Question

Distance of the point $(x_l,y_l,z_l)$ from the line $\frac{x-x_2}{l}=\frac{y-y_2}{m}=\frac{z-z_2}{n}$, where I, m and n are the direction cosines of line is

• A.
  
• B.
  
• C.
  
• D. None of these

Answer 1

The correct option is A

Let $r_l=(x_2-x_1)i+(y_2-y_1)j+(z_2-z_1)k{r}_{2}li+mj+nk\therefore cos\theta =\frac{r_2{r}_1}{|r_1||r_2|}Also,d=|r_1|sin\theta ,d^2=|r_1{|}^{2}si{n}^2\theta \Rightarrow d^2=|r_1{|}^2(1-co{s}^2\theta )\Rightarrow d^2=|r_1{|}^2(1-\frac{r_1,r_2}{|r_1{|}^2|r_2{|}^2})\Rightarrow d^2=|r_1{|}^2-(r_1,r_2{)}^2,\{where|r_2|=1\}\Rightarrow d=\sqrt{|r_1{|}^2}-(r_1{r}_2{)}^2$
Therefore,distance of the point $(x_1,y_1,z_1)$ from the line is
$d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2+(z_2-z_1{)}^2-[l(x_2-x_1)+m(y_2-y_1)+n(z_2-z_1){]}^2}$