Question Number 54418 by gunawan last updated on 03/Feb/19
$$\mathrm{The}\:\mathrm{projection}\:\mathrm{of}\:\mathrm{the}\:\mathrm{vector}\:\boldsymbol{\mathrm{a}}=\mathrm{4}\boldsymbol{\mathrm{i}}−\mathrm{3}\boldsymbol{\mathrm{j}}+\mathrm{2}\boldsymbol{\mathrm{k}} \\ $$$$\mathrm{on}\:\mathrm{the}\:\mathrm{axis}\:\mathrm{making}\:\mathrm{equal}\:\mathrm{acute}\:\mathrm{angles} \\ $$$$\mathrm{with}\:\mathrm{the}\:\mathrm{coordinate}\:\mathrm{axes}\:\mathrm{is} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19
$$\boldsymbol{{a}}=\mathrm{4}\boldsymbol{{i}}−\mathrm{3}\boldsymbol{{j}}+\mathrm{2}\boldsymbol{{k}}\:\:\:{so}\:\:\mid\boldsymbol{{a}}\mid=\sqrt{\mathrm{4}^{\mathrm{2}} +\left(−\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{2}\right)^{\mathrm{2}} }\:=\sqrt{\mathrm{29}}\: \\ $$$${projection}\:{along}\:{x}\:{axis}=\boldsymbol{{i}}\sqrt{\mathrm{29}}\:\boldsymbol{{cos}}\alpha= \\ $$$$ \\ $$$${along}\:{y}\:{axus}={j}\sqrt{\mathrm{29}}\:{cos}\beta \\ $$$$\:{along}\:\:{z}\:{axis}={k}\sqrt{\mathrm{29}}\:{cos}\gamma \\ $$$${cos}^{} \alpha={cos}\beta={cos}\gamma={l} \\ $$$${l}^{\mathrm{2}} +{l}^{\mathrm{2}} +{l}^{\mathrm{2}} =\mathrm{1}\:\:\:{so}\:{l}=\pm\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}\:} \\ $$$${so}\:{projection}\:{along}\:{x}\:{axus}=\pm\frac{\sqrt{\mathrm{29}}}{\:\sqrt{\mathrm{3}}}{i} \\ $$$$\:\:{projection}\:{along}\:{y}\:{axis}=\pm\frac{\sqrt{\mathrm{29}}}{\:\sqrt{\mathrm{3}}}{j}\:\: \\ $$$${projection}\:{z}\:{axis}=\pm\frac{\sqrt{\mathrm{29}}}{\:\sqrt{\mathrm{3}}}{k} \\ $$$${others}\:{pls}\:{check}… \\ $$