Question Number 54421 by gunawan last updated on 03/Feb/19
$$\mathrm{If}\:\mathrm{the}\:\mathrm{vector}\:\boldsymbol{\mathrm{c}},\:\boldsymbol{\mathrm{a}}={x}\boldsymbol{\mathrm{i}}+{y}\boldsymbol{\mathrm{j}}+{z}\boldsymbol{\mathrm{k}}\:\mathrm{and}\:\boldsymbol{\mathrm{b}}=\boldsymbol{\mathrm{j}} \\ $$$$\mathrm{are}\:\mathrm{such}\:\mathrm{that}\:\boldsymbol{\mathrm{a}},\:\boldsymbol{\mathrm{c}}\:\mathrm{and}\:\boldsymbol{\mathrm{b}}\:\mathrm{form}\:\mathrm{a}\:\mathrm{right} \\ $$$$\mathrm{handed}\:\mathrm{system},\:\mathrm{then}\:\boldsymbol{\mathrm{c}}\:\mathrm{is} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 04/Feb/19
$$\left.{p}\left.{ls}\:{carify}\:{i}\right)\left({a}\:,{b},{c}\:\right)\:\:\:{ii}\right)\left({a},{c},{b}\right)\:{which}\:{one}\:{form} \\ $$$${right}\:{handed}\:{system} \\ $$$$\:\:\:\:\:\:\: \\ $$
Commented by gunawan last updated on 05/Feb/19
$$\mathrm{option}\:\mathrm{to}\:\mathrm{equation} \\ $$$${a}.\:{z}\boldsymbol{\mathrm{i}}−{x}\boldsymbol{\mathrm{k}} \\ $$$${b}.\:\overset{\rightarrow} {\mathrm{0}} \\ $$$$\mathrm{c}.\:\mathrm{y}\boldsymbol{\mathrm{j}} \\ $$$$\mathrm{d}.\:−{z}\boldsymbol{{i}}+{x}\boldsymbol{{k}} \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{sorry}\:\mathrm{Sir}\:\mathrm{If}\:\mathrm{question}\:\mathrm{not}\:\mathrm{clearly} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19
![c=ip+jq+kr b=j a=ix+jy+kz a×c=b [i j k] [x y z] [p q r] i(yr−qz)−j(xr−pz)+k(xq−py)=j [given] yr−qz=0 (y/q)=(z/r)....(1)eqn −(xr−pz)=1.....(2)eqn (xq−py)=0 (x/p)=(y/q) so (x/p)=(y/q)=(z/r)=(1/α) p=xα q=yα r=zα −(xr−pz)=1 −x(zα)+(xα)z=1 some thing wrong pls check question...](https://www.tinkutara.com/question/Q54432.png)
$$\boldsymbol{{c}}={ip}+{jq}+{kr} \\ $$$$\boldsymbol{{b}}={j} \\ $$$$\boldsymbol{{a}}={ix}+{jy}+{kz} \\ $$$$\boldsymbol{{a}}×\boldsymbol{{c}}=\boldsymbol{{b}} \\ $$$$\left[\boldsymbol{{i}}\:\:\:\boldsymbol{{j}}\:\:\:\boldsymbol{{k}}\right] \\ $$$$\left[\boldsymbol{{x}}\:\:\:\boldsymbol{{y}}\:\:\:\boldsymbol{{z}}\right] \\ $$$$\left[\boldsymbol{{p}}\:\:\boldsymbol{{q}}\:\:\:\:\boldsymbol{{r}}\right] \\ $$$${i}\left({yr}−{qz}\right)−{j}\left({xr}−{pz}\right)+{k}\left({xq}−{py}\right)={j}\:\:\left[{given}\right] \\ $$$${yr}−{qz}=\mathrm{0} \\ $$$$\frac{{y}}{{q}}=\frac{{z}}{{r}}….\left(\mathrm{1}\right){eqn} \\ $$$$−\left({xr}−{pz}\right)=\mathrm{1}…..\left(\mathrm{2}\right){eqn} \\ $$$$\left({xq}−{py}\right)=\mathrm{0} \\ $$$$\frac{{x}}{{p}}=\frac{{y}}{{q}}\:\:{so}\:\frac{{x}}{{p}}=\frac{{y}}{{q}}=\frac{{z}}{{r}}=\frac{\mathrm{1}}{\alpha} \\ $$$${p}={x}\alpha\:\:\:{q}={y}\alpha\:\:\:\:{r}={z}\alpha \\ $$$$−\left({xr}−{pz}\right)=\mathrm{1} \\ $$$$−{x}\left({z}\alpha\right)+\left({x}\alpha\right){z}=\mathrm{1} \\ $$$$\boldsymbol{{some}}\:\boldsymbol{{thing}}\:\boldsymbol{{wrong}}\:\boldsymbol{{pls}}\:\boldsymbol{{check}}\:\boldsymbol{{question}}… \\ $$$$ \\ $$
Answered by ajfour last updated on 03/Feb/19
$$\bar {{c}}=\bar {{a}}×\bar {{b}}\:=\:{x}\hat {{k}}−{z}\hat {{i}} \\ $$