Triangle vector problem
$begingroup$
I consider how to draw this because I have no idea with the Origin,
is the 3.1 part we can solve with :
triangle QAB: triangle QBC so,
AQ : QC = 2:3 ?
But I can't solve the 3.2 part, can anyone show the picture of the triangle and steps to solve this problem ?
vectors triangle vector-fields
edited Dec 18 '18 at 4:02
David G. Stork
11.1k41432
asked Dec 18 '18 at 3:40
Aster ZenAster Zen
438
$endgroup$
add a comment |
$begingroup$
I consider how to draw this because I have no idea with the Origin,
is the 3.1 part we can solve with :
triangle QAB: triangle QBC so,
AQ : QC = 2:3 ?
But I can't solve the 3.2 part, can anyone show the picture of the triangle and steps to solve this problem ?
vectors triangle vector-fields
edited Dec 18 '18 at 4:02
David G. Stork
11.1k41432
asked Dec 18 '18 at 3:40
Aster ZenAster Zen
438
$endgroup$
add a comment |
$begingroup$
I consider how to draw this because I have no idea with the Origin,
is the 3.1 part we can solve with :
triangle QAB: triangle QBC so,
AQ : QC = 2:3 ?
But I can't solve the 3.2 part, can anyone show the picture of the triangle and steps to solve this problem ?
vectors triangle vector-fields
edited Dec 18 '18 at 4:02
David G. Stork
11.1k41432
asked Dec 18 '18 at 3:40
Aster ZenAster Zen
438
$endgroup$
I consider how to draw this because I have no idea with the Origin,
is the 3.1 part we can solve with :
triangle QAB: triangle QBC so,
AQ : QC = 2:3 ?
But I can't solve the 3.2 part, can anyone show the picture of the triangle and steps to solve this problem ?
vectors triangle vector-fields
vectors triangle vector-fields
edited Dec 18 '18 at 4:02
David G. Stork
11.1k41432
asked Dec 18 '18 at 3:40
Aster ZenAster Zen
438
edited Dec 18 '18 at 4:02
David G. Stork
11.1k41432
asked Dec 18 '18 at 3:40
Aster ZenAster Zen
438
edited Dec 18 '18 at 4:02
David G. Stork
11.1k41432
edited Dec 18 '18 at 4:02
David G. Stork
11.1k41432
edited Dec 18 '18 at 4:02
David G. Stork
11.1k41432
11.1k41432
asked Dec 18 '18 at 3:40
Aster ZenAster Zen
438
asked Dec 18 '18 at 3:40
Aster ZenAster Zen
438
asked Dec 18 '18 at 3:40
Aster ZenAster Zen
438
438
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
For simplicity, just draw three points somewhere in the first quadrant, and call them $ABC$. Now locate the centroid of the triangle, let's call it $M$. Areas of triangles $ABM$, $BCM$, and $CAM$ are all equal. So your point $P$ is farther from $AC$ then $M$, such as the area of $PCA$ is half of the area of the triangle $ABC$. Don't worry if your figure is slightly off.
answered Dec 18 '18 at 6:44
AndreiAndrei
13.1k21230
$endgroup$
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
For simplicity, just draw three points somewhere in the first quadrant, and call them $ABC$. Now locate the centroid of the triangle, let's call it $M$. Areas of triangles $ABM$, $BCM$, and $CAM$ are all equal. So your point $P$ is farther from $AC$ then $M$, such as the area of $PCA$ is half of the area of the triangle $ABC$. Don't worry if your figure is slightly off.
answered Dec 18 '18 at 6:44
AndreiAndrei
13.1k21230
$endgroup$
add a comment |
$begingroup$
For simplicity, just draw three points somewhere in the first quadrant, and call them $ABC$. Now locate the centroid of the triangle, let's call it $M$. Areas of triangles $ABM$, $BCM$, and $CAM$ are all equal. So your point $P$ is farther from $AC$ then $M$, such as the area of $PCA$ is half of the area of the triangle $ABC$. Don't worry if your figure is slightly off.
answered Dec 18 '18 at 6:44
AndreiAndrei
13.1k21230
$endgroup$
add a comment |
$begingroup$
For simplicity, just draw three points somewhere in the first quadrant, and call them $ABC$. Now locate the centroid of the triangle, let's call it $M$. Areas of triangles $ABM$, $BCM$, and $CAM$ are all equal. So your point $P$ is farther from $AC$ then $M$, such as the area of $PCA$ is half of the area of the triangle $ABC$. Don't worry if your figure is slightly off.
answered Dec 18 '18 at 6:44
AndreiAndrei
13.1k21230
$endgroup$
For simplicity, just draw three points somewhere in the first quadrant, and call them $ABC$. Now locate the centroid of the triangle, let's call it $M$. Areas of triangles $ABM$, $BCM$, and $CAM$ are all equal. So your point $P$ is farther from $AC$ then $M$, such as the area of $PCA$ is half of the area of the triangle $ABC$. Don't worry if your figure is slightly off.
answered Dec 18 '18 at 6:44
AndreiAndrei
13.1k21230
answered Dec 18 '18 at 6:44
AndreiAndrei
13.1k21230
answered Dec 18 '18 at 6:44
AndreiAndrei
13.1k21230
answered Dec 18 '18 at 6:44
AndreiAndrei
13.1k21230
13.1k21230
add a comment |
add a comment |
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