Question and Answers Forum

All Questions      Topic List

Vector Calculus Questions

Previous in All Question      Next in All Question      

Previous in Vector Calculus      Next in Vector Calculus      

Question Number 135054 by rexford last updated on 09/Mar/21

Commented by mr W last updated on 10/Mar/21

there are three possible answers: in plane: ∣a∣=((√6)/2) in space: ∣a∣=1 or (√3)

$${there}\:{are}\:{three}\:{possible}\:{answers}: \\ $$$${in}\:{plane}:\:\mid{a}\mid=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}} \\ $$$${in}\:{space}:\:\mid{a}\mid=\mathrm{1}\:{or}\:\sqrt{\mathrm{3}} \\ $$

Commented by liberty last updated on 10/Mar/21

i think this question wrong sir

$$\mathrm{i}\:\mathrm{think}\:\mathrm{this}\:\mathrm{question}\:\mathrm{wrong}\:\mathrm{sir} \\ $$

Commented by mr W last updated on 10/Mar/21

why wrong?

$${why}\:{wrong}? \\ $$

Answered by EDWIN88 last updated on 09/Mar/21

∣a^→ +b^→ +c^→ ∣=(√(∣a^→ ∣^2 +∣b^→ ∣^2 +∣c^→ ∣^2 −2(∣a^→ ∣∣b^→ ∣.cos (π/3)+∣a^→ ∣∣c^→ ∣.cos (π/3)+∣b^→ ∣∣c^→ ∣.cos (π/3))))

$$\mid\overset{\rightarrow} {{a}}+\overset{\rightarrow} {{b}}+\overset{\rightarrow} {{c}}\mid=\sqrt{\mid\overset{\rightarrow} {{a}}\mid^{\mathrm{2}} +\mid\overset{\rightarrow} {{b}}\mid^{\mathrm{2}} +\mid\overset{\rightarrow} {{c}}\mid^{\mathrm{2}} −\mathrm{2}\left(\mid\overset{\rightarrow} {{a}}\mid\mid\overset{\rightarrow} {{b}}\mid.\mathrm{cos}\:\frac{\pi}{\mathrm{3}}+\mid\overset{\rightarrow} {{a}}\mid\mid\overset{\rightarrow} {{c}}\mid.\mathrm{cos}\:\frac{\pi}{\mathrm{3}}+\mid\overset{\rightarrow} {{b}}\mid\mid\overset{\rightarrow} {{c}}\mid.\mathrm{cos}\:\frac{\pi}{\mathrm{3}}\right)} \\ $$$$ \\ $$

Answered by mr W last updated on 10/Mar/21

this question may have more than one solution. in 3D space there are basically two cases. case 1: all three vectors are in “same“ direction

$${this}\:{question}\:{may}\:{have}\:{more}\:{than} \\ $$$${one}\:{solution}. \\ $$$$ \\ $$$${in}\:\mathrm{3}\boldsymbol{{D}}\:\boldsymbol{{space}}\:{there}\:{are}\:{basically}\:{two} \\ $$$${cases}. \\ $$$$\boldsymbol{{case}}\:\mathrm{1}:\:{all}\:{three}\:{vectors}\:{are}\:{in}\:``{same}`` \\ $$$${direction} \\ $$

Commented by mr W last updated on 10/Mar/21

Commented by mr W last updated on 10/Mar/21

in this case: ∣a+b+c∣=3(√(1^2 −(((√3)/3))^2 ))∣a∣=(√6)∣a∣=(√6) ⇒∣a∣=1 case 2: two vectors are in the “same” direction, the third one in “opposite” direction.

$${in}\:{this}\:{case}: \\ $$$$\mid{a}+{b}+{c}\mid=\mathrm{3}\sqrt{\mathrm{1}^{\mathrm{2}} −\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{3}}\right)^{\mathrm{2}} }\mid{a}\mid=\sqrt{\mathrm{6}}\mid{a}\mid=\sqrt{\mathrm{6}} \\ $$$$\Rightarrow\mid{a}\mid=\mathrm{1} \\ $$$$ \\ $$$$\boldsymbol{{case}}\:\mathrm{2}:\:{two}\:{vectors}\:{are}\:{in}\:{the}\:``{same}'' \\ $$$${direction},\:{the}\:{third}\:{one}\:{in}\:``{opposite}'' \\ $$$${direction}. \\ $$

Commented by mr W last updated on 10/Mar/21

Commented by mr W last updated on 10/Mar/21

in this case, ∣a+b+c∣=(√(1^2 +((1/( (√3))))^2 +((√(2/3)))^2 ))∣a∣=(√6) ⇒∣a∣=(√3)

$${in}\:{this}\:{case}, \\ $$$$\mid{a}+{b}+{c}\mid=\sqrt{\mathrm{1}^{\mathrm{2}} +\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\left(\sqrt{\frac{\mathrm{2}}{\mathrm{3}}}\right)^{\mathrm{2}} }\mid{a}\mid=\sqrt{\mathrm{6}} \\ $$$$\Rightarrow\mid{a}\mid=\sqrt{\mathrm{3}} \\ $$

Commented by mr W last updated on 10/Mar/21

in 2D plane there are basically two cases. case 1: all three vectors are in the same “direction”, clockwise or counterclockwise.

$${in}\:\mathrm{2}\boldsymbol{{D}}\:\boldsymbol{{plane}}\:{there}\:{are}\:{basically}\:{two} \\ $$$${cases}. \\ $$$$\boldsymbol{{case}}\:\mathrm{1}:\:{all}\:{three}\:{vectors}\:{are}\:{in}\:{the} \\ $$$${same}\:``{direction}'',\:{clockwise}\:{or} \\ $$$${counterclockwise}. \\ $$

Commented by mr W last updated on 10/Mar/21

Commented by mr W last updated on 10/Mar/21

in this case: a+b+c=0 but since ∣a+b+c∣=(√6), so this case is not possible. case 2: two vectors are in the same “direction” and the third one is in “opposite direction”.

$${in}\:{this}\:{case}:\:\boldsymbol{{a}}+\boldsymbol{{b}}+\boldsymbol{{c}}=\mathrm{0} \\ $$$${but}\:{since}\:\mid{a}+{b}+{c}\mid=\sqrt{\mathrm{6}},\:{so}\:{this}\:{case} \\ $$$${is}\:{not}\:{possible}. \\ $$$$ \\ $$$$\boldsymbol{{case}}\:\mathrm{2}:\:{two}\:{vectors}\:{are}\:{in}\:{the}\:{same} \\ $$$$``{direction}''\:{and}\:{the}\:{third}\:{one}\:{is}\:{in} \\ $$$$``{opposite}\:{direction}''. \\ $$

Commented by mr W last updated on 10/Mar/21

Commented by mr W last updated on 10/Mar/21

in this case: ∣a+b+c∣=2∣a∣=(√6) ⇒∣a∣=((√6)/2) ■

$${in}\:{this}\:{case}: \\ $$$$\mid{a}+{b}+{c}\mid=\mathrm{2}\mid{a}\mid=\sqrt{\mathrm{6}} \\ $$$$\Rightarrow\mid{a}\mid=\frac{\sqrt{\mathrm{6}}}{\mathrm{2}} \\ $$$$\blacksquare \\ $$

Commented by bramlexs22 last updated on 10/Mar/21

the definition of the angle of two vectors is that the vector must meet the starting point with the base or the end point with the end point

$$ \\ $$the definition of the angle of two vectors is that the vector must meet the starting point with the base or the end point with the end point

Commented by bramlexs22 last updated on 10/Mar/21

from the original problem that each vector pair forms an angle of 60 degrees means that the three starting points of the vector must be the same or the three end points of the vector must be the same

$$ \\ $$from the original problem that each vector pair forms an angle of 60 degrees means that the three starting points of the vector must be the same or the three end points of the vector must be the same

Commented by mr W last updated on 10/Mar/21

a vector has NO fixed starting point or ending point! or can you tell me where the starting point and the ending point from the vector a^(→) =i+2j+3k are? and where the vectors a^(→) =i+2j+3k and b^→ =−i+3j+2k meet?

$${a}\:{vector}\:{has}\:\boldsymbol{{NO}}\:{fixed}\:{starting}\:{point} \\ $$$${or}\:{ending}\:{point}! \\ $$$${or}\:{can}\:{you}\:{tell}\:{me}\:{where}\:{the}\:{starting} \\ $$$${point}\:{and}\:{the}\:{ending}\:{point}\:{from}\:{the} \\ $$$${vector}\:\overset{\rightarrow} {\boldsymbol{{a}}}=\boldsymbol{{i}}+\mathrm{2}\boldsymbol{{j}}+\mathrm{3}\boldsymbol{{k}}\:{are}?\:{and}\:{where} \\ $$$${the}\:{vectors}\:\overset{\rightarrow} {\boldsymbol{{a}}}=\boldsymbol{{i}}+\mathrm{2}\boldsymbol{{j}}+\mathrm{3}\boldsymbol{{k}}\:{and} \\ $$$$\overset{\rightarrow} {\boldsymbol{{b}}}=−\boldsymbol{{i}}+\mathrm{3}\boldsymbol{{j}}+\mathrm{2}\boldsymbol{{k}}\:{meet}? \\ $$

Commented by bramlexs22 last updated on 10/Mar/21

we need to shift one of the vectors or take another vector parallel to it so that they meet at the same starting point

$$ \\ $$we need to shift one of the vectors or take another vector parallel to it so that they meet at the same starting point

Commented by bramlexs22 last updated on 10/Mar/21

Commented by mr W last updated on 10/Mar/21

i know. i used “two vectors make an angle θ”, that means they make also an angle π−θ. if exactly following the definition of angle between two vectors, then there is only the case 1 in 3D space which satisfies the question. in a plane you can not find three vectors with equal magnitude and the angle between two of them is 60°.

$${i}\:{know}. \\ $$$${i}\:{used}\:``{two}\:{vectors}\:{make}\:{an}\:{angle}\:\theta'', \\ $$$${that}\:{means}\:{they}\:{make}\:{also}\:{an}\:{angle} \\ $$$$\pi−\theta. \\ $$$${if}\:{exactly}\:{following}\:{the}\:{definition}\:{of} \\ $$$${angle}\:{between}\:{two}\:{vectors},\:{then}\:{there} \\ $$$${is}\:{only}\:{the}\:{case}\:\mathrm{1}\:{in}\:\mathrm{3}{D}\:{space}\:{which} \\ $$$${satisfies}\:{the}\:{question}.\:{in}\:{a}\:{plane}\:{you} \\ $$$${can}\:{not}\:{find}\:{three}\:{vectors}\:{with}\:{equal} \\ $$$${magnitude}\:{and}\:{the}\:{angle}\:{between} \\ $$$${two}\:{of}\:{them}\:{is}\:\mathrm{60}°. \\ $$

Commented by mr W last updated on 10/Mar/21

Commented by mr W last updated on 10/Mar/21

following definition the angle between a and b is θ, and only θ.

$${following}\:{definition}\:{the}\:{angle} \\ $$$${between}\:\boldsymbol{{a}}\:{and}\:\boldsymbol{{b}}\:{is}\:\theta,\:{and}\:{only}\:\theta. \\ $$

Commented by bramlexs22 last updated on 10/Mar/21

thank you sir for this discussion

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}\:\mathrm{for}\:\mathrm{this}\:\mathrm{discussion} \\ $$

Commented by rexford last updated on 10/Mar/21

The answer is 1 but i still don′t get the approach u used. please,can u explain it further

$${The}\:{answer}\:{is}\:\mathrm{1} \\ $$$${but}\:{i}\:{still}\:{don}'{t}\:{get}\:{the}\:{approach}\:{u}\:{used}. \\ $$$${please},{can}\:{u}\:{explain}\:{it}\:{further} \\ $$

Commented by mr W last updated on 10/Mar/21

Commented by mr W last updated on 10/Mar/21

AD=BD=CD=1 AC=BC=CA=1 OA=(2/3)×((√3)/2)=(1/( (√3))) cos θ=((OA)/(AD))=(1/( (√3))) ∣a+b+c∣=∣a∣ sin θ+∣b∣ sin θ+∣c∣ sin θ =3∣a∣ sin θ =3∣a∣(√(1−((1/( (√3))))^2 ))=(√6)∣a∣=(√6) ⇒∣a∣=1

$${AD}={BD}={CD}=\mathrm{1} \\ $$$${AC}={BC}={CA}=\mathrm{1} \\ $$$${OA}=\frac{\mathrm{2}}{\mathrm{3}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mathrm{cos}\:\theta=\frac{{OA}}{{AD}}=\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$$\mid{a}+{b}+{c}\mid=\mid{a}\mid\:\mathrm{sin}\:\theta+\mid{b}\mid\:\mathrm{sin}\:\theta+\mid{c}\mid\:\mathrm{sin}\:\theta \\ $$$$=\mathrm{3}\mid{a}\mid\:\mathrm{sin}\:\theta \\ $$$$=\mathrm{3}\mid{a}\mid\sqrt{\mathrm{1}−\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} }=\sqrt{\mathrm{6}}\mid{a}\mid=\sqrt{\mathrm{6}} \\ $$$$\Rightarrow\mid{a}\mid=\mathrm{1} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com