question_answer

If the equation of a line through a point a and parallel to vector b is \[\mathbf{r}=\mathbf{a}+t\,\mathbf{b},\] where t is a parameter, then its perpendicular distance from the point c is [MP PET 1998]

A)            \[|(\mathbf{c}-\mathbf{b})\times \mathbf{a}|\div |\mathbf{a}|\]

B)            \[|(\mathbf{c}-\mathbf{a})\times \mathbf{b}|\div |\mathbf{b}|\]

C)            \[|(\mathbf{a}-\mathbf{b})\times \mathbf{c}|\div |\mathbf{c}|\]

D)            \[|(\mathbf{a}-\mathbf{b})\times \mathbf{c}|\div |\mathbf{a}+\mathbf{c}|\]

Correct Answer: B

Solution :

           For point \[P\] on the line \[r=\mathbf{a}+t\mathbf{b}\]            \[\therefore \,\,\,\overrightarrow{PC}=(\mathbf{c}-\mathbf{a})-t\mathbf{b}\], \[\because \,\,\,\overrightarrow{PC}\,\bot \,\mathbf{b}\]            \[\therefore \,\,\,|(\mathbf{c}-\mathbf{a})-t\mathbf{b}|\,.\,\mathbf{b}=0\] or \[t=\frac{(\mathbf{c}-\mathbf{a})\,.\,\mathbf{b}}{{{\mathbf{b}}^{2}}}\]          ?..(i)            Distance of \[\mathbf{c}\] from line \[|\overrightarrow{PC}|\ =\]\[d=|\mathbf{c}-\mathbf{a}-t\mathbf{b}|\]            \[d=\left| \mathbf{c}-\mathbf{a}-\frac{(\mathbf{c}-\mathbf{a})\,.\,\mathbf{bb}}{{{\mathbf{b}}^{2}}} \right|=\left| \frac{(\mathbf{c}-\mathbf{a})\,\mathbf{b}\,.\,\mathbf{b}-(\mathbf{c}-\mathbf{a})\,.\,\mathbf{bb}}{{{\mathbf{b}}^{2}}} \right|\]            \[d=\left| \frac{\mathbf{b}\times (\mathbf{c}-\mathbf{a})\times \mathbf{b}}{{{\mathbf{b}}^{2}}} \right|=\frac{|\mathbf{b}||(\mathbf{c}-\mathbf{a})\times \mathbf{b}|\sin 90{}^\circ }{|\mathbf{b}{{|}^{2}}}\],                                                                 \[(\because \,\mathbf{b}\,\bot \,(\mathbf{c}-\mathbf{a})\times \mathbf{b})\]            \[d=\frac{|(\mathbf{c}-\mathbf{a})\times \mathbf{b}|}{|\mathbf{b}|}\].