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Question Number 51520 by Tawa1 last updated on 27/Dec/18

Commented by Tawa1 last updated on 27/Dec/18

Thanks but no idea

Thanksbutnoidea

Answered by tanmay.chaudhury50@gmail.com last updated on 28/Dec/18

n=ai+2aj−2ak m=2i−j+2k−ai−2aj+2ak [since m=q−n] m=i(2−a)−j(1+2a)+k(2+2a) since m⊥p so[x_1 x_2 +y_1 y_2 +z_1 z_2 =0] (2−a)×1−2(1+2a)−2(2+2a)=0 2−a−2−4a−4−4a=0 −9a=4 a=((−4)/9) so n=((−4)/9)(i+2j−2k) m=i(2−a)−j(1+2a)+k(2+2a) m=i(2+(4/9))−j(1−(8/9))+k(2−(8/9)) m=i(((22)/9))−j((1/9))+k(((10)/9))

n=ai+2aj−2akm=2i−j+2k−ai−2aj+2ak[sincem=q−n]m=i(2−a)−j(1+2a)+k(2+2a)sincem⊥pso[x1x2+y1y2+z1z2=0](2−a)×1−2(1+2a)−2(2+2a)=02−a−2−4a−4−4a=0−9a=4a=−49son=−49(i+2j−2k)m=i(2−a)−j(1+2a)+k(2+2a)m=i(2+49)−j(1−89)+k(2−89)m=i(229)−j(19)+k(109)

Commented by Tawa1 last updated on 29/Dec/18

God bless you sir

Godblessyousir

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