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Show that the straight lines whose direction cosinesare given by the equations `a l+b m+c n=0a n dul^2+z m^2=v n^2+w n^2=0`are parallel or perpendicular as`(a^2)/u+(b^2)/v+(c^2)/w=0ora^2(v+w)+b^2(w+u)+c^2(u+v)=0.` |
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Answer» The given equation are `al+bm+cn=0" "...(i)` `"ul"^(2)+vm^(2)+wn^(2)=0." "...(ii)` Putting `l=(-(bm+cn))/(a)` from (i) in (ii), we get `(u(bm+cn)^(2))/(a^(2))+vm^(2)+wn^(2)=0` `rArr(b^(2)u+a^(2)v)m^(2)+2ubcmn+(c^(2)u+a^(2)w)n^(2)=0` `rArr (b^(2)u+a^(2)v)((m)/(n))^(2)+2ubc((m)/(n))+(c^(2)u+a^(2)w)=0." "...(iii)` Let `(m_(1))/(n_(1))and (m_(2))/(n_(2))` be the roots of (iii). Then,`(m_(1))/(n_(1))*(m_(2))/(n_(2))=(c^(2)u+a^(2)w)/(b^(2)u+a^(2)v)` `rArr(m_(1)m_(2))/(c^(2)u+a^(2))=(n_(1)n_(2))/(b^(2)u+a^(2)v)=(l_(1)l_(2))/(b^(2)w+c^(2)v)=k` [by symmetry]. `thereforel_(1)l_(2)+m_(1)m_(2)+n_(1)n_(2)=k(b^(2)w+c^(2)v+c^(2)u+a^2w+b^(2)u+a^(2)v)`. The given lines are mutually perpendicular `iffl_(1)l_(2)+m_(1)m_(2)+n_(1)n_(2)=0` `iffa^(2)(v+w)+b^(2)(w+u)+c^(2)(u+v)=0` For the given lines to be parallel, the direction cosines must be equal. `therefore` the roots of (iii) must be equal. `therefore 4u^(2)b^(2)c^(2)-4(b^(2)u+a^(2)v)(c^(2)u+a^(2)w)=0iff(a^(2))/(u)+(b^(2))/(v)+(c^(2))/(w)=0`. |
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