A pyramid $OABC$ has vectors $vec{OA}=a$, $vec{OB}=b$ and $vec{OC}=c$.
$begingroup$
A pyramid $OABC$ has vectors $vec{OA}=a$, $vec{OB}=b$ and $vec{OC}=c$.
The vectors $v_1,v_2,v_3$ and $v_4$ are perpendicular to each of the faces of and of magnitude equal to the area of the face. Show that $v_1+v_2+v_3+v_4=0$.
I have no idea where to even start with this so some help would be appreciated.
geometry vectors 3d cross-product solid-geometry
asked Dec 18 '18 at 14:25
H.LinkhornH.Linkhorn
35213
$endgroup$
add a comment |
$begingroup$
A pyramid $OABC$ has vectors $vec{OA}=a$, $vec{OB}=b$ and $vec{OC}=c$.
The vectors $v_1,v_2,v_3$ and $v_4$ are perpendicular to each of the faces of and of magnitude equal to the area of the face. Show that $v_1+v_2+v_3+v_4=0$.
I have no idea where to even start with this so some help would be appreciated.
geometry vectors 3d cross-product solid-geometry
asked Dec 18 '18 at 14:25
H.LinkhornH.Linkhorn
35213
$endgroup$
add a comment |
$begingroup$
A pyramid $OABC$ has vectors $vec{OA}=a$, $vec{OB}=b$ and $vec{OC}=c$.
The vectors $v_1,v_2,v_3$ and $v_4$ are perpendicular to each of the faces of and of magnitude equal to the area of the face. Show that $v_1+v_2+v_3+v_4=0$.
I have no idea where to even start with this so some help would be appreciated.
geometry vectors 3d cross-product solid-geometry
asked Dec 18 '18 at 14:25
H.LinkhornH.Linkhorn
35213
$endgroup$
A pyramid $OABC$ has vectors $vec{OA}=a$, $vec{OB}=b$ and $vec{OC}=c$.
The vectors $v_1,v_2,v_3$ and $v_4$ are perpendicular to each of the faces of and of magnitude equal to the area of the face. Show that $v_1+v_2+v_3+v_4=0$.
I have no idea where to even start with this so some help would be appreciated.
geometry vectors 3d cross-product solid-geometry
geometry vectors 3d cross-product solid-geometry
asked Dec 18 '18 at 14:25
H.LinkhornH.Linkhorn
35213
asked Dec 18 '18 at 14:25
H.LinkhornH.Linkhorn
35213
edited Dec 18 '18 at 16:03
user593746
asked Dec 18 '18 at 14:25
H.LinkhornH.Linkhorn
35213
asked Dec 18 '18 at 14:25
H.LinkhornH.Linkhorn
35213
asked Dec 18 '18 at 14:25
H.LinkhornH.Linkhorn
35213
35213
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint: Start by showing that the vector $v_1$ for the face $OAB$ is given by
$$
v_1 = frac12 a times b,
$$
at least for one choice of orientation.
(By the way, the claim as stated is false; replacing $v_1$ with $-v_1$ keeps it perpendicular to the face, and with the same magnitude. If the claim were true, then we'd also have
$$
(-v_1) + v_2 + v_3 + v_4 = 0,
$$
from which we could conclude that $v_1 = 0$. But not every pyramid has all faces with area zero.)
answered Dec 18 '18 at 14:37
John HughesJohn Hughes
63.1k24090
$endgroup$
$begingroup$
Not that it matters, but $vec{v_1}=frac12vec atimesvec b$
$endgroup$
– Shubham Johri
Dec 18 '18 at 14:46
$begingroup$
D'oh! Thanks -- fixed.
$endgroup$
– John Hughes
Dec 18 '18 at 14:47
add a comment |
$begingroup$
You have $$vec v_1=frac 12 vec atimes vec b$$
You have similar expressions for $vec v_2$ and $vec v_3$. The missing thing is $vec v_4$. For that, you can see that that particular face is a triangle, with sides $vec b-vec a$ and $vec c-vec a$. Note however that we need to account for the direction of the vector. If you make a quick drawing of the $ABC$ triangle with $A,B,C$ in counter clocwise order, and put $O$ in the middle (as seen from above), you have $vec atimesvec b$ pointing outwards. Similarly $vec btimesvec c$ and $vec ctimesvec a$. Now the $vec b-vec a $ points from $A$ to $B$, and $vec c -vec a$ points from $A$ to $C$. Their cross product points "up" so inwards. To make it point outward, you can just change the sign of one of the vectors. Now all you need is to put everything into a single equation and use $vec btimes vec a=-vec atimes vec b$.
answered Dec 18 '18 at 14:59
AndreiAndrei
11.7k21026
$endgroup$
2
$begingroup$
..and fiddle with the signs to actually make things work out. :(
$endgroup$
– John Hughes
Dec 18 '18 at 15:20
$begingroup$
Yes. But you can get these if you do a simple sketch. I drew the $ABC$ triangle, and I've put $O$ in the middle. then $vec v_{1,2,3}$ point outwards. The $(vec b-vec a)times(vec c-vec a)$ points inwards, so I've changed the sign on the first
$endgroup$
– Andrei
Dec 18 '18 at 15:34
$begingroup$
"But you can get these..." Hmm. If "you" here refers to me, John Hughes, then I agree. But can the OP understand and get these? I'm less sure of that.
$endgroup$
– John Hughes
Dec 18 '18 at 16:46
$begingroup$
@JohnHughes Thanks for pointing that out. I will edit my answer.
$endgroup$
– Andrei
Dec 18 '18 at 16:48
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
oldest
votes
$begingroup$
Hint: Start by showing that the vector $v_1$ for the face $OAB$ is given by
$$
v_1 = frac12 a times b,
$$
at least for one choice of orientation.
(By the way, the claim as stated is false; replacing $v_1$ with $-v_1$ keeps it perpendicular to the face, and with the same magnitude. If the claim were true, then we'd also have
$$
(-v_1) + v_2 + v_3 + v_4 = 0,
$$
from which we could conclude that $v_1 = 0$. But not every pyramid has all faces with area zero.)
answered Dec 18 '18 at 14:37
John HughesJohn Hughes
63.1k24090
$endgroup$
$begingroup$
Not that it matters, but $vec{v_1}=frac12vec atimesvec b$
$endgroup$
– Shubham Johri
Dec 18 '18 at 14:46
$begingroup$
D'oh! Thanks -- fixed.
$endgroup$
– John Hughes
Dec 18 '18 at 14:47
add a comment |
$begingroup$
Hint: Start by showing that the vector $v_1$ for the face $OAB$ is given by
$$
v_1 = frac12 a times b,
$$
at least for one choice of orientation.
(By the way, the claim as stated is false; replacing $v_1$ with $-v_1$ keeps it perpendicular to the face, and with the same magnitude. If the claim were true, then we'd also have
$$
(-v_1) + v_2 + v_3 + v_4 = 0,
$$
from which we could conclude that $v_1 = 0$. But not every pyramid has all faces with area zero.)
answered Dec 18 '18 at 14:37
John HughesJohn Hughes
63.1k24090
$endgroup$
$begingroup$
Not that it matters, but $vec{v_1}=frac12vec atimesvec b$
$endgroup$
– Shubham Johri
Dec 18 '18 at 14:46
$begingroup$
D'oh! Thanks -- fixed.
$endgroup$
– John Hughes
Dec 18 '18 at 14:47
add a comment |
$begingroup$
Hint: Start by showing that the vector $v_1$ for the face $OAB$ is given by
$$
v_1 = frac12 a times b,
$$
at least for one choice of orientation.
(By the way, the claim as stated is false; replacing $v_1$ with $-v_1$ keeps it perpendicular to the face, and with the same magnitude. If the claim were true, then we'd also have
$$
(-v_1) + v_2 + v_3 + v_4 = 0,
$$
from which we could conclude that $v_1 = 0$. But not every pyramid has all faces with area zero.)
answered Dec 18 '18 at 14:37
John HughesJohn Hughes
63.1k24090
$endgroup$
Hint: Start by showing that the vector $v_1$ for the face $OAB$ is given by
$$
v_1 = frac12 a times b,
$$
at least for one choice of orientation.
(By the way, the claim as stated is false; replacing $v_1$ with $-v_1$ keeps it perpendicular to the face, and with the same magnitude. If the claim were true, then we'd also have
$$
(-v_1) + v_2 + v_3 + v_4 = 0,
$$
from which we could conclude that $v_1 = 0$. But not every pyramid has all faces with area zero.)
answered Dec 18 '18 at 14:37
John HughesJohn Hughes
63.1k24090
edited Dec 18 '18 at 14:47
answered Dec 18 '18 at 14:37
John HughesJohn Hughes
63.1k24090
answered Dec 18 '18 at 14:37
John HughesJohn Hughes
63.1k24090
answered Dec 18 '18 at 14:37
John HughesJohn Hughes
63.1k24090
63.1k24090
$begingroup$
Not that it matters, but $vec{v_1}=frac12vec atimesvec b$
$endgroup$
– Shubham Johri
Dec 18 '18 at 14:46
$begingroup$
D'oh! Thanks -- fixed.
$endgroup$
– John Hughes
Dec 18 '18 at 14:47
add a comment |
$begingroup$
Not that it matters, but $vec{v_1}=frac12vec atimesvec b$
$endgroup$
– Shubham Johri
Dec 18 '18 at 14:46
$begingroup$
D'oh! Thanks -- fixed.
$endgroup$
– John Hughes
Dec 18 '18 at 14:47
$begingroup$
Not that it matters, but $vec{v_1}=frac12vec atimesvec b$
$endgroup$
– Shubham Johri
Dec 18 '18 at 14:46
$begingroup$
Not that it matters, but $vec{v_1}=frac12vec atimesvec b$
$endgroup$
– Shubham Johri
Dec 18 '18 at 14:46
$begingroup$
D'oh! Thanks -- fixed.
$endgroup$
– John Hughes
Dec 18 '18 at 14:47
$begingroup$
D'oh! Thanks -- fixed.
$endgroup$
– John Hughes
Dec 18 '18 at 14:47
add a comment |
$begingroup$
You have $$vec v_1=frac 12 vec atimes vec b$$
You have similar expressions for $vec v_2$ and $vec v_3$. The missing thing is $vec v_4$. For that, you can see that that particular face is a triangle, with sides $vec b-vec a$ and $vec c-vec a$. Note however that we need to account for the direction of the vector. If you make a quick drawing of the $ABC$ triangle with $A,B,C$ in counter clocwise order, and put $O$ in the middle (as seen from above), you have $vec atimesvec b$ pointing outwards. Similarly $vec btimesvec c$ and $vec ctimesvec a$. Now the $vec b-vec a $ points from $A$ to $B$, and $vec c -vec a$ points from $A$ to $C$. Their cross product points "up" so inwards. To make it point outward, you can just change the sign of one of the vectors. Now all you need is to put everything into a single equation and use $vec btimes vec a=-vec atimes vec b$.
answered Dec 18 '18 at 14:59
AndreiAndrei
11.7k21026
$endgroup$
2
$begingroup$
..and fiddle with the signs to actually make things work out. :(
$endgroup$
– John Hughes
Dec 18 '18 at 15:20
$begingroup$
Yes. But you can get these if you do a simple sketch. I drew the $ABC$ triangle, and I've put $O$ in the middle. then $vec v_{1,2,3}$ point outwards. The $(vec b-vec a)times(vec c-vec a)$ points inwards, so I've changed the sign on the first
$endgroup$
– Andrei
Dec 18 '18 at 15:34
$begingroup$
"But you can get these..." Hmm. If "you" here refers to me, John Hughes, then I agree. But can the OP understand and get these? I'm less sure of that.
$endgroup$
– John Hughes
Dec 18 '18 at 16:46
$begingroup$
@JohnHughes Thanks for pointing that out. I will edit my answer.
$endgroup$
– Andrei
Dec 18 '18 at 16:48
add a comment |
$begingroup$
You have $$vec v_1=frac 12 vec atimes vec b$$
You have similar expressions for $vec v_2$ and $vec v_3$. The missing thing is $vec v_4$. For that, you can see that that particular face is a triangle, with sides $vec b-vec a$ and $vec c-vec a$. Note however that we need to account for the direction of the vector. If you make a quick drawing of the $ABC$ triangle with $A,B,C$ in counter clocwise order, and put $O$ in the middle (as seen from above), you have $vec atimesvec b$ pointing outwards. Similarly $vec btimesvec c$ and $vec ctimesvec a$. Now the $vec b-vec a $ points from $A$ to $B$, and $vec c -vec a$ points from $A$ to $C$. Their cross product points "up" so inwards. To make it point outward, you can just change the sign of one of the vectors. Now all you need is to put everything into a single equation and use $vec btimes vec a=-vec atimes vec b$.
answered Dec 18 '18 at 14:59
AndreiAndrei
11.7k21026
$endgroup$
2
$begingroup$
..and fiddle with the signs to actually make things work out. :(
$endgroup$
– John Hughes
Dec 18 '18 at 15:20
$begingroup$
Yes. But you can get these if you do a simple sketch. I drew the $ABC$ triangle, and I've put $O$ in the middle. then $vec v_{1,2,3}$ point outwards. The $(vec b-vec a)times(vec c-vec a)$ points inwards, so I've changed the sign on the first
$endgroup$
– Andrei
Dec 18 '18 at 15:34
$begingroup$
"But you can get these..." Hmm. If "you" here refers to me, John Hughes, then I agree. But can the OP understand and get these? I'm less sure of that.
$endgroup$
– John Hughes
Dec 18 '18 at 16:46
$begingroup$
@JohnHughes Thanks for pointing that out. I will edit my answer.
$endgroup$
– Andrei
Dec 18 '18 at 16:48
add a comment |
$begingroup$
You have $$vec v_1=frac 12 vec atimes vec b$$
You have similar expressions for $vec v_2$ and $vec v_3$. The missing thing is $vec v_4$. For that, you can see that that particular face is a triangle, with sides $vec b-vec a$ and $vec c-vec a$. Note however that we need to account for the direction of the vector. If you make a quick drawing of the $ABC$ triangle with $A,B,C$ in counter clocwise order, and put $O$ in the middle (as seen from above), you have $vec atimesvec b$ pointing outwards. Similarly $vec btimesvec c$ and $vec ctimesvec a$. Now the $vec b-vec a $ points from $A$ to $B$, and $vec c -vec a$ points from $A$ to $C$. Their cross product points "up" so inwards. To make it point outward, you can just change the sign of one of the vectors. Now all you need is to put everything into a single equation and use $vec btimes vec a=-vec atimes vec b$.
answered Dec 18 '18 at 14:59
AndreiAndrei
11.7k21026
$endgroup$
You have $$vec v_1=frac 12 vec atimes vec b$$
You have similar expressions for $vec v_2$ and $vec v_3$. The missing thing is $vec v_4$. For that, you can see that that particular face is a triangle, with sides $vec b-vec a$ and $vec c-vec a$. Note however that we need to account for the direction of the vector. If you make a quick drawing of the $ABC$ triangle with $A,B,C$ in counter clocwise order, and put $O$ in the middle (as seen from above), you have $vec atimesvec b$ pointing outwards. Similarly $vec btimesvec c$ and $vec ctimesvec a$. Now the $vec b-vec a $ points from $A$ to $B$, and $vec c -vec a$ points from $A$ to $C$. Their cross product points "up" so inwards. To make it point outward, you can just change the sign of one of the vectors. Now all you need is to put everything into a single equation and use $vec btimes vec a=-vec atimes vec b$.
answered Dec 18 '18 at 14:59
AndreiAndrei
11.7k21026
edited Dec 18 '18 at 16:56
answered Dec 18 '18 at 14:59
AndreiAndrei
11.7k21026
answered Dec 18 '18 at 14:59
AndreiAndrei
11.7k21026
answered Dec 18 '18 at 14:59
AndreiAndrei
11.7k21026
11.7k21026
2
$begingroup$
..and fiddle with the signs to actually make things work out. :(
$endgroup$
– John Hughes
Dec 18 '18 at 15:20
$begingroup$
Yes. But you can get these if you do a simple sketch. I drew the $ABC$ triangle, and I've put $O$ in the middle. then $vec v_{1,2,3}$ point outwards. The $(vec b-vec a)times(vec c-vec a)$ points inwards, so I've changed the sign on the first
$endgroup$
– Andrei
Dec 18 '18 at 15:34
$begingroup$
"But you can get these..." Hmm. If "you" here refers to me, John Hughes, then I agree. But can the OP understand and get these? I'm less sure of that.
$endgroup$
– John Hughes
Dec 18 '18 at 16:46
$begingroup$
@JohnHughes Thanks for pointing that out. I will edit my answer.
$endgroup$
– Andrei
Dec 18 '18 at 16:48
add a comment |
2
$begingroup$
..and fiddle with the signs to actually make things work out. :(
$endgroup$
– John Hughes
Dec 18 '18 at 15:20
$begingroup$
Yes. But you can get these if you do a simple sketch. I drew the $ABC$ triangle, and I've put $O$ in the middle. then $vec v_{1,2,3}$ point outwards. The $(vec b-vec a)times(vec c-vec a)$ points inwards, so I've changed the sign on the first
$endgroup$
– Andrei
Dec 18 '18 at 15:34
$begingroup$
"But you can get these..." Hmm. If "you" here refers to me, John Hughes, then I agree. But can the OP understand and get these? I'm less sure of that.
$endgroup$
– John Hughes
Dec 18 '18 at 16:46
$begingroup$
@JohnHughes Thanks for pointing that out. I will edit my answer.
$endgroup$
– Andrei
Dec 18 '18 at 16:48
2
2
$begingroup$
..and fiddle with the signs to actually make things work out. :(
$endgroup$
– John Hughes
Dec 18 '18 at 15:20
$begingroup$
..and fiddle with the signs to actually make things work out. :(
$endgroup$
– John Hughes
Dec 18 '18 at 15:20
$begingroup$
Yes. But you can get these if you do a simple sketch. I drew the $ABC$ triangle, and I've put $O$ in the middle. then $vec v_{1,2,3}$ point outwards. The $(vec b-vec a)times(vec c-vec a)$ points inwards, so I've changed the sign on the first
$endgroup$
– Andrei
Dec 18 '18 at 15:34
$begingroup$
Yes. But you can get these if you do a simple sketch. I drew the $ABC$ triangle, and I've put $O$ in the middle. then $vec v_{1,2,3}$ point outwards. The $(vec b-vec a)times(vec c-vec a)$ points inwards, so I've changed the sign on the first
$endgroup$
– Andrei
Dec 18 '18 at 15:34
$begingroup$
"But you can get these..." Hmm. If "you" here refers to me, John Hughes, then I agree. But can the OP understand and get these? I'm less sure of that.
$endgroup$
– John Hughes
Dec 18 '18 at 16:46
$begingroup$
"But you can get these..." Hmm. If "you" here refers to me, John Hughes, then I agree. But can the OP understand and get these? I'm less sure of that.
$endgroup$
– John Hughes
Dec 18 '18 at 16:46
$begingroup$
@JohnHughes Thanks for pointing that out. I will edit my answer.
$endgroup$
– Andrei
Dec 18 '18 at 16:48
$begingroup$
@JohnHughes Thanks for pointing that out. I will edit my answer.
$endgroup$
– Andrei
Dec 18 '18 at 16:48
add a comment |
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