Question Number 51520 by Tawa1 last updated on 27/Dec/18
Commented by Tawa1 last updated on 27/Dec/18
$$\mathrm{Thanks}\:\mathrm{but}\:\mathrm{no}\:\mathrm{idea} \\ $$
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))](https://www.tinkutara.com/question/Q51537.png)
$${n}={ai}+\mathrm{2}{aj}−\mathrm{2}{ak} \\ $$$${m}=\mathrm{2}{i}−{j}+\mathrm{2}{k}−{ai}−\mathrm{2}{aj}+\mathrm{2}{ak}\:\left[{since}\:{m}={q}−{n}\right] \\ $$$${m}={i}\left(\mathrm{2}−{a}\right)−{j}\left(\mathrm{1}+\mathrm{2}{a}\right)+{k}\left(\mathrm{2}+\mathrm{2}{a}\right) \\ $$$${since}\:{m}\bot{p}\:{so}\left[{x}_{\mathrm{1}} {x}_{\mathrm{2}} +{y}_{\mathrm{1}} {y}_{\mathrm{2}} +{z}_{\mathrm{1}} {z}_{\mathrm{2}} =\mathrm{0}\right] \\ $$$$\left(\mathrm{2}−{a}\right)×\mathrm{1}−\mathrm{2}\left(\mathrm{1}+\mathrm{2}{a}\right)−\mathrm{2}\left(\mathrm{2}+\mathrm{2}{a}\right)=\mathrm{0} \\ $$$$\mathrm{2}−{a}−\mathrm{2}−\mathrm{4}{a}−\mathrm{4}−\mathrm{4}{a}=\mathrm{0} \\ $$$$−\mathrm{9}{a}=\mathrm{4}\:\:\:{a}=\frac{−\mathrm{4}}{\mathrm{9}} \\ $$$${so}\:{n}=\frac{−\mathrm{4}}{\mathrm{9}}\left({i}+\mathrm{2}{j}−\mathrm{2}{k}\right) \\ $$$${m}={i}\left(\mathrm{2}−{a}\right)−{j}\left(\mathrm{1}+\mathrm{2}{a}\right)+{k}\left(\mathrm{2}+\mathrm{2}{a}\right) \\ $$$${m}={i}\left(\mathrm{2}+\frac{\mathrm{4}}{\mathrm{9}}\right)−{j}\left(\mathrm{1}−\frac{\mathrm{8}}{\mathrm{9}}\right)+{k}\left(\mathrm{2}−\frac{\mathrm{8}}{\mathrm{9}}\right) \\ $$$${m}={i}\left(\frac{\mathrm{22}}{\mathrm{9}}\right)−{j}\left(\frac{\mathrm{1}}{\mathrm{9}}\right)+{k}\left(\frac{\mathrm{10}}{\mathrm{9}}\right) \\ $$
Commented by Tawa1 last updated on 29/Dec/18
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\ $$