Find the equations of the two lines through the origin which intersect the line $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}$ at angles of $\frac{\pi}{3}$ each.
Given equation of the line is $\frac{x-3}{2}=\frac{y-3}{1}=\frac{z}{1}=\lambda\quad\text{.... (i)}$
So, DR's of the line are 2, 1, 1 and DC's of the given line are $\frac{2}{\sqrt{6}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{16}}$.
Also, the required lines make angle $\frac{\pi}{3}$ with the given line.
$$\begin{array}{lrl} \text { From Eq. (i), } & x & =(2 \lambda+3), y=(\lambda+3) \text { and } z=\lambda \\ \because & \cos \theta & =\frac{a_1 a_2+b_1 b_2+c_1 c_2}{\sqrt{a_1^2+b_1^2+c_1^2} \sqrt{a_2^2+b_2^2+c_2^2}} \\ \therefore & \cos \frac{\pi}{3} & =\frac{(4 \lambda+6)+(\lambda+3)+(\lambda)}{\sqrt{6} \sqrt{(2 \lambda+3)^2+(\lambda+3)^2+\lambda^2}} \end{array}$$
$$\begin{array}{ll} \Rightarrow & \frac{1}{2}=\frac{6 \lambda+9}{\sqrt{6} \sqrt{\left(4 \lambda^2+9+12 \lambda+\lambda^2+9+6 \lambda+\lambda^2\right)}} \\ \Rightarrow & \frac{\sqrt{6}}{2}=\frac{6 \lambda+9}{\sqrt{6 \lambda^2+18 \lambda+18}} \end{array}$$
$$\begin{array}{lrl} \Rightarrow & 6 \sqrt{\left(\lambda^2+3 \lambda+3\right)} & =2(6 \lambda+9) \\ \Rightarrow & 36\left(\lambda^2+3 \lambda+3\right) & =36\left(4 \lambda^2+9+12 \lambda\right) \\ \Rightarrow & \lambda^2+3 \lambda+3 & =4 \lambda^2+9+12 \lambda \\ \Rightarrow & 3 \lambda^2+9 \lambda+6 & =0 \\ \Rightarrow & \lambda^2+3 \lambda+2 & =0 \\ \Rightarrow & \lambda(\lambda+2)+1(\lambda+2) & =0 \\ \Rightarrow & (\lambda+1)(\lambda+2) & =0 \end{array}$$
$\therefore \quad \lambda=-1,-2$
So, the DC's are $1,2,-1$ and $-1,1,-2$.
Also, both the required lines passes through origin.
So, the equations of required lines are $\frac{x}{1}=\frac{y}{2}=\frac{z}{-1}$ and $\frac{x}{-1}=\frac{y}{1}=\frac{z}{-2}$.
Find the angle between the lines whose direction cosines are given by the equation $l+m+n=0$ and $l^2+m^2-n^2=0$.
$$\begin{aligned} &\text { Eliminating } n \text { from both the equations, we have }\\ &\begin{aligned} & l^2+m^2-(l-m)^2=0 \\ & \Rightarrow \quad l^2+m^2-l^2-m^2+2 m l=0 \quad \Rightarrow \quad 2 l m=0 \end{aligned} \end{aligned}$$
$$ \begin{aligned} & \Rightarrow \quad l m=0 \quad \Rightarrow \quad(-m-n) m=0 \quad[\because l=-m-n] \\ & \Rightarrow \quad(m+n) m=0 \\ & \Rightarrow \quad m=-n \Rightarrow m=0 \\ & \Rightarrow \quad l=0, l=-n \end{aligned}$$
Thus, Dr's two lines are proportional to $0,-n, n$ and $-n, 0, n$ i.e., $0,-1,1$ and $-1,0,1$.
So, the vector parallel to these given lines are $\overrightarrow{\mathbf{a}}=-\hat{\mathbf{j}}+\hat{\mathbf{k}}$ and $\overrightarrow{\mathbf{b}}=-\hat{\mathbf{i}}+\hat{\mathbf{k}}$
$$ \begin{aligned} &\text { Now, }\\ &\begin{aligned} \cos \theta & =\frac{\vec{a} \vec{b}}{|\vec{a}||\vec{b}|}=\frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} \Rightarrow \cos \theta=\frac{1}{2} \\ \therefore\quad \theta & =\frac{\pi}{3} \quad\left[\because \cos \frac{\pi}{3}=\frac{1}{2}\right] \end{aligned} \end{aligned}$$
If a variable line in two adjacent positions has direction cosines $l, m, n$ and $l+\delta l, m+\delta m, n+\delta n$, then show that the small angle $\delta \theta$ between the two positions is given by $\delta \theta^2=\delta l^2+\delta m^2+\delta n^2$.
We have $l, m, n$ and $l+\delta l, m+\delta m, n+\delta n$ as direction cosines of a variable line in two different positions.
$$\begin{array}{lr} \therefore & l^2+m^2+n^2=1 \quad\text{.... (i)}\\ \text { and } & (l+\delta l)^2+(m+\delta m)^2+(n+\delta n)^2=1 \quad\text{.... (ii)} \end{array}$$
$$\begin{array}{lcc} \Rightarrow & l^2+m^2+n^2+\delta l^2+\delta m^2+\delta n^2+2(l \delta l+m \delta m+n \delta n)=1 \\ \Rightarrow & \delta l^2+\delta m^2+\delta n^2=-2(l \delta l+m \delta m+n \delta n) & \quad\left[\because l^2+m^2+n^2=1\right] \\ \Rightarrow & l \delta l+m \delta m+n \delta n=\frac{-1}{2}\left(\delta l^2+\delta m^2+\delta n^2\right) & \ldots(\text { iii }) \end{array}$$
Now, $\overrightarrow{\mathbf{a}}$ and $\overrightarrow{\mathrm{b}}$ are unit vectors along a line with direction cosines $l, m, n$ and $(l+\delta l),(m+\delta m),(n+\delta n)$, respectively.
$$\begin{array}{ll} \therefore & \overrightarrow{\mathbf{a}}=l \hat{\mathbf{i}}+m \hat{\mathbf{j}}+n \hat{\mathbf{k}} \text { and } \overrightarrow{\mathbf{b}}=(l+\delta l) \hat{\mathbf{i}}+(m+\delta m) \hat{\mathbf{j}}+(n+\delta n) \hat{\mathbf{k}} \\ \Rightarrow & \cos \delta \theta=\frac{\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}}}{|\overrightarrow{\mathbf{a}}||\overrightarrow{\mathbf{b}}|}=\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{b}} \quad[\because|\hat{\mathbf{a}}|=|\hat{\mathbf{b}}|=1] \end{array}$$
$$ \begin{aligned} &\begin{aligned} \Rightarrow \quad \cos \delta \theta & =l(l+\delta l)+m(m+\delta m)+n(n+\delta n) \\ & =\left(l^2+m^2+n^2\right)+(l \delta l+m \delta m+n \delta n) \\ & =1-\frac{1}{2}\left(\delta l^2+\delta m^2+\delta n^2\right) \text { [using Eq. (iii)] } \end{aligned}\\ \end{aligned}$$
$$\begin{array}{lcc} \Rightarrow & 2(1-\cos \delta \theta)=\left(\delta l^2+\delta m^2+\delta n^2\right) \\ \Rightarrow & 2 \cdot 2 \sin ^2 \frac{\delta \theta}{2}=\delta l^2+\delta m^2+\delta n^2 & \quad\left[\because 1-\cos \theta=2 \sin ^2 \frac{\theta}{2}\right] \\ \Rightarrow & 4\left(\frac{\delta \theta}{2}\right)^2=\delta l^2+\delta m^2+\delta n^2 & {\left[\text { since, } \frac{\delta \theta}{2} \text { is small, then } \sin \frac{\delta \theta}{2}=\frac{\delta \theta}{2}\right]}\\ \therefore & \delta \theta^2=\delta l^2+\delta m^2+\delta n^2 \end{array}$$
If 0 is the origin and $A$ is $(a, b, c)$, then find the direction cosines of the line $O A$ and the equation of plane through $A$ at right angle to $O A$.
Since, DC's of line OA are $\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}$ and $\frac{c}{\sqrt{a^2+b^2+c^2}}$.
Also, $$\overrightarrow{\mathbf{n}}=\overrightarrow{\mathrm{OA}}=\overrightarrow{\mathbf{a}}=a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}$$
The equation of plane passes through $(a, b, c)$ and perpendicular to $O A$ is given by
$$\begin{array}{rlrl} & {[\overrightarrow{\mathbf{r}}-\overrightarrow{\mathbf{a}}] \cdot \vec{n}} =0 \\ \Rightarrow & \overrightarrow{\mathbf{r}} \cdot \overrightarrow{\mathbf{n}} =\overrightarrow{\mathbf{a}} \cdot \overrightarrow{\mathbf{n}} \\ \Rightarrow & {[(x \hat{\mathbf{i}}+y \hat{\mathbf{j}}+z \hat{\mathbf{k}}) \cdot(a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}})]} =(a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}) \cdot(a \hat{\mathbf{i}}+b \hat{\mathbf{j}}+c \hat{\mathbf{k}}) \\ \Rightarrow & a x+b y+c z =a^2+b^2+c^2 \end{array}$$
Two systems of rectangular axis have the same origin. If a plane cuts them at distances $a, b, c$ and $a^{\prime}, b^{\prime}, c^{\prime}$, respectively from the origin, then prove that $\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^{\prime 2}}+\frac{1}{b^{\prime 2}}+\frac{1}{c^{\prime 2}}$.
Consider OX, OY, OZ and ox, oy, oz are two system of rectangular axes.
Let their corresponding equation of plane be
$$\begin{aligned} & \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 \quad\text{.... (i)}\\ \text{and}\quad & \frac{x}{a^{\prime}}+\frac{y}{b^{\prime}}+\frac{z}{c^{\prime}}=1\quad\text{.... (ii)} \end{aligned}$$
Also, the length of perpendicular from origin to Eqs. (i) and (ii) must be same.
$$\therefore \quad \frac{\frac{0}{a}+\frac{0}{b}+\frac{0}{c}-1}{\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}}=\frac{\frac{0}{a^{\prime}}+\frac{0}{b^{\prime}}+\frac{0}{c^{\prime}}-1}{\sqrt{\frac{1}{a^{\prime 2}}+\frac{1}{b^{\prime 2}}+\frac{1}{c^{\prime 2}}}}$$
$$\begin{array}{ll} \Rightarrow & \sqrt{\frac{1}{a^{2^2}}+\frac{1}{b^{\prime 2}}+\frac{1}{c^{\prime^2}}}=\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}} \\ \Rightarrow & \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\frac{1}{a^{\prime^2}}+\frac{1}{b^{2^2}}+\frac{1}{c^{\prime 2}} \end{array}$$