Mixing
Vector proof question for pre-calculus
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A vector $mathbfv$ is called a unit vector if $|mathbfv| = 1$.
Let $mathbfa$, $mathbfb$, and $mathbfc$ be unit vectors, such that $mathbfa + mathbfb + mathbfc = mathbf0$. Show that the angle between any two of these vectors is $120^circ$. I think I should use law of cosines.
I would like to involve vectors.
Also, on the side, I am confused with this question:
If $mathbfa$ and $mathbfb$ are vectors such that $|mathbfa| = 4$, $|mathbfb| = 5$, and $|mathbfa + mathbfb| = 7$, then find $|2 mathbfa - 3 mathbfb|$. I think you are supposed to use vectors inside of the dot operations. I don't know how to solve.
algebra-precalculus
edited 48 mins ago
N. F. Taussig
39.8k93153
asked 1 hour ago
ilikepi314
606
add a comment |Â
up vote
1
down vote
favorite
A vector $mathbfv$ is called a unit vector if $|mathbfv| = 1$.
Let $mathbfa$, $mathbfb$, and $mathbfc$ be unit vectors, such that $mathbfa + mathbfb + mathbfc = mathbf0$. Show that the angle between any two of these vectors is $120^circ$. I think I should use law of cosines.
I would like to involve vectors.
Also, on the side, I am confused with this question:
If $mathbfa$ and $mathbfb$ are vectors such that $|mathbfa| = 4$, $|mathbfb| = 5$, and $|mathbfa + mathbfb| = 7$, then find $|2 mathbfa - 3 mathbfb|$. I think you are supposed to use vectors inside of the dot operations. I don't know how to solve.
algebra-precalculus
edited 48 mins ago
N. F. Taussig
39.8k93153
asked 1 hour ago
ilikepi314
606
for the second question, How should I approach it?
– ilikepi314
52 mins ago
Polarization identity aka the parallelogram law.
– amd
14 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
A vector $mathbfv$ is called a unit vector if $|mathbfv| = 1$.
Let $mathbfa$, $mathbfb$, and $mathbfc$ be unit vectors, such that $mathbfa + mathbfb + mathbfc = mathbf0$. Show that the angle between any two of these vectors is $120^circ$. I think I should use law of cosines.
I would like to involve vectors.
Also, on the side, I am confused with this question:
If $mathbfa$ and $mathbfb$ are vectors such that $|mathbfa| = 4$, $|mathbfb| = 5$, and $|mathbfa + mathbfb| = 7$, then find $|2 mathbfa - 3 mathbfb|$. I think you are supposed to use vectors inside of the dot operations. I don't know how to solve.
algebra-precalculus
edited 48 mins ago
N. F. Taussig
39.8k93153
asked 1 hour ago
ilikepi314
606
A vector $mathbfv$ is called a unit vector if $|mathbfv| = 1$.
Let $mathbfa$, $mathbfb$, and $mathbfc$ be unit vectors, such that $mathbfa + mathbfb + mathbfc = mathbf0$. Show that the angle between any two of these vectors is $120^circ$. I think I should use law of cosines.
I would like to involve vectors.
Also, on the side, I am confused with this question:
If $mathbfa$ and $mathbfb$ are vectors such that $|mathbfa| = 4$, $|mathbfb| = 5$, and $|mathbfa + mathbfb| = 7$, then find $|2 mathbfa - 3 mathbfb|$. I think you are supposed to use vectors inside of the dot operations. I don't know how to solve.
algebra-precalculus
algebra-precalculus
edited 48 mins ago
N. F. Taussig
39.8k93153
asked 1 hour ago
ilikepi314
606
edited 48 mins ago
N. F. Taussig
39.8k93153
asked 1 hour ago
ilikepi314
606
edited 48 mins ago
N. F. Taussig
39.8k93153
edited 48 mins ago
N. F. Taussig
39.8k93153
edited 48 mins ago
N. F. Taussig
39.8k93153
39.8k93153
asked 1 hour ago
ilikepi314
606
asked 1 hour ago
ilikepi314
606
asked 1 hour ago
ilikepi314
606
606
for the second question, How should I approach it?
– ilikepi314
52 mins ago
Polarization identity aka the parallelogram law.
– amd
14 mins ago
add a comment |Â
for the second question, How should I approach it?
– ilikepi314
52 mins ago
Polarization identity aka the parallelogram law.
– amd
14 mins ago
for the second question, How should I approach it?
– ilikepi314
52 mins ago
for the second question, How should I approach it?
– ilikepi314
52 mins ago
Polarization identity aka the parallelogram law.
– amd
14 mins ago
Polarization identity aka the parallelogram law.
– amd
14 mins ago
add a comment |Â
7 Answers
7
active
oldest
votes
up vote
2
down vote
Any two vectors define a plane. The third vector must lie in that plane if the sum of the three is to equal $bf 0$. Equilateral triangles (which are in a plane) have angles $120^circ$.
answered 1 hour ago
David G. Stork
8,22321232
add a comment |Â
up vote
2
down vote
We have: $0 = acdot (b+c+a) = 1 + cos B + cos C$, and similarly $cos A + cos C = -1, cos A + cos B = -1 $. Thus $cos A = cos B = cos C = -1/2implies A = B = C = 120^circ$
answered 52 mins ago
DeepSea
69.4k54285
add a comment |Â
up vote
2
down vote
Given $a+b+c=0$
$$(a+b+c)^2=|a|^2+|b|^2+|c|^2+2(acdot b+bcdot c+ccdot a)=0$$
Since it is given that they are unit vectors,
begingather|a|+|b|+|c|=1 \[4px]
1+1+1+2(acdot b+bcdot c+ccdot a)=0\[4px]
2(acdot b+bcdot c+ccdot a)=-3\[4px]
acdot b+bcdot c+ccdot a=-dfrac32
endgather
which implies that $$acdot b=bcdot c=ccdot a=-dfrac12$$.
So, the angle between any two vectors is $120^circ$
edited 48 mins ago
egreg
167k1281190
answered 57 mins ago
Key Flex
5,021628
wait so you do not need law of cosine?
– ilikepi314
56 mins ago
@ilikepi314 If $theta$ is the angle between the unit vectors $a,b$ then $costheta=acdot b$
– Key Flex
54 mins ago
oh i pulled a silly.
– ilikepi314
53 mins ago
2
@KeyFlex : Not trying to downvote your post, but I don't see a clear implication from $a cdot b + bcdot c + ccdot a = -3/2 implies acdot b = -1/2$. This line is what you have to prove, and you have not proved it !
– DeepSea
43 mins ago
Yeah there is a big jump here where you assume that $a cdot b$, $b cdot c$, and $c cdot a$ are all equal.
– Morgan Rodgers
33 mins ago
add a comment |Â
up vote
0
down vote
Solution for the rest of your question:
$$
langle a,brangle=frac^22=frac49-16-252=4.
$$
$$
|2a-3b|^2=4|a|^2+9|b|^2-12langle a,brangle=4.16+9.25-12.4=241.
$$
Hence $|2a-3b|=sqrt241.$
answered 50 mins ago
Blind
451217
add a comment |Â
up vote
0
down vote
With reference to the triangle with vertices $A,B,C$ such that $OA=vec a$ $OB=vec b$ and $OC=vec c$ we have that the centroid coincides with the origin indeed
$$OG=fracvec a+vec b+vec c3=vec 0$$
and the origin also coincides with the circumcenter, therefore the $triangle ABC$ is equilateral and the angle between any two of $vec a,vec b,vec c$ is $120°$.
answered 14 mins ago
gimusi
74.3k73889
add a comment |Â
up vote
0
down vote
Big hint for the first part: If $mathbf a+mathbf b+mathbf c=0$ then $mathbf acdot(mathbf a+mathbf b+mathbf c)=0$. Now use linearity of the dot product and the identity $mathbf vcdotmathbf w = |mathbf v|,|mathbf w|costheta$, where $theta$ is the angle between $mathbf v$ and $mathbf w$.
answered 8 mins ago
amd
27k21046
add a comment |Â
up vote
-1
down vote
As pointed out firstly by David Stork, we have that the sum of three unit vectors is equal to zero if and only if they are on the side of a equilateral triangle therefore the the angle between any two of these vectors is $120°$.
answered 38 mins ago
gimusi
74.3k73889
@MorganRodgers I did'n see that! I think it is the best answer. The OP can see that mine was given later. Anyway I've also added a sketch.
– gimusi
27 mins ago
@MorganRodgers I've added another answer with a different derivation of this fact.
– gimusi
13 mins ago
add a comment |Â
7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Any two vectors define a plane. The third vector must lie in that plane if the sum of the three is to equal $bf 0$. Equilateral triangles (which are in a plane) have angles $120^circ$.
answered 1 hour ago
David G. Stork
8,22321232
add a comment |Â
up vote
2
down vote
Any two vectors define a plane. The third vector must lie in that plane if the sum of the three is to equal $bf 0$. Equilateral triangles (which are in a plane) have angles $120^circ$.
answered 1 hour ago
David G. Stork
8,22321232
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Any two vectors define a plane. The third vector must lie in that plane if the sum of the three is to equal $bf 0$. Equilateral triangles (which are in a plane) have angles $120^circ$.
answered 1 hour ago
David G. Stork
8,22321232
Any two vectors define a plane. The third vector must lie in that plane if the sum of the three is to equal $bf 0$. Equilateral triangles (which are in a plane) have angles $120^circ$.
answered 1 hour ago
David G. Stork
8,22321232
answered 1 hour ago
David G. Stork
8,22321232
answered 1 hour ago
David G. Stork
8,22321232
answered 1 hour ago
David G. Stork
8,22321232
8,22321232
add a comment |Â
add a comment |Â
up vote
2
down vote
We have: $0 = acdot (b+c+a) = 1 + cos B + cos C$, and similarly $cos A + cos C = -1, cos A + cos B = -1 $. Thus $cos A = cos B = cos C = -1/2implies A = B = C = 120^circ$
answered 52 mins ago
DeepSea
69.4k54285
add a comment |Â
up vote
2
down vote
We have: $0 = acdot (b+c+a) = 1 + cos B + cos C$, and similarly $cos A + cos C = -1, cos A + cos B = -1 $. Thus $cos A = cos B = cos C = -1/2implies A = B = C = 120^circ$
answered 52 mins ago
DeepSea
69.4k54285
add a comment |Â
up vote
2
down vote
up vote
2
down vote
We have: $0 = acdot (b+c+a) = 1 + cos B + cos C$, and similarly $cos A + cos C = -1, cos A + cos B = -1 $. Thus $cos A = cos B = cos C = -1/2implies A = B = C = 120^circ$
answered 52 mins ago
DeepSea
69.4k54285
We have: $0 = acdot (b+c+a) = 1 + cos B + cos C$, and similarly $cos A + cos C = -1, cos A + cos B = -1 $. Thus $cos A = cos B = cos C = -1/2implies A = B = C = 120^circ$
answered 52 mins ago
DeepSea
69.4k54285
answered 52 mins ago
DeepSea
69.4k54285
answered 52 mins ago
DeepSea
69.4k54285
answered 52 mins ago
DeepSea
69.4k54285
69.4k54285
add a comment |Â
add a comment |Â
up vote
2
down vote
Given $a+b+c=0$
$$(a+b+c)^2=|a|^2+|b|^2+|c|^2+2(acdot b+bcdot c+ccdot a)=0$$
Since it is given that they are unit vectors,
begingather|a|+|b|+|c|=1 \[4px]
1+1+1+2(acdot b+bcdot c+ccdot a)=0\[4px]
2(acdot b+bcdot c+ccdot a)=-3\[4px]
acdot b+bcdot c+ccdot a=-dfrac32
endgather
which implies that $$acdot b=bcdot c=ccdot a=-dfrac12$$.
So, the angle between any two vectors is $120^circ$
edited 48 mins ago
egreg
167k1281190
answered 57 mins ago
Key Flex
5,021628
wait so you do not need law of cosine?
– ilikepi314
56 mins ago
@ilikepi314 If $theta$ is the angle between the unit vectors $a,b$ then $costheta=acdot b$
– Key Flex
54 mins ago
oh i pulled a silly.
– ilikepi314
53 mins ago
2
@KeyFlex : Not trying to downvote your post, but I don't see a clear implication from $a cdot b + bcdot c + ccdot a = -3/2 implies acdot b = -1/2$. This line is what you have to prove, and you have not proved it !
– DeepSea
43 mins ago
Yeah there is a big jump here where you assume that $a cdot b$, $b cdot c$, and $c cdot a$ are all equal.
– Morgan Rodgers
33 mins ago
add a comment |Â
up vote
2
down vote
Given $a+b+c=0$
$$(a+b+c)^2=|a|^2+|b|^2+|c|^2+2(acdot b+bcdot c+ccdot a)=0$$
Since it is given that they are unit vectors,
begingather|a|+|b|+|c|=1 \[4px]
1+1+1+2(acdot b+bcdot c+ccdot a)=0\[4px]
2(acdot b+bcdot c+ccdot a)=-3\[4px]
acdot b+bcdot c+ccdot a=-dfrac32
endgather
which implies that $$acdot b=bcdot c=ccdot a=-dfrac12$$.
So, the angle between any two vectors is $120^circ$
edited 48 mins ago
egreg
167k1281190
answered 57 mins ago
Key Flex
5,021628
wait so you do not need law of cosine?
– ilikepi314
56 mins ago
@ilikepi314 If $theta$ is the angle between the unit vectors $a,b$ then $costheta=acdot b$
– Key Flex
54 mins ago
oh i pulled a silly.
– ilikepi314
53 mins ago
2
@KeyFlex : Not trying to downvote your post, but I don't see a clear implication from $a cdot b + bcdot c + ccdot a = -3/2 implies acdot b = -1/2$. This line is what you have to prove, and you have not proved it !
– DeepSea
43 mins ago
Yeah there is a big jump here where you assume that $a cdot b$, $b cdot c$, and $c cdot a$ are all equal.
– Morgan Rodgers
33 mins ago
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Given $a+b+c=0$
$$(a+b+c)^2=|a|^2+|b|^2+|c|^2+2(acdot b+bcdot c+ccdot a)=0$$
Since it is given that they are unit vectors,
begingather|a|+|b|+|c|=1 \[4px]
1+1+1+2(acdot b+bcdot c+ccdot a)=0\[4px]
2(acdot b+bcdot c+ccdot a)=-3\[4px]
acdot b+bcdot c+ccdot a=-dfrac32
endgather
which implies that $$acdot b=bcdot c=ccdot a=-dfrac12$$.
So, the angle between any two vectors is $120^circ$
edited 48 mins ago
egreg
167k1281190
answered 57 mins ago
Key Flex
5,021628
Given $a+b+c=0$
$$(a+b+c)^2=|a|^2+|b|^2+|c|^2+2(acdot b+bcdot c+ccdot a)=0$$
Since it is given that they are unit vectors,
begingather|a|+|b|+|c|=1 \[4px]
1+1+1+2(acdot b+bcdot c+ccdot a)=0\[4px]
2(acdot b+bcdot c+ccdot a)=-3\[4px]
acdot b+bcdot c+ccdot a=-dfrac32
endgather
which implies that $$acdot b=bcdot c=ccdot a=-dfrac12$$.
So, the angle between any two vectors is $120^circ$
edited 48 mins ago
egreg
167k1281190
answered 57 mins ago
Key Flex
5,021628
edited 48 mins ago
egreg
167k1281190
edited 48 mins ago
egreg
167k1281190
edited 48 mins ago
egreg
167k1281190
167k1281190
answered 57 mins ago
Key Flex
5,021628
answered 57 mins ago
Key Flex
5,021628
answered 57 mins ago
Key Flex
5,021628
5,021628
wait so you do not need law of cosine?
– ilikepi314
56 mins ago
@ilikepi314 If $theta$ is the angle between the unit vectors $a,b$ then $costheta=acdot b$
– Key Flex
54 mins ago
oh i pulled a silly.
– ilikepi314
53 mins ago
2
@KeyFlex : Not trying to downvote your post, but I don't see a clear implication from $a cdot b + bcdot c + ccdot a = -3/2 implies acdot b = -1/2$. This line is what you have to prove, and you have not proved it !
– DeepSea
43 mins ago
Yeah there is a big jump here where you assume that $a cdot b$, $b cdot c$, and $c cdot a$ are all equal.
– Morgan Rodgers
33 mins ago
add a comment |Â
wait so you do not need law of cosine?
– ilikepi314
56 mins ago
@ilikepi314 If $theta$ is the angle between the unit vectors $a,b$ then $costheta=acdot b$
– Key Flex
54 mins ago
oh i pulled a silly.
– ilikepi314
53 mins ago
2
@KeyFlex : Not trying to downvote your post, but I don't see a clear implication from $a cdot b + bcdot c + ccdot a = -3/2 implies acdot b = -1/2$. This line is what you have to prove, and you have not proved it !
– DeepSea
43 mins ago
Yeah there is a big jump here where you assume that $a cdot b$, $b cdot c$, and $c cdot a$ are all equal.
– Morgan Rodgers
33 mins ago
wait so you do not need law of cosine?
– ilikepi314
56 mins ago
wait so you do not need law of cosine?
– ilikepi314
56 mins ago
@ilikepi314 If $theta$ is the angle between the unit vectors $a,b$ then $costheta=acdot b$
– Key Flex
54 mins ago
@ilikepi314 If $theta$ is the angle between the unit vectors $a,b$ then $costheta=acdot b$
– Key Flex
54 mins ago
oh i pulled a silly.
– ilikepi314
53 mins ago
oh i pulled a silly.
– ilikepi314
53 mins ago
2
2
@KeyFlex : Not trying to downvote your post, but I don't see a clear implication from $a cdot b + bcdot c + ccdot a = -3/2 implies acdot b = -1/2$. This line is what you have to prove, and you have not proved it !
– DeepSea
43 mins ago
@KeyFlex : Not trying to downvote your post, but I don't see a clear implication from $a cdot b + bcdot c + ccdot a = -3/2 implies acdot b = -1/2$. This line is what you have to prove, and you have not proved it !
– DeepSea
43 mins ago
Yeah there is a big jump here where you assume that $a cdot b$, $b cdot c$, and $c cdot a$ are all equal.
– Morgan Rodgers
33 mins ago
Yeah there is a big jump here where you assume that $a cdot b$, $b cdot c$, and $c cdot a$ are all equal.
– Morgan Rodgers
33 mins ago
add a comment |Â
up vote
0
down vote
Solution for the rest of your question:
$$
langle a,brangle=frac^22=frac49-16-252=4.
$$
$$
|2a-3b|^2=4|a|^2+9|b|^2-12langle a,brangle=4.16+9.25-12.4=241.
$$
Hence $|2a-3b|=sqrt241.$
answered 50 mins ago
Blind
451217
add a comment |Â
up vote
0
down vote
Solution for the rest of your question:
$$
langle a,brangle=frac^22=frac49-16-252=4.
$$
$$
|2a-3b|^2=4|a|^2+9|b|^2-12langle a,brangle=4.16+9.25-12.4=241.
$$
Hence $|2a-3b|=sqrt241.$
answered 50 mins ago
Blind
451217
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Solution for the rest of your question:
$$
langle a,brangle=frac^22=frac49-16-252=4.
$$
$$
|2a-3b|^2=4|a|^2+9|b|^2-12langle a,brangle=4.16+9.25-12.4=241.
$$
Hence $|2a-3b|=sqrt241.$
answered 50 mins ago
Blind
451217
Solution for the rest of your question:
$$
langle a,brangle=frac^22=frac49-16-252=4.
$$
$$
|2a-3b|^2=4|a|^2+9|b|^2-12langle a,brangle=4.16+9.25-12.4=241.
$$
Hence $|2a-3b|=sqrt241.$
answered 50 mins ago
Blind
451217
answered 50 mins ago
Blind
451217
answered 50 mins ago
Blind
451217
answered 50 mins ago
Blind
451217
451217
add a comment |Â
add a comment |Â
up vote
0
down vote
With reference to the triangle with vertices $A,B,C$ such that $OA=vec a$ $OB=vec b$ and $OC=vec c$ we have that the centroid coincides with the origin indeed
$$OG=fracvec a+vec b+vec c3=vec 0$$
and the origin also coincides with the circumcenter, therefore the $triangle ABC$ is equilateral and the angle between any two of $vec a,vec b,vec c$ is $120°$.
answered 14 mins ago
gimusi
74.3k73889
add a comment |Â
up vote
0
down vote
With reference to the triangle with vertices $A,B,C$ such that $OA=vec a$ $OB=vec b$ and $OC=vec c$ we have that the centroid coincides with the origin indeed
$$OG=fracvec a+vec b+vec c3=vec 0$$
and the origin also coincides with the circumcenter, therefore the $triangle ABC$ is equilateral and the angle between any two of $vec a,vec b,vec c$ is $120°$.
answered 14 mins ago
gimusi
74.3k73889
add a comment |Â
up vote
0
down vote
up vote
0
down vote
With reference to the triangle with vertices $A,B,C$ such that $OA=vec a$ $OB=vec b$ and $OC=vec c$ we have that the centroid coincides with the origin indeed
$$OG=fracvec a+vec b+vec c3=vec 0$$
and the origin also coincides with the circumcenter, therefore the $triangle ABC$ is equilateral and the angle between any two of $vec a,vec b,vec c$ is $120°$.
answered 14 mins ago
gimusi
74.3k73889
With reference to the triangle with vertices $A,B,C$ such that $OA=vec a$ $OB=vec b$ and $OC=vec c$ we have that the centroid coincides with the origin indeed
$$OG=fracvec a+vec b+vec c3=vec 0$$
and the origin also coincides with the circumcenter, therefore the $triangle ABC$ is equilateral and the angle between any two of $vec a,vec b,vec c$ is $120°$.
answered 14 mins ago
gimusi
74.3k73889
answered 14 mins ago
gimusi
74.3k73889
answered 14 mins ago
gimusi
74.3k73889
answered 14 mins ago
gimusi
74.3k73889
74.3k73889
add a comment |Â
add a comment |Â
up vote
0
down vote
Big hint for the first part: If $mathbf a+mathbf b+mathbf c=0$ then $mathbf acdot(mathbf a+mathbf b+mathbf c)=0$. Now use linearity of the dot product and the identity $mathbf vcdotmathbf w = |mathbf v|,|mathbf w|costheta$, where $theta$ is the angle between $mathbf v$ and $mathbf w$.
answered 8 mins ago
amd
27k21046
add a comment |Â
up vote
0
down vote
Big hint for the first part: If $mathbf a+mathbf b+mathbf c=0$ then $mathbf acdot(mathbf a+mathbf b+mathbf c)=0$. Now use linearity of the dot product and the identity $mathbf vcdotmathbf w = |mathbf v|,|mathbf w|costheta$, where $theta$ is the angle between $mathbf v$ and $mathbf w$.
answered 8 mins ago
amd
27k21046
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Big hint for the first part: If $mathbf a+mathbf b+mathbf c=0$ then $mathbf acdot(mathbf a+mathbf b+mathbf c)=0$. Now use linearity of the dot product and the identity $mathbf vcdotmathbf w = |mathbf v|,|mathbf w|costheta$, where $theta$ is the angle between $mathbf v$ and $mathbf w$.
answered 8 mins ago
amd
27k21046
Big hint for the first part: If $mathbf a+mathbf b+mathbf c=0$ then $mathbf acdot(mathbf a+mathbf b+mathbf c)=0$. Now use linearity of the dot product and the identity $mathbf vcdotmathbf w = |mathbf v|,|mathbf w|costheta$, where $theta$ is the angle between $mathbf v$ and $mathbf w$.
answered 8 mins ago
amd
27k21046
answered 8 mins ago
amd
27k21046
answered 8 mins ago
amd
27k21046
answered 8 mins ago
amd
27k21046
27k21046
add a comment |Â
add a comment |Â
up vote
-1
down vote
As pointed out firstly by David Stork, we have that the sum of three unit vectors is equal to zero if and only if they are on the side of a equilateral triangle therefore the the angle between any two of these vectors is $120°$.
answered 38 mins ago
gimusi
74.3k73889
@MorganRodgers I did'n see that! I think it is the best answer. The OP can see that mine was given later. Anyway I've also added a sketch.
– gimusi
27 mins ago
@MorganRodgers I've added another answer with a different derivation of this fact.
– gimusi
13 mins ago
add a comment |Â
up vote
-1
down vote
As pointed out firstly by David Stork, we have that the sum of three unit vectors is equal to zero if and only if they are on the side of a equilateral triangle therefore the the angle between any two of these vectors is $120°$.
answered 38 mins ago
gimusi
74.3k73889
@MorganRodgers I did'n see that! I think it is the best answer. The OP can see that mine was given later. Anyway I've also added a sketch.
– gimusi
27 mins ago
@MorganRodgers I've added another answer with a different derivation of this fact.
– gimusi
13 mins ago
add a comment |Â
up vote
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up vote
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As pointed out firstly by David Stork, we have that the sum of three unit vectors is equal to zero if and only if they are on the side of a equilateral triangle therefore the the angle between any two of these vectors is $120°$.
answered 38 mins ago
gimusi
74.3k73889
As pointed out firstly by David Stork, we have that the sum of three unit vectors is equal to zero if and only if they are on the side of a equilateral triangle therefore the the angle between any two of these vectors is $120°$.
answered 38 mins ago
gimusi
74.3k73889
edited 13 mins ago
answered 38 mins ago
gimusi
74.3k73889
answered 38 mins ago
gimusi
74.3k73889
answered 38 mins ago
gimusi
74.3k73889
74.3k73889
@MorganRodgers I did'n see that! I think it is the best answer. The OP can see that mine was given later. Anyway I've also added a sketch.
– gimusi
27 mins ago
@MorganRodgers I've added another answer with a different derivation of this fact.
– gimusi
13 mins ago
add a comment |Â
@MorganRodgers I did'n see that! I think it is the best answer. The OP can see that mine was given later. Anyway I've also added a sketch.
– gimusi
27 mins ago
@MorganRodgers I've added another answer with a different derivation of this fact.
– gimusi
13 mins ago
@MorganRodgers I did'n see that! I think it is the best answer. The OP can see that mine was given later. Anyway I've also added a sketch.
– gimusi
27 mins ago
@MorganRodgers I did'n see that! I think it is the best answer. The OP can see that mine was given later. Anyway I've also added a sketch.
– gimusi
27 mins ago
@MorganRodgers I've added another answer with a different derivation of this fact.
– gimusi
13 mins ago
@MorganRodgers I've added another answer with a different derivation of this fact.
– gimusi
13 mins ago
add a comment |Â
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for the second question, How should I approach it?
– ilikepi314
52 mins ago
Polarization identity aka the parallelogram law.
– amd
14 mins ago