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
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 \[{{d}^{2}}=\,|{{\mathbf{r}}_{1}}{{|}^{2}}\left( 1-\frac{{{\mathbf{r}}_{1}}.{{\mathbf{r}}_{2}}}{|{{\mathbf{r}}_{1}}{{|}^{2}}|{{\mathbf{r}}_{2}}{{|}^{2}}} \right)\] 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=\[l{{l}_{1}}+m{{m}_{1}}+n{{n}_{1}}=0\].
You need to login to perform this action.
You will be redirected in
3 sec
