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Question Number 54418 by gunawan last updated on 03/Feb/19

$$\mathrm{The}\mathrm{projection}\mathrm{of}\mathrm{the}\mathrm{vector}\mathbf{a}=4\mathbf{i}-3\mathbf{j}+2\mathbf{k}$$$$\mathrm{on}\mathrm{the}\mathrm{axis}\mathrm{making}\mathrm{equal}\mathrm{acute}\mathrm{angles}$$$$\mathrm{with}\mathrm{the}\mathrm{coordinate}\mathrm{axes}\mathrm{is}$$

Answered by tanmay.chaudhury50@gmail.com last updated on 03/Feb/19

$$\mathit{a}=4\mathit{i}-3\mathit{j}+2\mathit{k}so\mid \mathit{a}\mid =\sqrt{4^2+{(-3)}^2+{\left(2\right)}^2}=\sqrt{29}$$$$projectionalongxaxis=\mathit{i}\sqrt{29}\mathit{c}\mathit{o}\mathit{s}\alpha =$$$$$$$$alongyaxus=j\sqrt{29}cos\beta$$$$alongzaxis=k\sqrt{29}cos\gamma$$$${cos}^{}\alpha =cos\beta =cos\gamma =l$$$$l^2+l^2+l^2=1sol=\pm \frac{1}{\sqrt{3}}$$$$soprojectionalongxaxus=\pm \frac{\sqrt{29}}{\sqrt{3}}i$$$$projectionalongyaxis=\pm \frac{\sqrt{29}}{\sqrt{3}}j$$$$projectionzaxis=\pm \frac{\sqrt{29}}{\sqrt{3}}k$$$$othersplscheck...$$

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