+1}}{\text{7}}\text{=}\frac{\text{y+1}}{\text{-6}}\text{=}\frac{\text{z+1}}{\text{1}}$ and $\frac{\text{x-3}}{\text{1}}\text{=}\frac{\text{y-5}}{\text{-2}}\text{=}\frac{\text{z-7}}{\text{1}}$

Ans: As we know that the shortest distance can be found by,

$\text{d=}\frac{\left| \begin{matrix} {{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\ {{a}_{1}} & {{b}_{1}} & {{c}_{1}} \\ {{a}_{2}} & {{b}_{2}} & {{c}_{2}} \\ \end{matrix} \right|}{\sqrt{{{\left( {{b}_{1}}{{c}_{2}}-{{b}_{2}}{{c}_{1}} \right)}^{2}}+{{\left( {{c}_{1}}{{a}_{2}}-{{c}_{2}}{{a}_{1}} \right)}^{2}}+{{\left( {{a}_{1}}{{b}_{2}}-{{a}_{2}}{{b}_{1}} \right)}^{2}}}}$

Now, from the given lines we got that

${{\text{x}}_{\text{1}}}\text{=-1,}{{\text{y}}_{\text{1}}}\text{=-1,}{{\text{z}}_{\text{1}}}\text{=-1,}{{\text{a}}_{\text{1}}}\text{=7,}{{\text{b}}_{\text{1}}}\text{=-6,}{{\text{c}}_{\text{1}}}\text{=1}$

${{\text{x}}_{\text{2}}}\text{=3,}{{\text{y}}_{\text{2}}}\text{=5,}{{\text{z}}_{\text{2}}}\text{=7,}{{\text{a}}_{\text{2}}}\text{=1,}{{\text{b}}_{\text{2}}}\text{=-2,}{{\text{c}}_{\text{2}}}\text{=1}$

And,

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

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

$\text{=-16-36-64}{=}-\text{116}$

And,

$\sqrt{{{\left( {{\text{b}}_{\text{1}}}{{\text{c}}_{\text{2}}}\text{-}{{\text{b}}_{\text{2}}}{{\text{c}}_{\text{1}}} \right)}^{\text{2}}}\text{+}{{\left( {{\text{c}}_{\text{1}}}{{\text{a}}_{\text{2}}}\text{-}{{\text{c}}_{\text{2}}}{{\text{a}}_{\text{1}}} \right)}^{\text{2}}}\text{+}{{\left( {{\text{a}}_{\text{1}}}{{\text{b}}_{\text{2}}}\text{-}{{\text{a}}_{\text{2}}}{{\text{b}}_{\text{1}}} \right)}^{\text{2}}}}\text{=}\sqrt{{{\left( \text{-6+2} \right)}^{\text{2}}}\text{+}{{\left( \text{1+7} \right)}^{\text{2}}}\text{+}{{\left( \text{-14+6} \right)}^{\text{2}}}}$

$\sqrt{{{\left( {{\text{b}}_{\text{1}}}{{\text{c}}_{\text{2}}}\text{-}{{\text{b}}_{\text{2}}}{{\text{c}}_{\text{1}}} \right)}^{\text{2}}}\text{+}{{\left( {{\text{c}}_{\text{1}}}{{\text{a}}_{\text{2}}}\text{-}{{\text{c}}_{\text{2}}}{{\text{a}}_{\text{1}}} \right)}^{\text{2}}}\text{+}{{\left( {{\text{a}}_{\text{1}}}{{\text{b}}_{\text{2}}}\text{-}{{\text{a}}_{\text{2}}}{{\text{b}}_{\text{1}}} \right)}^{\text{2}}}}\text{=2}\sqrt{\text{29}}$

Putting all the values, we get

$\text{d=}\frac{\text{-116}}{2\sqrt{29}}$

$\text{d=}\frac{\text{-58}}{\sqrt{\text{29}}}\Rightarrow \frac{\text{-58}\sqrt{\text{29}}}{\text{29}}$

$\text{d=}\frac{\text{-58}}{\sqrt{\text{29}}}\Rightarrow \left| \text{d} \right|\text{=2}\sqrt{\text{29}}$

Therefore, the distance between the lines is $\text{2}\sqrt{\text{29}}$ units.

14. Find the shortest distance between the lines whose vector equations are ${\vec{r}}$=$\hat{i}+2\hat{j}+3\hat{k}+ \!\!\lambda\!\!\text{ }\left( {\hat{i}-3\hat{j}+2\hat{k}} \right)$ and $\vec{r}$=$4\hat{i}+5\hat{j}+6\hat{k}+ \!\!\mu\!\!\text{ }\left( {2\hat{i}+3\hat{j}+\hat{k}} \right)$

Ans: We have been given lines $\vec{r}$=$\hat{i}+2\hat{j}+3\hat{k}+ \!\!\lambda\!\!\text{ }\left( {\hat{i}-3\hat{j}+2\hat{k}} \right)$ and $\vec{r}$=$4\hat{i}+5\hat{j}+6\hat{k}+ \!\!\mu\!\!\text{ }\left( {2\hat{i}+3\hat{j}+\hat{k}} \right)$

As we know that the shortest distance between the lines can be found by,

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

Now, from the given lines, we got

${{{\vec{a}}}_{\text{1}}}{=\hat{i}+2\hat{j}+3\hat{k}, }{{{\vec{b}}}_{\text{1}}}{=\hat{i}-3\hat{j}+2\hat{k}}$

${{{\vec{a}}}_{\text{2}}}{=4\hat{i}+5\hat{j}+6\hat{k}, }{{{\vec{b}}}_{\text{2}}}{=2\hat{i}+3\hat{j}+\hat{k}}$

${{{\vec{a}}}_{\text{2}}}\text{-}{{{\vec{a}}}_{\text{1}}}\text{=}\left( {4\hat{i}+5\hat{j}+6\hat{k}} \right)\text{-}\left( {\hat{i}+2\hat{j}+3\hat{k}} \right)$

${=3\hat{i}+3\hat{j}+3\hat{k}}$

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

$\Rightarrow {{{\vec{b}}}_{\text{1}}}\text{ }\!\!\times\!\!\text{ }{{{\vec{b}}}_{\text{2}}}{=-9\hat{i}+3\hat{j}+9\hat{k}}$

$\left| {{{{\vec{b}}}}_{\text{1}}}\text{ }\!\!\times\!\!\text{ }{{{{\vec{b}}}}_{\text{2}}} \right|\text{=}\sqrt{{{\left( \text{-9} \right)}^{\text{2}}}\text{+}{{\text{3}}^{\text{2}}}\text{+}{{\text{9}}^{\text{2}}}}\text{=3}\sqrt{\text{19}}$

Now, putting all the values, we get

$\text{d=}\left| \frac{\text{9}}{\text{3}\sqrt{\text{19}}} \right|\text{=}\frac{\text{3}}{\sqrt{\text{19}}}$

Therefore, the shortest distance between the lines is $\frac{\text{3}}{\sqrt{\text{19}}}$ units.

15. Find the shortest distance between the lines whose vector equations are ${\vec{r}=}\left( \text{1-t} \right){\hat{i}+}\left( \text{t-2} \right){\hat{j}+}\left( \text{3-2t} \right){\hat{k}}$ and ${\vec{r}=}\left( \text{s+1} \right){\hat{i}+}\left( \text{2s-1} \right){\hat{j}-}\left( \text{2s+1} \right){\hat{k}}$.

Ans: We have been given lines ${\vec{r}=}\left( \text{1-t} \right){\hat{i}+}\left( \text{t-2} \right){\hat{j}+}\left( \text{3-2t} \right){\hat{k}}$ and ${\vec{r}=}\left( \text{s+1} \right){\hat{i}+}\left( \text{2s-1} \right){\hat{j}-}\left( \text{2s+1} \right){\hat{k}}$

\[\Rightarrow \overrightarrow{r}=\widehat{i}-2\widehat{j}+3\widehat{k}+t\left( -\widehat{i}+\widehat{j}-2\widehat{k} \right)\] and ${\vec{r}=\hat{i}-\hat{j}+\hat{k}+s}\left( {\hat{i}+2\hat{j}-2\hat{k}} \right)$

Now, the shortest distance can be found by,

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

Now, from the given lines we got,

\[{{\overrightarrow{a}}_{1}}=\widehat{i}-2\widehat{j}+3\widehat{k},{{\overrightarrow{b}}_{1}}=-\widehat{i}+\widehat{j}-2\widehat{k}\],

${{{\vec{a}}}_{\text{2}}}{=\hat{i}-\hat{j}-\hat{k}, }{{{\vec{b}}}_{\text{2}}}{=\hat{i}+2\hat{j}-2\hat{k}}$

${{{\vec{a}}}_{\text{2}}}\text{-}{{{\vec{a}}}_{\text{1}}}\text{=}\left( {\hat{i}-\hat{j}-\hat{k}} \right)\text{-}\left( {\hat{i}+2\hat{j}+3\hat{k}} \right){=\hat{j}-4\hat{k}}$,

${{{\vec{b}}}_{\text{1}}}\text{ }\!\!\times\!\!\text{ }{{{\vec{b}}}_{\text{2}}}\text{=}\left| \begin{matrix} {{\hat{i}}} & {{\hat{j}}} & {{\hat{k}}} \\ \text{-1} & \text{1} & \text{-2} \\ \text{1} & \text{2} & \text{-2} \\ \end{matrix} \right|$

$\Rightarrow {{\overrightarrow{b}}_{1}}\times {{\overrightarrow{b}}_{2}}=2\widehat{i}-4\widehat{j}-3\widehat{k}$

$\left| {{{{\vec{b}}}}_{\text{1}}}\text{ }\!\!\times\!\!\text{ }{{{{\vec{b}}}}_{\text{2}}} \right|\text{=}\sqrt{{{\left( 2 \right)}^{\text{2}}}\text{+}{{\left( -4 \right)}^{\text{2}}}\text{+}{{\left( -3 \right)}^{\text{2}}}}\text{=}\sqrt{\text{29}}$

\[\left( {{\overrightarrow{b}}_{1}}\times {{\overrightarrow{b}}_{2}} \right)\times \left( {{\overrightarrow{a}}_{2}}\times {{\overrightarrow{a}}_{1}} \right)=\left( 2\widehat{i}-4\widehat{j}-3\widehat{k} \right)\left( \widehat{j}-4\widehat{k} \right)\]

$=-\text{4+12}=8$

Putting all the values, we get

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

Therefore, the shortest distance between the lines is $\frac{\text{8}}{\sqrt{\text{29}}}$ units.

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 $\frac{\text{x-1}}{\text{3k}}\text{=}\frac{\text{y-1}}{\text{1}}\text{=}\frac{\text{z-6}}{\text{-5}}$and $\frac{\text{x-1}}{\text{-3}}\text{=}\frac{\text{y-2}}{\text{2k}}\text{=}\frac{\text{z-3}}{\text{2}}$ 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 $\overrightarrow{r}=6\widehat{i}+2\widehat{j}+2\widehat{k}+\lambda \left( 2\widehat{i}-2\widehat{j}+2\widehat{k} \right)$

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

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

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

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

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

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

Therefore, the vector equation will be ${\vec{r}=}\left( {\hat{i}+2\hat{j}-4\hat{k}} \right)\text{+ }\!\!\lambda\!\!\text{ }\left( {2\hat{i}+3\hat{j}+6\hat{k}} \right)$.

Chapter 11 – Three-Dimensional Geometry  

11.1 Introduction

The CH 11 Maths Class 12 will take you on a revision tour about Analytical Geometry in two dimensions and the three-dimensional geometry and uses of Cartesian methods. This chapter will talk about and also revisit basic concepts of vectors and how to use vector algebra to three-dimensional geometry. You will also study the direction, e-direction cosines and direction ratios of a line joining two points. Also, the chapter will help you learn the equations of lines and planes in space under different conditions, the angle between two lines, two planes, a line and a plane, the shortest distance between two skew lines and distance of a point from a plane.

11.2 Direction Cosines and Direction Ratios of a Line

In Chapter 11 Class 12 Maths, you will learn and observe direction cosines and direction ratios of a line by an example in the beginning. Moving on, you will also understand how the line in space does not pass through the origin, and then, to find its direction cosines, a line is drawn through the origin and parallel to the given line.

Further, into the 3D Geometry Class 12 NCERT Solutions, you will study the relationship between the direction cosines of a line which will require you to observe the example provided in this section for better understanding. You will also study the direction cosines of a line passing through two points. This too will require you to observe the example provided to understand the concept. This section will be full of relatable examples of different types to help you follow all the ideas.

11.3 Equation of a Line Space

In the 3D Geometry Class 12 Solutions, you will study vector and cartesian equations of a line in space. You will explore how a line is uniquely determined if it passes through a given point and has given direction or it passes through two given points. Further, in this section, you will learn about the equation of a line through a given point and parallel to a given vector. You will study how to derive a cartesian form from vector form. You will be given some examples to understand the concept in detail. Then you will learn about the equation of lines, passing through two given points. You will learn how to derive cartesian form from vector form under this study. You will be given several more examples to understand this concept. 

11.4 Angle between Two Lines

In this NCERT Solutions Class 12 Maths Chapter 11, you will be given an example even before you start on with the concept of the angle between two lines in three-dimensional geometry. This section does have a lot of theoretical concepts in it, but you will have to look deeply into the examples provided in this section to grasp these concepts. 

11.5 Shortest Distance between Two Lines

In this 3 Dimensional Geometry Class 12, you will study and understand how if two lines in space are parallel, the shortest distance between them will be the perpendicular distance, that is, the length of the perpendicular drawn from a point on one line onto the other line. Further, you will also study how and why the shortest distance between two lines, which means we mean the join of a point in one line with one point on the other line so that the length of the segment obtained is the smallest.

You will also study how for skew lines, the line of the shortest distance will be perpendicular to both the lines. You will learn more about the distance between two skew lines and the distance between parallel lines. You will be given several examples related to these concepts, and these examples will help you understand.

11.6 Plane

In this Three Dimensional Geometry Class 12 NCERT PDF download, you will study planes and how a plane is uniquely determined if at all by the normal to the plane and its distance from the origin. You will understand the equation of a plane in a standard form, or it passes through a point and is perpendicular to a given direction, or even it passes through three given non-collinear points. You will further study the equation of a plane in normal form and then you will study the examples provided to understand the concept appropriately. Later, in this section, you will study the equation of the plane perpendicular to a given vector and passing through a given point. You will also study the equation of a plane passing through three non-collinear points. Again, you will go through the examples to understand the concept in and out and how to apply them. You will also learn about the intercept form of the equation of a plane, and you will be required to go through the examples before you start solving the exercise. You will also learn about a plane passing through the intersection of two given planes and some more examples to understand this.

11.7 Coplanarity of Two Lines  

In NCERT Solutions for Class 12 Maths Chapter 11 PDF Download, you will study the coplanarity of two lines in three-dimensional geometry. Again, this section does have a lot of theoretical concepts in it, but you will have to look deeply into the examples provided in this section to grasp these concepts. This is more of a concept which will help you understand the next section and its concepts.

11.8 Angle between Two Planes

In this Class 12 Maths Chapter 11 Solutions, at first, you will study a new definition which is, the angle between two planes is defined as the angle between their normal—followed by several examples for you to understand and hold on to the definition. This section helps you in better understanding of the questions so that it becomes easier for you to solve them.

11.9 Distance of a Point from a Plane

In this Class 12th Maths Chapter 11, like the previous section, you will study the distance of a point from a plane both from a vector form as well as cartesian form. You will be given examples for both of these forms. You will be given many examples for you to understand this concept. Without these concepts, it will get tricky for you to grasp these concepts.

11.10 Angle between a Line and a Plane

In this section, you will learn another definition which is, the angle between a line and a plane is the complement of the angle between the line and normal to the plane. You will be given a figure to understand this definition followed by few examples on how to apply this definition in three-dimensional geometry. You will study this definition and concept from vector form followed by another example to grasp this concept. 

Summary of Three Dimensional Geometry

• Direction cosines of a line are the cosines of the angles made by the lime with the positive directions of the coordinate axes

• If $l, m n$ are the direction cosines of a line, then $l^2+m^2+n^2-1$.

• Direction cosines of a line joining two points $P\left(x_1, y_1, z_1\right)$ and $Q\left(x_2, y_2, z_2\right)$ are $\frac{x_1-x_1}{P Q}, \frac{y_1-y_1}{P Q}, \frac{z_2-z_1}{P Q}$.

where $P Q=\sqrt{\left(x_1-x_1\right)^2+\left(y_2-y_1\right)^2+\left(z_2-z_1\right)^2}$

• Direction ratios of a lime are the numbers which are proportional to the direction cosines of a line

• If $l, m, n$ are the direction cosines and $a b, c$ are the direction ratios of a line then $l=\frac{a}{\sqrt{a^2+b^2+c^2}} ; m=\frac{b}{\sqrt{a}+b^2+c^2} ; n=\frac{c}{\sqrt{a^2+b^2+c^2}}$

• Skew lines are lines in space which are neither parallel nor intersecting They lie in different planes

• Angle between skew lines is the angle between two intersecting lines drawn from any point (preferably through the origin) parallel to each of the skew lines $\cos \theta-\left|l_1+m_1 m_2+n_1 n_1\right|$

• Vector equation of a line that passes through the given point whose position vector is $\dot{a}$ and parallel to a given vector $b$ is $f=a+2 b$.

• Equation of a line through a point $\left(x_1, y_1, z_1\right)$ and having disction cosines $l, m n$ is $\frac{x-x_1}{l}=\frac{y-y_1}{m}=\frac{z-z_1}{n}$.

• The vector equation of a line which passes through two points whose position vectors are $\vec{a}$ and $b$ is $f=d+z(b-d)$.

• Cartesian equation of a line that passes through two points $\left(x_1, y_1, z_1\right)$ and $\left(x_2, y_2, z_1\right)$ is $\frac{x-x_1}{x_1-x_1}=\frac{y-y_1}{y_2-y_1}=\frac{z-z_1}{z_2-z_1}$.

- If $\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1}=\frac{z-z_1}{n_1}$ and $\frac{x-x_1}{l_1}=\frac{y-y_1}{m_2}=\frac{z-z_1}{n_1}$ are the equations of two lines, then the acute angle between the two limes is given by $\cos \theta-\left|k_1+m_1 m_2+n_1 n_2\right|$

• Shortest distance between two skew lines is the line segment perpendicular to both the lines

• Shortest distance between $f=a_1+2 b_1$ and $f=b_1+\mu_2$ is $\left|\frac{\left(b_1 \times b_1\right)-\left(a_2-a_1\right)}{\left|b_1 \times b_2\right|}\right|$.

• Shortest distance between the lines: $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_1}{a_1}=\frac{y-y_2}{b_1}=\frac{z-z_1}{c_2}$ is

$\frac{\left|\begin{array}{ccc}c_1-x_1 & y_2-y_1 & z_2-z_1 \\a_1 & b_1 & c_1 \\a_1 & b_1& c_2\end{array}\right|}{\sqrt{\left(b c_2-b_2 c_1\right)^2+\left(c a_2-c_2 a_1\right)^2+\left(a_1 b_2-a_2 b_1\right)^2}}$

• Distance between parallel lines $t=\bar{a}+2 b$ and $t=a_2+\mu b$ is $\left|\frac{\mid b \times\left(d \vec{a}-a_1\right)}{|b|}\right|$.

• In the vector form, the equation of a plane which is at a distance $d$ from the origin, and $d$ is the unit vector normal to the plane through the origin is $\gamma \cdot k=d$.

• Equation of a plane which is at a distance of $d$ from the origin and the direction cosines of the normal to the plane as $l, m, n$ is $i x+m y+n z=d$.

• The equation of a plane through a point whose position vector is a and perpendicular to the vector $\hat{N}$ is $(t-\bar{d}) \hat{N}-0$.

• Equation of a plane perpendicular to a given line with direction ratios $A, B, C$ and passing through a given point $\left(x_1, y_1, z_1\right)$ is $A\left(x-x_1\right)+B\left(y-y_1\right)+C\left(z-z_1\right)=0$

• Equation of a plane passing through three noncollinear points $\left(x_1, y_1, z_1\right) \quad\left(x_2, y_1, z_2\right)$ and $\left(x_2, y_3, z_1\right)$ is $\left|\begin{array}{ccc}x-x_1 & y-y_1 & z-z_1 \\ x_2-x_1 & y_2-y_1 & z_2-z_1 \\ x_1-x_1 & y_2-y_1 & z_2-z_1\end{array}\right|=0$.

• Vectors equation of a plane that contains three noncollinear points having position vectors $d, b$ and $\dot{b}$ is $\left(f^{\prime}-a\right) \cdot[(b-d) \times(c-d)]=0$

• Equation of a plane that cuts the coordinates axes at $(a, 0,0),(0, b, 0)$ and $(0,0, c)$ is $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$.

• Vector equation of a plane that passes through the intersection of planes $r-n_1-d_1$ and $n_2=d_2$ is $r-\left(n_1+2 n_2\right)=d+2 d_1$, where 2 is any nonzero constant

• Cartesian equation of a plane that passes through the intersection of two given planes $\left(A x+B_1 y+C_1 z-d_1\right)$

$+2\left(A_2 x+B_2 y+C_2 z-d_2\right)=0$

• Two line $t=a_1+2 b_1$ and $t=b_1+\mu b_1$ are coplanar if $\left(a_1-a_1\right)-\left(b_1 \times b_1\right)=0$.

• In the cartesian form two lines $-\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_1}{a_2}=\frac{y-y_2}{b_1}=\frac{z-z_2}{c_2}$ are coplanar if

$\left|\begin{array}{ccc}x_1-x_1 & y_2-y_1 & z_2-z_1 \\a_1 & b_1 & c_1 \\a_1 & b_1 & c_2\end{array}\right|=0.$

• In the vector form, if $\theta$ is the angle between the two planes, $t-n_1-d_1$ and $t-n_2-d_1$, then $\theta-\cos ^{-1} \frac{\left|n_1-n_2\right|}{\left|x_1\right|\left|n_2\right|}$

• The angle $\phi$ between the line $f-d+2 b$ and the plane $f \cdot b=d$ is $\sin \phi-\left|\frac{b-h}{|b||\hat{a}|}\right|$.

• The angle $\theta$ between the planes $A x+B_1 y+C_1 z+D=0$ and $A_2 x+B_2 y+C_2 z+D_2=0$ is given by $\cos \theta=\left|\frac{A A_1+B B_1+C C_2}{\mid C_1 A_1^2+B_1^2+C_1^2 \cdot A_1^2+B_2^2+C_2^2}\right|$

• The distance of a point whose position vector is a from the plane $\gamma-A-d$ is $|a-a-\hat{y}|$ The distance from a point $\left(x_1, y_1, z\right)$ to the plane $A x+B y+C z+D-0$ is $\left|\frac{A x_1+B y_1+C z+D}{\sqrt{-A}+B^2+C^2}\right|$

Overview of Deleted Syllabus for CBSE Class 12 Maths-Three Dimensional Geometry

Class 12 Maths Chapter 11 : Exercises Breakdown

Conclusion

In NCERT Solutions for chapter 11 maths class 12  by Vedantu, students explore the fundamental concepts of spatial geometry. The importance lies in understanding Cartesian coordinates, equations of lines and planes, and the angle between lines and planes. Focus on mastering the distance formula, direction cosines, and the cross product of vectors for accurate problem-solving. Previous year question papers typically include 8-10 questions, emphasizing the application of these concepts in real-world scenarios. By grasping these fundamentals, students lay a strong foundation for advanced topics in mathematics and its practical applications in various fields

Other Study Material for CBSE Class 12  Maths  Chapter 11-Three Dimensional Geometry

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.