Three Dimensional Geometry Question 203
Question: Distance of the point $ (x_1,y_1,z_1) $ from the line $ \frac{x-x_2}{l}=\frac{y-y_2}{m}=\frac{z-z_2}{n} $ , where $ l, $ m and n are the direction cosines of line is
Options:
A) $ \sqrt{{{(x_1-x_2)}^{2}}+{{(y_1-y_2)}^{2}}+{{(z_1-z_2)}^{2}}-{{[l(x_1-x_2)+m(y_1-y_2)+n(z_1-z_2)]}^{2}}} $
B) $ \sqrt{{{(x_2-x_1)}^{2}}+{{(y_2-y_1)}^{2}}+{{(z_2-z_1)}^{2}}} $
C) $ \sqrt{(x_2-x_1)l+(y_2-y_1)m+(z_2-z_1)n} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
Let $ {{\mathbf{r}}_1}=(x_2-x_1)\mathbf{i}+(y_2-y_1)\mathbf{j}+(z_2-z_1),\mathbf{k} $ $ {{\mathbf{r}}_2}=l\mathbf{i}+m\mathbf{j}+n\mathbf{k} $
$ \therefore $ $ \cos \theta =\frac{{{\mathbf{r}}_2}.{{\mathbf{r}}_1}}{|{{\mathbf{r}}_1}|,|{{\mathbf{r}}_2}|} $ Also, $ d=,|{{\mathbf{r}}_1}|\sin \theta $ , $ d^{2}=|{{\mathbf{r}}_1}{{|}^{2}}{{\sin }^{2}}\theta $
Þ $ d^{2}=|{{\mathbf{r}}_1}{{|}^{2}}(1-{{\cos }^{2}}\theta ) $
Þ $ d^{2}=,|{{\mathbf{r}}_1}{{|}^{2}}( 1-\frac{{{\mathbf{r}}_1}.{{\mathbf{r}}_2}}{|{{\mathbf{r}}_1}{{|}^{2}}|{{\mathbf{r}}_2}{{|}^{2}}} ) $
Þ $ d^{2}=|{{\mathbf{r}}_1}{{|}^{2}}-{{({{\mathbf{r}}_1}.{{\mathbf{r}}_2})}^{2}} $ , {where $ |{{\mathbf{r}}_2}|=1 $ }
Þ $ d=\sqrt{|{{\mathbf{r}}_1}{{|}^{2}}-{{({{\mathbf{r}}_1}.{{\mathbf{r}}_2})}^{2}}} $ Therefore, distance of the point ( $ x_1,,y_1,,z_1 $ ) from the line is d= $ ll_1+mm_1+nn_1=0 $ .
BETA
