Chapter 11: Three dimensional Geometry

Ans: Two lines with direction cosines $\left( {{l}_{1}},{{m}_{1}},{{n}_{1}} \right)$ and $\left( {{l}_{2}},{{m}_{2}},{{n}_{2}} \right)$ are perpendicular to each other if

${{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}=0$

(i) For the lines with direction cosines
$\dfrac{12}{13},\dfrac{-3}{13},\dfrac{-4}{13}$ and $\dfrac{4}{13},\dfrac{12}{13},\dfrac{3}{13}$, we obtain

$ {{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}=\dfrac{12}{13}\times \dfrac{4}{13}+\dfrac{-3}{13}\times \dfrac{12}{13}+\dfrac{-4}{13}\times \dfrac{3}{13} \\ $

$ =\dfrac{48}{169}-\dfrac{36}{169}-\dfrac{12}{169} \\ $

$ =0 \\ $

Therefore, the lines are perpendicular.

(ii) For the lines with direction cosines
$\dfrac{4}{13},\dfrac{12}{13},\dfrac{3}{13}$ and $\dfrac{3}{13},\dfrac{-4}{13},\dfrac{12}{13}$, we obtain

$ {{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}=\dfrac{4}{13}\times \dfrac{-3}{13}+\dfrac{12}{13}\times \dfrac{-4}{13}+\dfrac{3}{13}\times \dfrac{12}{13} \\ $

$ =\dfrac{12}{169}-\dfrac{48}{169}+\dfrac{36}{169} \\ $

$ =0 \\ $

Therefore, the lines are perpendicular.

(iii) For the lines with direction cosines
$\dfrac{3}{13},\dfrac{-4}{13},\dfrac{12}{13}$ and $\dfrac{12}{13},\dfrac{-3}{13},\dfrac{-4}{13}$, we obtain

$ {{l}_{1}}{{l}_{2}}+{{m}_{1}}{{m}_{2}}+{{n}_{1}}{{n}_{2}}=\dfrac{3}{13}\times \dfrac{12}{13}+\dfrac{-4}{13}\times \dfrac{-3}{13}+\dfrac{12}{13}\times \dfrac{-4}{13} \\ $

$ =\dfrac{36}{169}+\dfrac{12}{169}-\dfrac{48}{169} \\ $

$ =0 \\ $

Therefore, the lines are perpendicular.

Ans: Let AB be the line joining the points, \[\left( 1,-1,2 \right)\] and \[\left( 3,4,-2 \right)\], 

and CD be the line through the points \[\left( 0,3,2 \right)\] and \[\left( 3,5,6 \right)\]. 

The direction ratios ${{a}_{1}},{{b}_{1}},{{c}_{1}}$ of AB are 

\[\left( 3-1 \right),\left( 4\left( -1 \right) \right),\left( -2-2 \right)\] i.e.,

$2,5,-4$ 

The direction ratios ${{a}_{2}},{{b}_{2}},{{c}_{2}}$ of CD are 

\[\left( 3-0 \right),\left( 5-3 \right),\left( 6-2 \right)\] i.e.,

$3,2,4$ 

AB and CD will be perpendicular to each other if 

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

$ \left( 2\times 3 \right)+\left( 5\times 2 \right)+\left( -4\times 4 \right) \\ $

$ =6+10-16 \\ $

$ =0 \\ $

Therefore, AB and CD are perpendicular to each other.

Ans: Let AB be the line through the points \[\left( 4,7,8 \right)\]and \[\left( 2,3,4 \right)\],

CD be the line through the points, \[\left( -1,-2,1 \right)\]and \[\left( 1,2,5 \right)\]. 

The directions ratios ${{a}_{1}},{{b}_{1}},{{c}_{1}}$, of AB are 

$\left( 2-4 \right),\left( 3-7 \right),\left( 4-8 \right)$ i.e., 

$-2,-4,-4$

The direction ratios ${{a}_{2}},{{b}_{2}},{{c}_{2}}$ of CD are 

$\left( 1-\left( -1 \right) \right),\left( 2-\left( -2 \right) \right),\left( 5-1 \right)$ i.e., 

\[2,4,4\]

AB will be parallel to CD, if 

$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{{{c}_{1}}}{{{c}_{2}}}$

$\dfrac{{{a}_{1}}}{{{a}_{2}}}=\dfrac{-2}{2}=-1$…(1)

$\dfrac{{{b}_{1}}}{{{b}_{2}}}=\dfrac{-4}{4}=-1$…(2)

$\dfrac{{{c}_{1}}}{{{c}_{2}}}=\dfrac{-4}{4}=-1$…(3)

From Equations (1), (2) and (3), we get

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

Thus, AB is parallel to CD.

Ans: It is given that the line passes through the point A \[\left( 1,2,3 \right)\]. Therefore, the position vector through A is \[\vec{a}=\hat{i}+2\hat{j}+3\hat{k}\]

Also, \[\vec{b}=3\hat{i}+2\hat{j}-2\hat{k}\]

It is known that the line which passes through point A and parallel to $\vec{b}$ is given by $\vec{r}=\vec{a}+\lambda \vec{b}$ where is $\vec{r}$ a constant.

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

This is the required equation of the line.

Ans: It is given that the line passes through the point with positive vector 

$\vec{a}=2\hat{i}-\hat{j}+4\hat{k}$

\[\vec{b}=\hat{i}+2\hat{j}-\hat{k}\]

It is known that the line which passes through point A and parallel to $\vec{b}$ is given by $\vec{r}=\vec{a}+\lambda \vec{b}$ 

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

This is the required equation in the vector form

$\vec{r}=\left( 2+\lambda  \right)\hat{i}+\left( 2\lambda -1 \right)\hat{j}+\left( 4-\lambda  \right)\hat{k}$

Eliminating $\lambda $, we obtain the Cartesian form of equation as

$\dfrac{x-2}{1}=\dfrac{y+1}{2}=\dfrac{z-4}{-1}$

This is the required equation of line in the cartesian form.

Ans: It is given that the line passes through the point \[(-2,4,-5)\] and is parallel to $\dfrac{x+3}{3}=\dfrac{y-4}{5}=\dfrac{z+8}{6}$

The direction ratios of the line, $\dfrac{x+3}{3}=\dfrac{y-4}{5}=\dfrac{z+8}{6}$ are $3,5,6$.

The required line is parallel to $\dfrac{x+3}{3}=\dfrac{y-4}{5}=\dfrac{z+8}{6}$

Therefore, its direction ratios are \[3k,5k,6k\], when \[k\ne 0\]

It is known that the equation of the line through the point $\left( {{x}_{1}},{{y}_{1}},{{z}_{1}} \right)$ and with direction ratios $\left( a,b,c \right)$, is given by

$\dfrac{x-{{x}_{1}}}{a}=\dfrac{y-{{y}_{1}}}{b}=\dfrac{z-{{z}_{1}}}{c}$

Therefore, the equation of the required line is

$ \dfrac{x+2}{3k}=\dfrac{y-4}{5k}=\dfrac{z+5}{6k} \\ $ 

$ \Rightarrow \dfrac{x+2}{3}=\dfrac{y-4}{5}=\dfrac{z+5}{6}=k \\ $

Ans: The Cartesian equation of the line is \[\dfrac{x-5}{3}=\dfrac{y+4}{7}=\dfrac{z-6}{2}\].

The given line passes through the point \[\left( 5,-4,6 \right)\]. The positive vector of this point is $\vec{a}=5\hat{i}-4\hat{j}+6\hat{k}$

Also, the direction ratios of the given line are \[3,7,2\].

This means that the line is in the direction of the vector $\vec{b}=3\hat{i}+7\hat{j}+2\hat{k}$

It is known that the line which passes through point A and parallel to $\vec{b}$ is given by $\vec{r}=\vec{a}+\lambda \vec{b}$

$\vec{r}=\left( 5\hat{i}-4\hat{j}+6\hat{k} \right)+\lambda \left( 3\hat{i}+7\hat{j}+2\hat{k} \right)$ 

This is the required equation of the given line in vector form.

Ans: (i) Let Q be the angle between the given lines. 

The angle between the given pairs of lines is given by, $\cos Q=\left. \left| \dfrac{{{{\vec{b}}}_{1}}\cdot {{{\vec{b}}}_{2}}}{|{{{\vec{b}}}_{1}}|\cdot |{{{\vec{b}}}_{2}}|} \right. \right|$

The given lines are parallel to the vectors, ${{\vec{b}}_{1}}=3\hat{i}-2\hat{j}+6\hat{k}$ and ${{\vec{b}}_{2}}=\hat{i}+2\hat{j}+2\hat{k}$, respectively.

$ |{{b}_{1}}|=\sqrt{{{3}^{2}}+{{2}^{2}}+{{6}^{2}}}=7 \\$ 

$ |{{b}_{2}}|=\sqrt{{{1}^{2}}+{{2}^{2}}+{{2}^{2}}}=3 \\ $

$ {{b}_{1}}\cdot {{b}_{2}}=\left( 3\hat{i}-2\hat{j}+6\hat{k} \right)\cdot \left( \hat{i}+2\hat{j}+2\hat{k} \right) \\ $

$ =3\times 1+2\times 2+6\times 2 \\ $

$ =3+4+12 \\ $

$ =19 \\ $

$\cos Q=\dfrac{19}{7\times 3} \\ $

$ \Rightarrow Q={{\cos }^{-1}}\left( \dfrac{19}{21} \right) \\$

(ii) Let Q be the angle between the given lines. 

The angle between the given pairs of lines is given by, $\cos Q=\left. \left| \dfrac{{{{\vec{b}}}_{1}}\cdot {{{\vec{b}}}_{2}}}{|{{{\vec{b}}}_{1}}|\cdot |{{{\vec{b}}}_{2}}|} \right. \right|$

The given lines are parallel to the vectors, ${{\vec{b}}_{1}}=\hat{i}-\hat{j}-2\hat{k}$ and ${{\vec{b}}_{2}}=3\hat{i}-5\hat{j}-4\hat{k}$, respectively.

$ |{{b}_{1}}|=\sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}+{{\left( -2 \right)}^{2}}}=\sqrt{6} \\ $

$ |{{b}_{2}}|=\sqrt{{{3}^{2}}+{{\left( -5 \right)}^{2}}+{{\left( -4 \right)}^{2}}}=\sqrt{50}=5\sqrt{2} \\ $

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

$ =1\times 3+\left( -1 \right)\times \left( -5 \right)+\left( -2 \right)\times \left( -4 \right) \\ $

$ =3+5+8 \\ $

$ =16 \\ $

$ \cos Q=\dfrac{16}{\sqrt{6}\times 5\sqrt{2}} \\ $

$ \Rightarrow Q={{\cos }^{-1}}\left( \dfrac{8}{5\sqrt{3}} \right) \\ $

Ans: (i) Let Q be the angle between the given lines. 

The angle between the given pairs of lines is given by, $\cos Q=\left. \left| \dfrac{{{{\vec{b}}}_{1}}\cdot {{{\vec{b}}}_{2}}}{|{{{\vec{b}}}_{1}}|\cdot |{{{\vec{b}}}_{2}}|} \right. \right|$

The given lines are parallel to the vectors, ${{\vec{b}}_{1}}=2\hat{i}+5\hat{j}-3\hat{k}$ and ${{\vec{b}}_{2}}=-\hat{i}+8\hat{j}+4\hat{k}$, respectively.

$ |{{b}_{1}}|=\sqrt{{{2}^{2}}+{{5}^{2}}+{{\left( -3 \right)}^{2}}}=\sqrt{38} \\ $

$ |{{b}_{2}}|=\sqrt{{{\left( -1 \right)}^{2}}+{{8}^{2}}+{{4}^{2}}}=9 \\ $

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

$ =2\times \left( -1 \right)+5\times 8+\left( -3 \right)\times 4 \\ $

$ =-2+40-12 \\ $

$ =26 \\ $

$ \cos Q=\dfrac{26}{9\sqrt{38}} \\ $

$ \Rightarrow Q={{\cos }^{-1}}\left( \dfrac{26}{9\sqrt{38}} \right) \\ $

(i) Let Q be the angle between the given lines. 

The angle between the given pairs of lines is given by, $\cos Q=\left. \left| \dfrac{{{{\vec{b}}}_{1}}\cdot {{{\vec{b}}}_{2}}}{|{{{\vec{b}}}_{1}}|\cdot |{{{\vec{b}}}_{2}}|} \right. \right|$

The given lines are parallel to the vectors, ${{\vec{b}}_{1}}=2\hat{i}+2\hat{j}+\hat{k}$ and ${{\vec{b}}_{2}}=8\hat{i}+\hat{j}+8\hat{k}$, respectively.

$ |{{b}_{1}}|=\sqrt{{{2}^{2}}+{{2}^{2}}+{{1}^{2}}}=3 \\ $

$ |{{b}_{2}}|=\sqrt{{{4}^{2}}+{{1}^{2}}+{{8}^{2}}}=9 \\  $

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

$ =2\times 4+2\times 1+1\times 8 \\ $

$ =8+2+8 \\ $

$ =18 \\ $

$ \cos Q=\dfrac{18}{3\times 9} \\ $

$ \Rightarrow Q={{\cos }^{-1}}\left( \dfrac{2}{3} \right) \\ $

Ans: The given equations can be written in the standard form as

\[\dfrac{x-1}{-3}=\dfrac{y-2}{\dfrac{2p}{7}}=\dfrac{z-3}{2}\] and

\[\dfrac{x-1}{-\dfrac{3p}{7}}=\dfrac{y-5}{1}=\dfrac{z-6}{-5}\]

The direction ratios of the lines are

$-3,\dfrac{2p}{7},2$ and

$-\dfrac{3p}{7},1,-5$

Two lines will be perpendicular to each other if 

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

$ \left( -3\times \dfrac{-3p}{7} \right)+\left( \dfrac{2p}{7}\times 1 \right)+\left( 2\times \left( -5 \right) \right)=0 \\ $

$ \Rightarrow \dfrac{9p}{7}+\dfrac{2p}{7}-10=0 \\ $

$ \Rightarrow 9p+2p-70=0 \\ $

$ \Rightarrow 11p-70=0 \\ $

$ \Rightarrow 11p=70 \\ $

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

Thus, the value of  $p=\dfrac{70}{11}$.