Three Dimensional Geometry

Given, lines

$ \begin{aligned} \mathbf{r} & =(-\hat{\mathbf{i}}+3 \hat{\mathbf{k}})+t(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \\ \text { and } \mathbf{r} & =(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+s(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \end{aligned} $

Shortest distance between lines is

$ \begin{aligned} & ((3+1) \hat{\mathbf{i}}+\hat{\mathbf{j}}-(\hat{\mathbf{k}}+3) \hat{\mathbf{k}}) \\ S D & =\frac{-[(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}}) \times(2 \hat{\mathbf{i}}-\mathbf{j}+2 \mathbf{k})]}{|(2 \hat{\mathbf{i}}+3 \hat{\mathbf{j}}+6 \hat{\mathbf{k}})| \times(2 \hat{\mathbf{i}}-\hat{\mathbf{j}}+2 \hat{\mathbf{k}}) \mid} \\ S D & =\frac{(4 \hat{\mathbf{i}}+\hat{\mathbf{j}}-4 \hat{\mathbf{k}}) \cdot(12 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}-8 \hat{\mathbf{k}})}{|12 \hat{\mathbf{i}}+8 \hat{\mathbf{j}}-8 \hat{\mathbf{k}}|} \\ S D & =\frac{48+8+32}{\sqrt{144+64+64}}=\frac{88}{4 \sqrt{17}}=\frac{22}{\sqrt{17}} \end{aligned} $

Direction ratio of diagonal $d_1$ i.e., direction of $A E$ is $a \hat{\mathbf{i}}-b \hat{\mathbf{j}}$

Direction ratio of diagonal $d_2$ i.e., direction of $A D a \hat{\mathbf{i}}-c \hat{\mathbf{k}}$

Angle between diagonal $d_1$ and diagonal $d_2$

$ \begin{aligned} \cos \theta & =\frac{(a \hat{\mathbf{i}}-b \hat{\mathbf{j}}) \cdot(a \hat{\mathbf{i}}-c \hat{\mathbf{k}})}{\sqrt{a^2+b^2} \sqrt{a^2+c^2}} \\ & =\frac{a^2}{\sqrt{a^2+b^2} \sqrt{a^2+c^2}} \\ \Rightarrow \quad \theta & =\cos ^{-1}\left(\frac{a^2}{\sqrt{a^2+b^2} \sqrt{a^2+c^2}}\right) \end{aligned} $

$ \begin{aligned} & \text {We have, } a+b+c=0 \\ & \text { and } 2 a b+2 a c-b c=0 \\ & \quad a=-(b+c) \\ & \therefore \quad-2(b+c) b+-2(b+c) c-b c=0 \\ & \Rightarrow \quad-2 b^2-2 b c-2 b c-2 c^2-b c=0 \\ & \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad 2 b^2+5 b c+2 c^2=0 \\ & \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad 2 b^2+4 b c+b c+2 c^2=0 \\ & \Rightarrow\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \quad(2 b+c)(b+2 c)=0 \\ & \Rightarrow \quad b=-2 c, b=-c / 2 \\ & \quad a=-(-2 c+c)=c \\ & \Rightarrow \quad a=-\left(\frac{-c}{2}+c\right)=\frac{-c}{2} \\ & \therefore a_1=1, b_1=-2, c_1=1, \\ & \quad a_2=\frac{-1}{2}, b_2=\frac{-1}{2}, c_2=1 \\ & \quad a_1=1, b_1=-2, c_1=1, \\ & a_2=1, b_2=1, c_2=-2 \\ & \cos \theta=\frac{(1)(1)+(-2)(1)+(1)(-2)}{\sqrt{1+4+1} \sqrt{1+1+4}} \\ & \cos \theta=\frac{-1}{2} \Rightarrow \theta=\frac{2 \pi}{3} \end{aligned} $

Let the equation of plane is, $\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1$

where $a, b, c$ are $x, y$ and $z$ intercepts ∴ Centroid of $\triangle A B C$ is $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)$.

Given, $\left(\frac{a}{3}, \frac{b}{3}, \frac{c}{3}\right)=(2,3,5)$

$ \therefore \quad a=6, b=9, c=15 $

Equation of plane is, $\frac{x}{6}+\frac{y}{9}+\frac{z}{15}=1$

$ 15 x+10 y+6 z=90 $

Equation of plane ( $\pi$ ) containing the points

$ \begin{aligned} & (0,-5,-1),(1,-2,5),(-3,5,0) \\ & \text { is }\left|\begin{array}{ccc} x-0 & y+5 & z+1 \\ 1-0 & -2+5 & 5+1 \\ -3-0 & 5+5 & 0+1 \end{array}\right|=0 \\ & \Rightarrow \quad\left|\begin{array}{ccc} x & y+5 & z+1 \\ 1 & 3 & 6 \\ -3 & 10 & 1 \end{array}\right|=0 \\ & \Rightarrow \quad 3 x+y-z+4=0 \\ & \Rightarrow \quad \mathbf{r}(3 \hat{\mathbf{i}}+\hat{\mathbf{j}}-\hat{\mathbf{k}})+4=0 \\ & \Rightarrow \quad \mathbf{r}(-3 \hat{\mathbf{i}}-\hat{\mathbf{j}}+\hat{\mathbf{k}})=4 \\ & \Rightarrow \mathbf{r}\left(\frac{-3}{\sqrt{11}}-\frac{\hat{\mathbf{j}}}{\sqrt{11}}+\frac{\hat{\mathbf{k}}}{\sqrt{11}}\right)=\frac{4}{\sqrt{11}} \end{aligned} $

∴ Unit normal vector to plane $\pi$ is

$ \hat{\mathbf{n}}=\frac{-3}{\sqrt{11}} \hat{\mathbf{i}}-\frac{\hat{\mathbf{j}}}{\sqrt{11}}+\frac{\hat{\mathbf{k}}}{\sqrt{11}} $

∴ length of the projection of the unit normal vector $\hat{\mathbf{n}}$ to plane $\pi$ on the line $L$ is

$ \left|\frac{\left(\frac{-3}{\sqrt{11}} \hat{\mathbf{i}} \frac{-\hat{\mathbf{j}}}{\sqrt{11}} \frac{+\hat{\mathbf{k}}}{\sqrt{11}}\right) \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}-\hat{\mathbf{k}})}{\sqrt{1+25+36}}\right| $

$ =\left|\frac{\frac{-3}{\sqrt{11}} \frac{-5}{\sqrt{11}} \frac{-6}{\sqrt{11}}}{\sqrt{62}}\right|=\left|\frac{-14}{\sqrt{11} \sqrt{62}}\right|=\frac{14}{\sqrt{682}} $

Equation of line passing through the points $(a, 2,-4)$ and $(5,3, b)$ is given by

$ \begin{aligned} \frac{x-a}{5-a} & =\frac{y-2}{3-2}=\frac{z+4}{b+4} \\ \Rightarrow \quad \frac{x-a}{5-a} & =\frac{y-2}{1}=\frac{z+4}{b+4} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\end{aligned} $

The equation of $Z X$-plane, $y=0$ Since, line (i) meets the $Z X$-plane.

$ \begin{array}{ll} \Rightarrow & \frac{x-a}{5-a}=\frac{0-2}{1}=\frac{z+4}{b+4} \\ \Rightarrow & \frac{x-a}{5-a}=-2 \text { and } \frac{z+4}{b+4}=-2 \\ \Rightarrow & x-a=-2(5-a) \\ \text { and } & z+4=-2(b+4) \end{array} $

$ \begin{aligned} &\begin{aligned} & \Rightarrow \quad x=a-10+2 a \\ & \text { and } z+4=-2 b-8 \\ & \Rightarrow \quad x=3 a-10 \text { and } z=-2 b-12 \\ & \text { and } y=0 \\ & \therefore \quad 3 a-10=-a+2 b \text { and } \\ & -2 b-12=a+b \\ & \Rightarrow \quad 4 a-2 b=10 \text { and } a+3 b=-12 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \Rightarrow \quad 2 a-b=5 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned}\\ &\text { On solving Eqs (i) and (ii), we get }\\ &\begin{aligned} a=\frac{3}{7} & \text { and } b=\frac{-29}{7} \\ \therefore \quad 14 a+7 b & =14\left(\frac{3}{7}\right)+7\left(\frac{-29}{7}\right) \\ & =6-29=-23 \end{aligned} \end{aligned} $

$\operatorname{Let} \mathbf{b}_1=\hat{\mathbf{i}}+2 \hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\mathbf{b}_2=4 \hat{\mathbf{i}}+5 \hat{\mathbf{j}}-3 \hat{\mathbf{k}}$

Normal vector to plane

$ \begin{aligned} & \mathbf{b}_1 \times \mathbf{b}_2=\left|\begin{array}{ccc} \hat{\mathbf{i}} & \hat{\mathbf{j}} & \hat{\mathbf{k}} \\ 1 & 2 & 1 \\ 4 & 5 & -3 \end{array}\right| \\ = & \hat{\mathbf{i}}(-6-5)-\hat{\mathbf{j}}(-3-4)+\hat{\mathbf{k}}(5-8) \\ = & -11 \hat{\mathbf{i}}+7 \hat{\mathbf{j}}-3 \hat{\mathbf{k}} \end{aligned} $

∴ Direction cosines of normal to the plane

$ \begin{aligned} & =\frac{-11}{\sqrt{(-11)^2+7^2+(-3)^2}}, \\ & \frac{7}{\sqrt{(-11)^2+7^2+(-3)^2}}, \\ & \frac{-3}{\sqrt{(-11)^2+7^2+(-3)^2}} \\ & =\frac{-11}{\sqrt{179}}, \frac{7}{\sqrt{179}}, \frac{-3}{\sqrt{179}} \end{aligned} $

We know that, foot of perpendicular from a point $\left(x_1, y_1, z_1\right)$ on the plane $a x+b y+c z+d=0$ is $(x, y, z)$ where

$ \begin{aligned} \frac{x-x_1}{a} & =\frac{y-y_1}{b}=\frac{z-z_1}{c} \\ & =\frac{-\left(a x_1+b y_1+c z_1+d\right)}{a^2+b^2+c^2} \end{aligned} $

Since, the foot of the perpendicular drawn from the point $(1,1,1)$ to the plane $\pi_1$ is $(1,3,5)$

$ \begin{aligned} & \therefore \frac{1-1}{a}=\frac{3-1}{b}=\frac{5-1}{c}=\frac{-(a+b+c+d)}{a^2+b^2+c^2} \\ & \Rightarrow \quad \frac{0}{a}=\frac{2}{b}=\frac{4}{c}=\frac{-(a+b+c+d)}{a^2+b^2+c^2} \\ & \Rightarrow \quad a=0, b=2 k, c=4 k . \end{aligned} $

Equation of plane $\pi_1$ through point $(1,3,5)$ and having directions of normal to plane are $a, b, c$ is

$ \begin{array}{rlr} & 0(x-1)+2 k(y-3)+4 k(z-5) =0 \\ \Rightarrow & (y-3)+2(z-5) =0 \\ \Rightarrow & y+2 z-13=0 & \ldots(\mathrm{i}) \tag{i} \end{array} $

And equation of plane $\pi_2$ is

$ \begin{aligned} & \quad\left|\begin{array}{ccc} x-2 & y-2 & z+1 \\ 3-2 & 4-2 & 2+1 \\ 3-2 & 3-2 & 0+1 \end{array}\right|=0 \\ & \Rightarrow \quad\left|\begin{array}{ccc} x-2 & y-2 & z+1 \\ 1 & 2 & 3 \\ 1 & 1 & 1 \end{array}\right|=0 \\ & \Rightarrow(x-2)(2-3)-(y-2)(1-3) \\ & \quad+(z+1)(1-2)=0 \\ & \Rightarrow(x-2)+(y-2)(2)+(z+1)=0 \\ & \Rightarrow x-2 y+z-3=0 \end{aligned} $

∴ Angle between plane $\pi_1$ and $\pi_2$

$ \begin{aligned} & \theta=\cos ^{-1}\left(\left|\frac{0 \times 0+(1) \times(-2)+(2) \times(1)}{\sqrt{0+1+4} \sqrt{1+4+1}}\right|\right) \\ & \Rightarrow \quad \theta=\cos ^{-1}(0)=\frac{\pi}{2} \end{aligned} $

Given,

Position vector of a point $P$ is

$ (2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) $

Equation of plane,

$ (\mathbf{r}-2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \cdot(\mathbf{a} \times \mathbf{b})=0 $

∴ Normal of plane $=\mathbf{a} \times \mathbf{b}$

Equation of line through

$\mathbf{P}(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})$ and normal to $\mathbf{b}$ and lying on the plane is

$ \begin{aligned} & \mathbf{r}=(2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}})+\lambda((\mathbf{a} \times \mathbf{b}) \times \mathbf{b}) \\ & \mathbf{r}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda((\mathbf{b} \cdot \mathbf{a}) \mathbf{b}-(\mathbf{b} . \mathbf{b}) \mathbf{a}) \\ & \mathbf{r}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}+ \\ & \lambda[(-5)(\hat{\mathbf{i}}+\hat{\mathbf{j}}+2 \hat{\mathbf{k}})-6(-\hat{\mathbf{i}}-2 \hat{\mathbf{k}})] \\ & \mathbf{r}=2 \hat{\mathbf{i}}+\hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda(-\hat{\mathbf{i}}+5 \hat{\mathbf{j}}-2 \hat{\mathbf{k}}) \end{aligned} $

We have,

$ \begin{aligned} & L: \mathbf{r}=2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}+\lambda(\hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \\ & P: \mathbf{r} \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}})=5 \end{aligned} $

$ \begin{aligned}Now, (\hat{\mathbf{i}}-\hat{\mathbf{j}}+4 \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}}) =1-5+4=0 \end{aligned} $

∴ Line is parallel to plane.

∴ Shortest distance between line and plane $=$ Distance of point $(2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}})$ from planer $\cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}})-5=0$

$ \begin{aligned} & =\left|\frac{(2 \hat{\mathbf{i}}-2 \hat{\mathbf{j}}+3 \hat{\mathbf{k}}) \cdot(\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}})-5}{|\hat{\mathbf{i}}+5 \hat{\mathbf{j}}+\hat{\mathbf{k}}|}\right| \\ & =\left|\frac{2-10+3-5}{\sqrt{1+25+1}}\right|=\frac{10}{\sqrt{27}}=\frac{10}{3 \sqrt{3}} \end{aligned} $

We have, $A(-1,0,7), B(3,2, t)$ and $C(5, k,-2)$ are collinear.

$ \begin{aligned} & \therefore \quad \frac{3+1}{5-3}=\frac{2-0}{k-2}=\frac{t-7}{-2-t} \\ & \Rightarrow \quad 2=\frac{2}{k-2}=\frac{t-7}{-2-t} \Rightarrow k=3, \\ & t=1 \\ & \therefore \quad B=(3,2,1), C=(5,3,-2) \text { and } \\ & P(1,1,4) \end{aligned} $

Let $P$ divides $B C$ in the ratio $\lambda: 1$

$ \begin{array}{ll} \therefore & \frac{5 \lambda+3}{\lambda+1}=1 \\ \Rightarrow & 5 \lambda+3=\lambda+1 \Rightarrow 4 \lambda=-2 \end{array} $

$ \begin{aligned} &\Rightarrow \quad \lambda=-1 / 2\\ &\text { ∴ } \text { Required ratio }=-1: 2 \end{aligned} $

We have,

$ \begin{aligned} & 3 l+m+5 n=0 \text { and } 6 m n-2 n l+5 l m=0 \\ & \Rightarrow \quad 6 n(-3 l-5 n)-2 n l+5 l(-3 l-5 n)=0 \\ & \Rightarrow \quad-18 n l-30 n^2-2 n l-15 l^2-25 n l=0 \\ & \Rightarrow \quad-15 l^2-30 n^2-45 n l=0 \\ & \Rightarrow \quad l^2+3 n l+2 n^2=0 \\ & \Rightarrow \quad(l+n)(l+2 n)=0 \\ & \Rightarrow \quad l=-n,-2 n \end{aligned} $

When $l=-n$

$ -3 n+m+5 n=0 \Rightarrow m=-2 n $

So, d'r are

$ -n,-2 n, n \text { or } 1,2,-1 $

When $l=-2 n$

$ -6 n+m+5 n=0 \Rightarrow m=n $

So, d'r are

$ \begin{aligned} & -2 n, n, n \\ & -2,1,1 \\ \therefore \cos \theta & =\frac{1 \times(-2)+2 \times 1+(-1) \times 1}{\sqrt{1+4+1} \sqrt{4+1+1}}=\frac{-1}{6} \\ \therefore|\cos \theta| & =\frac{1}{6} \end{aligned} $

Equation of plane $O A B$ is given by,

$ \begin{aligned} &\left|\begin{array}{ccc} x-0 & y-0 & z-0 \\ 1-0 & 2-0 & 1-0 \\ 2-0 & 1-0 & 3-0 \end{array}\right|=0 \\ & \Rightarrow\left|\begin{array}{lll} x & y & z \\ 1 & 2 & 1 \\ 2 & 1 & 3 \end{array}\right|=0 \\ & x(6-1)-y(3-2)+z(1-4)=0 \\ & 5 x-y-3 z=0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{aligned} $

Equation of plane $A B C$ is given by,

$ \begin{aligned} \left|\begin{array}{ccc} x-1 & y-2 & z-1 \\ 2-1 & 1-2 & 3-1 \\ -1-1 & 1-2 & 2-1 \end{array}\right| & =0 \\ \left|\begin{array}{ccc} x-1 & y-2 & z-1 \\ 1 & -1 & 2 \\ -2 & -1 & 1 \end{array}\right| & =0 \end{aligned} $

$ \begin{gathered} (x-1)(-1+2)-(y-2)(1+4) \\ +(z-1)(-1-2)=0 \\ x-1-5 y+10-3 z+3=0 \\ x-5 y-3 z+12=0 \quad \ldots(\mathrm{ii}) \end{gathered} $

Then angle between planes represented by Eqs. (i) and (ii) is given by,

$ \begin{aligned} \cos \theta & =\frac{(5)(1)+(-1)(-5)+(-3)(-3)}{\sqrt{25+1+9} \sqrt{1+25+9}} \\ & =\frac{19}{35} \end{aligned} $

We have P(3, 7, 1) and R(2, 5, 7)

DR's of PR = ($-$1, $-$2, 6)

Angle between line PQ and plane is

$\sin \theta = {{( - 1)(3) + ( - 2) \times 2 + 6 \times 11} \over {\sqrt {1 + 4 + 36} \sqrt {9 + 4 + 121} }}$

$\sin \theta = {{59} \over {\sqrt {41} \sqrt {134} }}$

$PM = \left| {{{3(3) + 2(7) + 11( \times 1) - 9} \over {\sqrt {9 + 4 + 121} }}} \right|$

$ = \left| {{{9 + 14 + 11 - 9} \over {\sqrt {134} }}} \right| = {{25} \over {\sqrt {134} }}$

$\sin \theta = {{PM} \over {PQ}}$

$ \Rightarrow PQ = {{PM} \over {\sin \theta }} = {{25} \over {\sqrt {134} }} \times {{\sqrt {41} \times \sqrt {134} } \over {59}} = {{25\sqrt {41} } \over {59}}$