Question Number 132303 by mohammad17 last updated on 13/Feb/21
Answered by EDWIN88 last updated on 13/Feb/21
$$\:\overset{\rightarrow} {\mathrm{a}}=\left(−\mathrm{1},\:\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\:,\:\mathrm{1}\right)\: \\ $$$$\:\overset{\rightarrow} {\mathrm{b}}=\left(−\mathrm{3},\:−\mathrm{5ln}\:\mathrm{2},\:−\mathrm{4}\right) \\ $$$$\:\overset{\rightarrow} {\mathrm{c}}=\:\left(\mathrm{4},\:−\mathrm{5},\:\mathrm{5}\:\right)\: \\ $$$$\Leftrightarrow\overset{\rightarrow} {\mathrm{a}}−\overset{\rightarrow} {\mathrm{b}}+\overset{\rightarrow} {\mathrm{c}}=\left(\mathrm{6},\:\mathrm{5}\left(\mathrm{ln}\:\mathrm{2}−\mathrm{1}\right)+\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\:;\:\mathrm{2}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\mathrm{6},\:\frac{\mathrm{10}\:\mathrm{ln}\:\left(\frac{\mathrm{2}}{\mathrm{e}}\right)+\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\:,\:\mathrm{2}\right) \\ $$$$\mathrm{unit}\:\mathrm{vector}\:\mathrm{is}\:\frac{\left(\mathrm{6},\frac{\mathrm{10ln}\:\left(\frac{\mathrm{2}}{\mathrm{e}}\right)+\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{2}}\:,\mathrm{2}\right)}{\:\sqrt{\mathrm{40}+\frac{\left(\mathrm{10ln}\:\left(\frac{\mathrm{2}}{\mathrm{e}}\right)+\mathrm{3}\sqrt{\mathrm{3}}\right)^{\mathrm{2}} }{\mathrm{4}}}} \\ $$