Question Number 46686 by peter frank last updated on 30/Oct/18
Commented by peter frank last updated on 30/Oct/18
$$\mathrm{ABCDEFGH}\:\mathrm{is}\:\mathrm{a}\:\mathrm{cube}.\mathrm{find}\:\mathrm{the} \\ $$$$\mathrm{projections}\:\mathrm{of} \\ $$$$\left.\mathrm{a}\right)\mathrm{AF}\:\mathrm{on}\:\:\mathrm{ABCD} \\ $$$$\left.\mathrm{b}\right)\mathrm{AG}\:\mathrm{on}\:\mathrm{ABCD} \\ $$$$\left.\mathrm{c}\right)\mathrm{FD}\:\:\mathrm{on}\:\mathrm{HGCD} \\ $$$$\left.\mathrm{d}\right)\:\mathrm{EC}\:\mathrm{on}\:\:\mathrm{FGCB} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 30/Oct/18
$${let}\:{side}\:{of}\:{cube}\:{of}\:{length}\:{a} \\ $$$${A}\:{be}\:{origin}\left(\mathrm{0},\mathrm{0},\mathrm{0}\right)\:\:{D}\left({a},\mathrm{0},\mathrm{0}\right)\:\:\:{F}\left(\mathrm{0},{a},{a}\right) \\ $$$${point}\:{B}\:{is}\:{the}\:{foot}\:{of}\:{perpendicular}\:{from}\:{point} \\ $$$${F}\:{on}\:{plane}\:{ABCD} \\ $$$$ \\ $$$${so}\:{projection}\:{of}\:{AF}\:{on}\:{ABCD}\:{is}\:{AB}\:\:\left({yaxis}\right) \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 30/Oct/18
$${ptojection}\:{of}\:{AG}\:{on}\:{ABCD}\:{is}\:{AC} \\ $$$${projection}\:{of}\:{FD}\:{on}\:{HGCD}\:{is}\:{DG} \\ $$$${prljection}\:{of}\:{EC}\:{on}\:{EGCB}\:{is}\:{FC} \\ $$
Commented by peter frank last updated on 30/Oct/18
$$\mathrm{thank}\:\mathrm{you} \\ $$