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Category: Vector

prove-that-curl-r-n-c-r-n-2-r-n-c-nr-n-2-r-c-where-c-is-the-constant-vector-

Question Number 85865 by subhankar10 last updated on 25/Mar/20 $$\mathrm{prove}\:\mathrm{that}\: \\ $$$$\mathrm{curl}\left(\mathrm{r}^{\mathrm{n}} \overset{\rightarrow} {\mathrm{c}}×\overset{\rightarrow} {\mathrm{r}}\right)=\left(\mathrm{n}+\mathrm{2}\right)\mathrm{r}^{\mathrm{n}} \overset{\rightarrow} {\mathrm{c}}−\mathrm{nr}^{\mathrm{n}−\mathrm{2}} \left(\overset{\rightarrow} {\mathrm{r}}.\overset{\rightarrow} {\mathrm{c}}\right)\:\:. \\ $$$$\mathrm{where}\:\mathrm{c}\:\mathrm{is}\:\mathrm{the}\:\mathrm{constant}\:\mathrm{vector}. \\ $$ Answered…

A-plane-is-drawn-through-the-midpoint-of-a-diagonal-of-a-cube-perpendicular-to-the-diagonal-Determine-the-area-of-the-figure-resulting-from-the-section-of-the-cube-cut-by-this-plane-if-the-edge-of-t

Question Number 20194 by ajfour last updated on 23/Aug/17 $${A}\:{plane}\:{is}\:{drawn}\:{through}\:{the}\: \\ $$$${midpoint}\:{of}\:{a}\:{diagonal}\:{of}\:{a}\:{cube} \\ $$$${perpendicular}\:{to}\:{the}\:{diagonal}. \\ $$$${Determine}\:{the}\:{area}\:{of}\:{the}\:{figure} \\ $$$${resulting}\:{from}\:{the}\:{section}\:{of}\:{the} \\ $$$${cube}\:{cut}\:{by}\:{this}\:{plane}\:{if}\:{the}\:{edge} \\ $$$${of}\:{the}\:{cube}\:{is}\:{equal}\:{to}\:\boldsymbol{{a}}. \\ $$ Commented…

Three-vectors-A-B-and-C-add-up-to-zero-Find-which-is-false-a-A-B-C-is-not-zero-unless-B-C-are-parallel-b-A-B-C-is-not-zero-unless-B-

Question Number 19976 by Tinkutara last updated on 19/Aug/17 $$\mathrm{Three}\:\mathrm{vectors}\:\overset{\rightarrow} {{A}},\:\overset{\rightarrow} {{B}}\:\mathrm{and}\:\overset{\rightarrow} {{C}}\:\mathrm{add}\:\mathrm{up}\:\mathrm{to} \\ $$$$\mathrm{zero}.\:\mathrm{Find}\:\mathrm{which}\:\mathrm{is}\:\mathrm{false}. \\ $$$$\left({a}\right)\:\left(\overset{\rightarrow} {{A}}×\overset{\rightarrow} {{B}}\right)×\overset{\rightarrow} {{C}}\:\mathrm{is}\:\mathrm{not}\:\mathrm{zero}\:\mathrm{unless}\:\overset{\rightarrow} {{B}},\:\overset{\rightarrow} {{C}} \\ $$$$\mathrm{are}\:\mathrm{parallel} \\…

Question-84641

Question Number 84641 by M±th+et£s last updated on 14/Mar/20 Commented by M±th+et£s last updated on 14/Mar/20 $${a}\:{square},\:{a}\:{circle}\:{and}\:{tow}\:{semicircles} \\ $$$${the}\:{are}\:{of}\:{the}\:{square}\:{is}\:\mathrm{4}.{what}\:{is}\:{the} \\ $$$${length}\:{of}\:{the}\:{blue}\:{lenght} \\ $$ Answered by…