Category: Vector

Question Number 146472 by nadovic last updated on 13/Jul/21 Commented by nadovic last updated on 13/Jul/21 $${OK} \\ $$ Terms of Service Privacy Policy Contact:…

Question Number 15358 by Tinkutara last updated on 09/Jun/17 $$\mathrm{If}\:\overset{\rightarrow} {{a}},\:\overset{\rightarrow} {{b}},\:\overset{\rightarrow} {{c}}\:\mathrm{are}\:\mathrm{mutually}\:\mathrm{perpendicular} \\ $$$$\mathrm{vectors}\:\mathrm{of}\:\mathrm{equal}\:\mathrm{magnitudes},\:\mathrm{show}\:\mathrm{that} \\ $$$$\mathrm{the}\:\mathrm{vector}\:\overset{\rightarrow} {{a}}\:+\:\overset{\rightarrow} {{b}}\:+\:\overset{\rightarrow} {{c}}\:\mathrm{is}\:\mathrm{equally}\:\mathrm{inclined} \\ $$$$\mathrm{to}\:\overset{\rightarrow} {{a}},\:\overset{\rightarrow} {{b}}\:\mathrm{and}\:\overset{\rightarrow} {{c}}\:.…

Question Number 15352 by Tinkutara last updated on 09/Jun/17 $$\mathrm{Let}\:\overset{\rightarrow} {{a}}\:=\:\overset{\wedge} {{i}}\:+\:\mathrm{4}\overset{\wedge} {{j}}\:+\:\mathrm{2}\overset{\wedge} {{k}},\:\overset{\rightarrow} {{b}}\:=\:\mathrm{3}\overset{\wedge} {{i}}\:−\:\mathrm{2}\overset{\wedge} {{j}}\:+\:\mathrm{7}\overset{\wedge} {{k}} \\ $$$$\mathrm{and}\:\overset{\rightarrow} {{c}}\:=\:\mathrm{2}\overset{\wedge} {{i}}\:−\:\overset{\wedge} {{j}}\:+\:\mathrm{4}\overset{\wedge} {{k}}\:.\:\mathrm{Find}\:\mathrm{a}\:\mathrm{vector}\:\overset{\rightarrow} {{d}}…

Question Number 146403 by gsk2684 last updated on 13/Jul/21 $${let}\:{f}\left({x},{y}\right)=\frac{{x}^{\mathrm{5}} {y}^{\mathrm{2}} }{\mathrm{10}}\:{then}\:{find}\: \\ $$$${D}_{{u}} {f}\left(−\mathrm{5},−\mathrm{3}\right)\:{in}\:{the}\:{direction}\:{of}\: \\ $$$${the}\:{vector}\:<\mathrm{0},−\mathrm{2}>? \\ $$ Commented by gsk2684 last updated on…

Question Number 146315 by mnjuly1970 last updated on 12/Jul/21 Answered by Olaf_Thorendsen last updated on 13/Jul/21 $$\sqrt{\frac{{x}\left({x}+\mathrm{5}\right)}{{x}+\mathrm{2}}}+\sqrt{{x}+\mathrm{3}}\:=\:\frac{\sqrt{\mathrm{16}{x}+\mathrm{32}}}{{x}+\mathrm{2}} \\ $$$$\sqrt{\frac{{x}\left({x}+\mathrm{5}\right)}{{x}+\mathrm{2}}}+\sqrt{{x}+\mathrm{3}}\:=\:\frac{\mathrm{4}}{\:\sqrt{{x}+\mathrm{2}}} \\ $$$$\sqrt{{x}\left({x}+\mathrm{5}\right)}+\sqrt{\left({x}+\mathrm{3}\right)\left({x}+\mathrm{2}\right)}\:=\:\mathrm{4} \\ $$$$\mathrm{Let}\:{u}\:=\:{x}+\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\sqrt{\left({u}−\frac{\mathrm{5}}{\mathrm{2}}\right)\left({u}+\frac{\mathrm{5}}{\mathrm{2}}\right)}+\sqrt{\left({u}+\frac{\mathrm{1}}{\mathrm{2}}\right)\left({u}−\frac{\mathrm{1}}{\mathrm{2}}\right)}\:=\:\mathrm{4}…