Three Dimensional Geometry

Any plane parallel to the plane,  $\overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)=2$ , is of the form  $\overrightarrow{r}.\left(\widehat{i}+\widehat{j}+\widehat{k}\right)=\lambda .........(1)$           

The plane passes through the point (a, b, c). Therefore, the position vector  $\overrightarrow{r}$  of this point is  $\overrightarrow{r}=a\widehat{i}+b\widehat{j}+c\widehat{k}$

Therefore, equation (1) becomes

$\begin{array}{l}\left(a\widehat{i}+b\widehat{j}+c\widehat{k}\right).\left(\widehat{i}+\widehat{j}+\widehat{k}\right)=\lambda \\ \Rightarrow a+b+c=\lambda \end{array}$

Substituting  $\lambda =a+b+c$ in equation (1), we obtain

$\overrightarrow{r}=\left(\widehat{i}+\widehat{j}+\widehat{k}\right)=a+b+c.........(2)$

This is the vector equation of the required plane.

Substituting  $\overrightarrow{r}=x\widehat{i}+y\widehat{j}+z\widehat{k}$  in equation (2), we obtain

$\begin{array}{l}\left(x\widehat{i}+y\widehat{j}+z\widehat{k}\right).\left(\widehat{i}+\widehat{j}+\widehat{k}\right)=a+b+c\\ \Rightarrow x+y+z=a+b+c\end{array}$

The position vector of the point $\left(1,2,3\right)$ is   $\overrightarrow{r}=\widehat{i}+2\widehat{j}+3\widehat{k}$

The direction ratios of the normal to the plane,   $\overrightarrow{r}=\left(\widehat{i}+2\widehat{j}-5\widehat{k}\right)+9=0$ , are $1,2,and-5$ and the normal vector is  $\overrightarrow{N}=\left(\widehat{i}+2\widehat{j}-5\widehat{k}\right)$

The equation of a line passing through a point and perpendicular to the given plane is given by,

$\overrightarrow{l}=\overrightarrow{r}+\lambda \overrightarrow{N},\lambda \in R\Rightarrow \overrightarrow{l}=\left(\widehat{i}+2\widehat{j}+3\widehat{k}\right)+\lambda \left(\widehat{i}+2\widehat{j}-5\widehat{k}\right)$

The direction of ratios of the lines,   $\frac{x-3}{-3}=\frac{y-2}{2k}=\frac{z-3}{2}\&\frac{x-1}{3k}=\frac{y-1}{1}=\frac{z-6}{-5}$ , are $-3,2k,2and3k,1,-5$ respectively.

It is known that two lines with direction ratios,   $a_1, b_1, c_1 and a_2, b_2,c_2$ , are perpendicular, if  $a_1{a}_2 + b_1{b}_2 + c_1{c}_2 =0$

$\begin{array}{l}\therefore -3(3k)+2k*1+2(-5)=0\\ \Rightarrow -9k+2k-10=0\\ \Rightarrow 7k=-10\\ \Rightarrow k=\frac{-10}{7}\end{array}$

Therefore, for k= -10/7, the given lines are perpendicular to each other.

The coordinates of $A,B,C,andDare\left(1,2,3\right),\left(4,5,7\right),\left(­-4,3,-6\right),$ and $\left(2,9,2\right)$ respectively.

The direction ratios of $ABare\left(4-1\right)=3,\left(5-2\right)=3,and\left(7-3\right)=4$

The direction ratios of $CDare\left(2-\left(-4\right)\right)=6,\left(9-3\right)=6,and\left(2-\left(-6\right)\right)=8$

It can be seen that,  $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=\frac{1}{2}$

Therefore, AB is parallel to CD.

Thus, the angle between $ABandCDiseither0°or18$$0$$°.$

The line parallel to x-axis and passing through the origin is x-axis itself.

Let A be a point on x-axis. Therefore, the coordinates of A are given by $\left(a,0,0\right),where a ?R.$ Direction ratios of $OAare\left(a ?0\right)= a,0,0$

The equation of OA is given by,

$\begin{array}{l}\frac{x?0}{a}=\frac{y?0}{0}=\frac{z?0}{0}\\ ?\frac{x}{1}=\frac{y}{0}=\frac{z}{0}=a\end{array}$

Thus, the equation of line parallel to x-axis and passing through origin is

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

It is given that  $l_1, m_1, n_1 and l_2, m_2, n_2$ are the direction cosines of two mutually perpendicular lines. Therefore,

$\begin{array}{l}{l}_1{l}_2+m_1{m}_2+n_1 n_2=0..........(1)\\ {l_1}^2+{m_1}^2+n_1{ }^2=1..........(2)\\ {l_2}^2+{m_2}^2+n_2{ }^2=1..........(3)\end{array}$

Let  $l, m, n$ be the direction cosines of the line which is perpendicular to the line with direction cosines  $l_1, m_1, n_1 and l_2, m_2, n_2.$

$\begin{array}{l}\therefore l{l}_1+ m{m}_1+ n{n}_1 =0\\ l l_2+m m_2+n n_2=0\\ \therefore \frac{l}{m_1{n}_2-m_2{n}_1}=\frac{m}{n_1{l}_2-n_2{l}_1}=\frac{n}{l_1{m}_2-l_2{m}_1}\\ \Rightarrow \frac{l^2}{{\left(m_1{n}_2-m_2{n}_1\right)}^2}=\frac{m^2}{{\left(n_1{l}_2-n_2{l}_1\right)}^2}=\frac{n^2}{{\left(l_1{m}_2-l_2{m}_1\right)}^2}\\ \Rightarrow \frac{l^2}{{\left(m_1{n}_2-m_2{n}_1\right)}^2}=\frac{m^2}{{\left(n_1{l}_2-n_2{l}_1\right)}^2}=\frac{n^2}{{\left(l_1{m}_2-l_2{m}_2\right)}^2}\\ =\frac{l^2+m^2+n^2}{{\left(m_1{n}_2-m_2{n}_1\right)}^2+{\left(n_1{l}_2-n_2{l}_1\right)}^2+{\left(l_1{m}_2-l_2{m}_2\right)}^2}..........(4)\end{array}$

$l, m, n$ are the direction cosines of the line.

$\therefore l^{2 }+ m^2 + n^2 =1\dots \left(5\right)$

It is known that,

$\begin{array}{l}\left({l_1}^2+{m_1}^2+n_1{ }^2\right)\left({l_2}^2+{m_2}^2+n_2{ }^2\right)-{\left(l_1{l}_2+m_1{m}_2+n_1 n_2\right)}^2\\ ={\left(m_1{n}_2-m_2{n}_1\right)}^2+{\left(n_1{l}_2-n_2{l}_1\right)}^2+{\left(l_1{m}_2-l_2{m}_1\right)}^2\\ From,(1),(2)\&(3),we.obtain\\ \Rightarrow 1.1-0={\left(m_1{n}_2-m_2{n}_1\right)}^2+{\left(n_1{l}_2-n_2{l}_1\right)}^2+{\left(l_1{m}_2-l_2{m}_1\right)}^2\\ \therefore {\left(m_1{n}_2-m_2{n}_1\right)}^2+{\left(n_1{l}_2-n_2{l}_1\right)}^2+{\left(l_1{m}_2-l_2{m}_1\right)}^2=1..........(6)\end{array}$

Substituting the values from equations (5) and (6) in equation (4), we obtain

$\frac{l^2}{{\left(m_1{n}_2-m_2{n}_1\right)}^2}=\frac{m^2}{{\left(n_1{l}_2-n_2{l}_1\right)}^2}=\frac{n^2}{{\left(l_1{m}_2-l_2{m}_1\right)}^2}=1$

Thus, the direction cosines of the required line are  $m_1{n}_2-m_2{n}_1,n_1{l}_2-n_2{l}_1,l_1{m}_2-l_2{m}_1$

Let OA be the line joining the origin,  $O\left(0,0,0\right),$ and the point,  $A\left(2,1,1\right).$

Also, let BC be the line joining the points,  $B\left(3,5,-1\right)andC\left(4,3,-1\right).$

The direction ratios of $OAare2,1,and1andofBCare\left(4-3\right)=1,\left(3-5\right)=-2,and\left(-1+1\right)=0$

OA is perpendicular to $BC,if a_1{a}_2 + b_1{b}_2 + c_1{c}_2 =0$

$\therefore a_1{a}_2 + b_1{b}_2 + c_1{c}_2 =2*1+1\left(-2\right)+1*0=2-2=0$

Thus, OA is perpendicular to BC.