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O is the origin and lines OA, OB and OC have direction cosines l_(r),m_(r)andn_(r)(r=1,2 and3). If lines OA', OB' and OC' bisect angles BOC, COA and AOB, respectively, prove that planes AOA', BOB' and COC' pass through the line (x)/(l_(2)+l_(2)+l_(3))=(y)/(m_(1)+m_(2)+m_(3))=(z)/(n_(1)+n_(2)+n_(3)). |
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Answer» Solution :If `theta` is the `angleBCO`, then the direction consines of OA' `("bisector of "angleBOC)" " are" "(l_(2)+l_(3))/(2cos(theta//2)),(m_(2)+m_(3))/(2cos(theta//2))` `(n_(2)+n_(3))/(2cos(theta//2))` or the direction ratios of OA' are `l_(2)+l_(3),m_(2)+m_(3)andn_(2)+n_(3)` ALSO, the direction cosines of OA are `l_(1),m_(1)andn_(1)`. Hence, the equation of plane AOA' is `|{:(X,y,z),(l_(2)+l_(3),m_(2)+m_(3),n_(2)+n_(3)),(l_(1),m_(1),n_(1)):}|=0` Applying `R_(2)toR_(2)+R_(3)`, we get the equation of plane AOA' as `|{:(x,y,z),(l_(1)+l_(2)+l_(3),m_(1)+m_(2)+m_(3),n_(1)+n_(2)+n_(3)),(l_(1),m_(1),n_(1)):}|=0` For all values of r, the point `(l_(1)+l_(2)+l_(3))r,(m_(1)+m_(2)+m_(3))and(n_(1)+n_(2)+n_(3))r)` LIES on plane AOA'. Hence, the line `(x)/(l_(1)+l_(2)+l_(3))=(y)/(m_(1)+m_(2)+m_(3))=(z)/(n_(1)+n_(2)+n_(3))=r` lies on plane AOA'. Similarly, this line lies on planes BOB' and COC' also. Hence, all the three planes, AOA' BOB' and COC', pass through the line. |
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